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Expectation Maximization for Poisson process

The problem

What if you want to run a regression to estimate the coefficients and retrieve the data generating process of some data BUT you are in a scenario with censored data? How does that affect my estimation? Can I do better?

Censored data

In short, you have censored data when some observation is unobserved or constrained because of some specific reason which is natural and unavoidable.

Unobserved

For instance, in survival analysis, if the subject is not yet dead (luckily) you can not observe the time of death yet and hence you don't know if he will die tomorrow or in two years. You can't observe yet the death date.

Constrained

For goods with limited demand, once you deplete your stock, by definition you can't sell more and you can't observe all the demand that you would have if more items were able to be sold. Or an Airline , when selling seats, can't observe the full demand after all the seats have been sold, maybe more people were willing to flight.

We will simulate an example of the latest case, for an airline with simple toy data.

import numpy as np
import statsmodels.api as sm
from scipy.stats import poisson
import matplotlib.pyplot as plt

We create some fake data, where seats sold are generated from a poisson distribution, with an intercept (2) and a coefficient related to how many days are left to the departure (-0.2), the further from the departure, the less seats are sold.

What we also do, is that on the last date, we constrain the demand. We don't get the full seats that would have been sold under a poisson distribution but the amount minus 3. Only ourselves know that because we want to tweak the data. In real life we would see data similar to the one generated and we couldn't see the actual sales that would have been present in a complete poisson generating process (that actually is how the tickets are sold in our example)

# Step 1: Generate data
np.random.seed(100)
n_samples = 20
n_features = 1
substract_sales = 3

# Generate features
X = np.random.normal(size=(n_samples, n_features))
X = np.array(range(n_samples+1, 1, -1))
X = sm.add_constant(X)  # Add intercept term

# True coefficients
beta_true = np.array([2, -0.2])

# Generate Poisson-distributed target values
linear_pred = np.dot(X, beta_true)
lambda_true = np.exp(linear_pred)
y = np.random.poisson(lambda_true)
# y

# Right-censoring threshold in the last date (when the fare is closed)
censor_threshold = np.concatenate([(y+1)[:-1],np.array([y[-1]-substract_sales])])
y_censored = np.where(y > censor_threshold, censor_threshold, y)
is_censored = (y > censor_threshold)
Maybe it's easier to visualize.
The green curve is the actual lambda parameter for each data, based on the intercept and the real coefficient multiplied by days to departure.
The blue dots are the actual data points we see for all dates except from the last one.
The grey dot is the data we should see if there was no limit of how many seats to sell, following the poisson distribution.
The red dot is the actual data point we have, the last day the airline sold all the remaining seats and couldn't fulfill the true demand and hence we see lower sales than the ones generated by the poisson distribution.

# Plot the data
plt.xlabel('Days to departure')
plt.ylabel('Bookings')
plt.title('Data Plot')
plt.plot(X[:-1, 1], y[:-1], 'o')
plt.plot(X[:, 1], lambda_true, label='Lambda True', color="green")
# plt.plot(X[:, 1], lambda_est, label='Lambda Estimated')
plt.scatter(X[-1, 1], y[-1:],  color="grey", label="Real unseen demand" )
plt.scatter(X[-1, 1], y[-1:]-substract_sales, color='red', label='Constrained')
plt.legend()
plt.show()

Image

Esimation
Poisson Regression

First we estimate the parameters using a poisson regression. We assume the data follows a poisson process and we don't do anything to manage the constrained data.
The results are kind of ok, we get some closeness to the true parameters but it is clearly biased.

model_censored = sm.GLM(y_censored, X, family=sm.families.Poisson())
results_censored = model_censored.fit()
print(f'Estimated Censored coefficients: {results_censored.params}')
## Estimated Censored coefficients: [ 1.86976922 -0.21124541]
Expectation Maximization

OK, here is our alternative. What we can do, a bit more sophisticated, is to apply EM algorithm, which is a iterative process, to estimate the coefficients of the poisson regression, including the knowledge we have, that there might be constrained data.

# Step 2: Initialize parameters
beta_est = np.zeros(n_features + 1)
tol = 1e-4
max_iter = 1000

for iteration in range(max_iter):
    # Step 3: E-step
    # Estimate the expected values of censored data
    lambda_est = np.exp(np.dot(X, beta_est))
    expected_y = np.where(
        is_censored,
        (censor_threshold + 1) / (1 - poisson.cdf(censor_threshold, lambda_est)),
        y_censored
    )
    # Ensure expected_y values are valid
    expected_y = np.nan_to_num(expected_y, nan=np.mean(y_censored), posinf=np.max(y_censored), neginf=0)
    # Step 4: M-step
    # Update parameter estimates using Poisson regression
    model = sm.GLM(expected_y, X, family=sm.families.Poisson())
    results = model.fit(method="lbfgs")
    # results = model.fit_regularized(L1_wt=0, alpha=0.1) 
    new_beta_est = results.params

    # Check convergence
    if np.linalg.norm(new_beta_est - beta_est) < tol:
        break

    beta_est = new_beta_est
## <string>:7: RuntimeWarning: divide by zero encountered in divide
print(f'Estimated coefficients: {beta_est}')

## Estimated coefficients: [ 2.11015996 -0.23573144]
We can see our estimates are of course not perfect but the intercept is closer to the true parameter.
Let's visualize the results to have a clear intuition.

uncensored_predictions = results.predict(X)
censored_predictions = results_censored.predict(X)
Promising!
For the furthest dates we see a bias in both approaches, we are not doing better than the Poisson regression, but neither worse.
As we move closer the the departure, the EM algorithm predictions get closer and closer to the true lambdas, while the regular poisson regression continues it's biased trajectory.
The EM procedure adjusted the coefficients to better match the constrained data.

plt.plot(X[:, 1], lambda_true, label='Lambda True', color="green")
plt.plot(X[:, 1], uncensored_predictions, label='Uncensored Predictions (EM)', color="blue")
plt.plot(X[:, 1], censored_predictions, label='Censored Predictions', color="red")
plt.xlabel('Days to departure')
plt.ylabel('Lambda - Poisson regression')
plt.title('Lambda Predictions')
plt.legend()
plt.show()

Image

Reflections on quitting my ML job

As I'm starting my sabbatical journey I am reading some posts of Jason Liu since he seems to have a career similar to something I would like to if I got into ML and LLM riding the hype wave. Furthermore I find his writing pleasant and his content could be useful even as a solopreneur.

One of my goals for this period of time is to write more. To take more notes since I struggle memorizing without them (or even with them but at least I can read a summary later) and I want to actually do more. More of everything, less thinking about and more actual doing. From shipping products, to really learning and that includes writing.

Reading his post and starting now my journey, a real action would be to take notes about his post, which I found useful, at least some bits of it. Not taking notes would be less writing, less doing, less memorizing.

Choosing

Right now, I feel more at this point

"This despair arises from the realization of one's absolute freedom and the responsibility for creating one's own essence and purpose."

Lately, last few years probably, I have started to realize that quote. We can go a long way by following the usual paths, at least usual to our surroundings. In my case, without major distress that turns everything upside down, it was high school, university, get a job, get married, retire. At least, that's something I (many?) thought as given, a fate that if didn't screw it, it would happen automatically.

And as I was applied and I did good in school, I got the first 3 steps quite easily. I never really thought of going out of that path and saw the ones doing so as outliers or with greater safety nets to fail. In some cases it could be true, in others it was probably me being short sighted, I guess it was not my fault but just lack of adult life.

As years go by, I start questioning the meaning of what I'm doing. Is my job meaningful? Does it create value or help anyone? Corporate world feels like a charade. Despite I was working in analytics and studying and building ML models, I couldn't see if that was worth the effort and time. Once you go past the hype of learning models and cool tech bits, looking over that it feels empty. I changed jobs. Still the same, I was dreading in boredom and rat race. Needing to "impress" people or feeling that I need to be there for work 9-5 each day without feeling motivated was awful. I quit after a few months willing to take some time off. What am I doing? I have no purpose and my day to day means nothing. I like seeing my friends and family, traveling, etc but the regular life has a lot more things and time to fill. I can't make this for 30 more years.

Fortunately, tech and ML pays high salaries so I felt safe. I could take some time off, I can always look for other job, etc. But the feeling of "I need to do something different" was there. I started looking for other opportunities, looking for purpose in jobs, looking for motivation. I was in despair as I understood that I needed to do my own way and no one was doing it for me.

Ironically, I took another job quit quickly because it paid much more and I could just try it, maybe I was just in the wrong place. Seemed to work better, it was less of a charade, I had more time for crafting and working, I got some motivation back as I felt at least more comfortable daily. Not that I had purpose really but working was at least with less pressure. The time zone differences helped because I didn't feel like 9 hours a day I was supposed to be there, with someone 1 click away of sending a slack message to me. I stayed two years. Earning good money compared to my spending, saving and "happy". Eventually everything started to decay, the job was less free, the business request were more urgent and I quickly lost motivation and started to fall in despair again. What am I doing? Who cares about this parquet file with funny numbers?

I quit, this time for good. Despair again, realization, I need to find something for myself, or try at least. I fear for its complexity, I fear if I can do it, if I can make a living in another way. I'm taking some time to rest but I know eventually I will need to figure things out. On my own, making my own way, and I see how no one will do it for me.

As a side note, finding purpose, finding your way, etc was also a thing when I thought about romantic relationships. Finding someone that you really care about and that cares about you is not easy at all and it doesn't happen magically as I thought as a kid. I am not going to expand about it here but it was another topic that made me realize about our own fate and efforts.

Side side note, retirement is another one. I don't see my national retirement plan as something you can live off. I see my family, that luckily does well but despite that nothing is granted and we can support the elder in my family because our own safety net. A younger me didn't know that but as my twenties went by I started to see a lot more of adult/real life. Eye opening period of my life.

How to be lucky

"Okay, I'm focused on getting X, but let's not forget to read the headlines."

High Agency and be the plumber

Reminder to myself, focus on bringing solutions and not the shiny new tool. I think I'm good at high agency or at being responsible and wanting to finish what I commit to.
I need to work on focusing on which solution am I bringing. Finding the right niche and actually bringing value. In my case with sportsjobs.online, not just throwing things into it, but making it clear what I am providing. For other topics related to ML and LLM I need to decide what I'll focus on this time coming in.

Impostor Syndrome I'm the classical example of not trusting myself and thinking stuff like this.

but at the end of the day, you must just think I have shit taste and that you've somehow tricked me into thinking you're good when you're an impostor? Right?

I need to stop with that. Quitting now I had plenty of nice words for every colleague, including the desire from managers that I stay or I come back if I get bored of not having a job. Of course, as I write that, I think in the back of my mind of exceptions and that technical colleagues were less effusive as my managers and etc. but all of that is probably not true. I'm going against all evidence and just guessing and making things in my head.

How to Be Good at Many Things Consistency. Every one says this. I need to do it the right way now that I'll have the time. No excuses.

Do things, practice, keep going for it. Everything will be easier and quicker.

And I need to be grateful for what I have and how lucky I am to be able to get a sabbatical time to try to change the path I'm going.

Weighted regression

Weighted regression consists on assigning different weights to each observation and hence more or less importance at the time of fitting the regression.

On way to look at it is to think as solving the regression problem minimizing Weighted Mean Squared Error(WSME) instead of Mean Squared Error(MSE)

\(\(WMSE(\beta, w) = \frac{1}{N} \sum_{i=1}^n w_i(y_i - \overrightarrow {x_i} \beta)^2\)\) Intuitively, we are looking fot the coefficients that minimize MSE but putting different weights to each observation. OLS is a particular case where all the \(w_i = 1\)

Why doing this? A few reasons (Shalizi 2015. Chapter 7.1)

  • Focusing Accuracy: We want to predict specially well some particular points or region of points, maybe because that's the focus for production or maybe because being wrong at those observations has a huge cost, etc. Using Weighted regression will do an extra effort to match that data.

  • Discount imprecision: OLS returns the maximum likelihood estimates when the residuals are independent, normal with mean 0 and with constant variance. When we face non constant variance OLS no longer returns the MLE. The logic behind using weighted regression is that makes no sense to pay equal attention to all the observations since some of them have higher variance and are less indicative of the conditional mean. We should put more emphasis on the regions of lower variance, predict it well and "expect to fit poorly where the noise is big".
    The weights that will return MLE are \(\frac{1}{\sigma_i^2}\)

  • Sampling bias: If we think or know that the observations in our data are not completely random and some subset of the population might be under-represented (in a survey for example or because of data availability) it might make sense to weight observation inversely to the probability of being included in the sample. Under-represented observations will get more weights and over-represented less weight.
    Another similar situation is related to covariate shift. If the distribution of variable x changes over time we can use a weight designed as the ratio of the probability density functions.

    "If the old pdf was p(x) and the new one is q(x), the weight we'd want to is \(w_i=q(x_i)/p(x_i)\)

  • Other: Related to GLM, when the conditional mean is a non linear function of a linear predictor. (Not further explained in the book at this point)

Is there a scenario where OLS is better than Weighted regression? Assuming we can compute the weights.

Example.

First we will see the impact of using weighted regression, using a simulated scenario where we actually know the variance of the error of each observation. This is not realistic but useful to see it in action.

library(tidyverse)

We generate 1000 datapoints with a linear relation between y and x. Intercept = 0, slope = 5. We let the variance of the error depend on the value of x. Higher values of x are associated with higher values of the variance of the error.

set.seed(23)
n=1000
x = runif(n,0,10)
error = rnorm(n,0, x/1.5)
df = data.frame(x)
df = df %>% mutate(y = 5*x + error)
Visually..

ggplot(data=df, aes(x=x, y=y)) +
  geom_point(alpha=0.3) + 
  # geom_smooth(color="blue") +
  # geom_smooth(method = "lm", mapping = aes(weight = (1/sqrt(x)^2)),
  #             color = "red", show.legend = FALSE) +
  NULL

Image

Linear regression
ols = lm(formula = "y~x", data=df)
summary(ols)

## 
## Call:
## lm(formula = "y~x", data = df)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -14.868  -1.720  -0.137   1.918  14.722 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  0.19192    0.24278   0.791    0.429    
## x            4.95585    0.04148 119.489   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.855 on 998 degrees of freedom
## Multiple R-squared:  0.9347, Adjusted R-squared:  0.9346 
## F-statistic: 1.428e+04 on 1 and 998 DF,  p-value: < 2.2e-16
We get an intercept of 0.19, non-significant when the actual value is 0 and a slope of 4.96 when the actual value is 5.

Weighted linear regression
wols = lm(formula = "y~x", data=df, weights = (1/sqrt(x)^2) )
summary(wols)
## 
## Call:
## lm(formula = "y~x", data = df, weights = (1/sqrt(x)^2))
## 
## Weighted Residuals:
##     Min      1Q  Median      3Q     Max 
## -4.8880 -0.8601 -0.0016  0.8936  4.6535 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 0.001483   0.030072   0.049    0.961    
## x           4.993473   0.021874 228.286   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.498 on 998 degrees of freedom
## Multiple R-squared:  0.9812, Adjusted R-squared:  0.9812 
## F-statistic: 5.211e+04 on 1 and 998 DF,  p-value: < 2.2e-16

We get an intercept of 0, non-significant too but much closer to 0 and with lower standard error and a slope of 4.99 also much closer to the actual value of 5 and with lower standard error.

Conclusion: if we know the right weights we can get better estimates from a linear regression in case of heteroscedasticity.

Inference is not valid in the dataset used for model selection.

Let's say we have a dataset and we want to fit a model to it and do some inference such as obtaining the coefficients and look for their confidence intervals.

For such a task we would first need to find a model that we think approximates to the real data generating process behind the phenomenon.
This will be the model selection step.
Then we would look at the output of our model and get the standard error of the coefficients or calculate the confidence interval or any other similar task. This will be the inference step.

The issue here is that, if we don't know the true model and we do model selection, our own model will be a random object. Why? Because the particular dataset we are using is also a set of random variables. Other datasets might return another model formula as the best between our options since that particular dataset would have other observations and particularities.

Main problem:

since we are selecting a model based on a particular dataset, the standard errors and p-values will be smaller than then actual ones.

"That means there is some extra randomness in your estimated parameters (and everything else), which isn't accounted for by formulas which assume a fixed model.
This is not just a problem with formal model-selection devices like cross-validation. If you do an initial, exploratory data analysis before deciding which model to use - and that's generally a good idea - you are, yourself, acting as a noisy, complicated model-selection device" (Sharizi 2017)

The most straightforward way to deal with this (if you are using independent observations) is to split the data, do model selection in one part and then fit the best model in the other part. Your second fit will be the one useful for inference.
You could fit the model to the full data but that would include the part used for model selection and you would still get false, overconfident standard errors.

Let's see an example.
We will generate data following a somewhat "complicated" model with interactions. We will split the data in two equal size parts. One for model selection and one for inference.
We will then fit a couple formulas to model selection part and pick the one with the minimum RMSE. We will compare the standard errors obtained in the model selection part and the ones obtained fitting that model to the inference part.

Thanks to BrunoRodrigues for this post that I used as guideline to fit models with Cross Validation in R.

We start by generating the data, including interactions.

set.seed(1)
N = 5000
b0 = 4
b1 = 1
b2 = 2
b3 = 3
b4 = 4
b5 = 5

x1 = runif(N, 0, 10)
x2 = rnorm(N, 20, 3)
x3 = runif(N, 20, 40)
error = rnorm(N, 0, 200)

y = b0 + b1*x1 + b2*x2 + b3*x3 + b4*x1*x2 + b5*x2*x3 + error


df = tibble(y, x1, x2, x3)

We do the first split, df_selection will be the one used to try different models and pick one.
df_inference will be used to do the actual inference given the model selected.

prop = 0.5

selection_inference_split = initial_split(df, prop=prop)

df_selection = training(selection_inference_split)
df_inference = testing(selection_inference_split)

To select a model using df_selection we will use Cross validation to try to get the model that best generalizes.
We will generate 30 split of 70% of the data and use the other 30% to calculate RMSE metric.

validation_data <- mc_cv(df_selection, prop = 0.7, times = 30)

We create two functions, my_lm() will run a linear regression for the training part of each split of CV and get the prediction for the testing part of each split. We will run this for a couple of formulas.
return_model will fit the model to the whole training data to extract the parameters and standard errors we get if we use the same dataset that was used to do model selection.

my_lm <- function(formula, split, id){

    analysis_set = analysis(split)  

    model <- lm(formula = formula, data=analysis_set)

    assessment_set <- assessment(split)


    pred = tibble::tibble("id" = id,
        "formula" = formula,
        "truth" = assessment_set$y,
        "prediction" = unlist(predict(model, newdata = assessment_set)))

}


return_model = function(formula){


    model <- lm(formula = formula, data=df_selection)
    output = list(model$coefficients, summary(model))

}

We will try 5 formulas. The first one is the actual data generating process and should the best in terms of RMSE. We will exclude that one for model selection since the aim of this is to simulate a scenario where we don't know the actual formula behind the data. We calculate it just for reference but we will pick one of the other 4 models for inference.

formulas = list("y ~ x1 + x2 +x3 + x1*x2 + x2*x3", 
                "y ~ .", 
                "y ~ x1 + x2", 
                "y ~ x1 + x2 + x3 + x1*x2",
                "y ~ x1 + x2 + x3 + x2*x3")
results = data.frame()

models = list()
for (formula in formulas){

results_selection <- map2_df(.x = validation_data$splits,
                           .y = validation_data$id,
                           ~my_lm(formula = formula, split = .x, id = .y))

model = return_model(formula)


results = rbind.data.frame(results, results_selection)
models = c(models, model )

}

We retrieve the mean RMSE across the splits, calculated in the test part of each split.
We can see that the real model is the best in terms of RMSE. Between the others, we can see that the one including the x2:x3 interaction is the best. So, we will keep that one as our "model selected"

results %>%
    group_by(id, formula) %>%
    rmse(truth, prediction) %>%
    group_by(formula) %>%
    summarise(mean_rmse = mean(.estimate)) %>%
    as.data.frame()
##                           formula mean_rmse
## 1                           y ~ .  219.4756
## 2                     y ~ x1 + x2  625.0173
## 3        y ~ x1 + x2 + x3 + x1*x2  217.3185
## 4        y ~ x1 + x2 + x3 + x2*x3  198.9802
## 5 y ~ x1 + x2 +x3 + x1*x2 + x2*x3  196.4747

We can check the parameters and the standard errors when fitted to the whole selection dataset.

## 
## Call:
## lm(formula = formula, data = df_selection)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -854.07 -132.91   -0.97  137.65  714.53 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -245.1084   138.9684  -1.764   0.0779 .  
## x1            82.7919     1.3540  61.148   <2e-16 ***
## x2            15.7118     6.8628   2.289   0.0221 *  
## x3            -1.5177     4.5626  -0.333   0.7394    
## x2:x3          5.1676     0.2256  22.906   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 199.1 on 2495 degrees of freedom
## Multiple R-squared:  0.947,  Adjusted R-squared:  0.9469 
## F-statistic: 1.115e+04 on 4 and 2495 DF,  p-value: < 2.2e-16

And let's see what happens if we fit the same model to the inference set.

model_test = lm(formula=formulas[[5]], data=df_inference)

summary(model_test)

## 
## Call:
## lm(formula = formulas[[5]], data = df_inference)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -656.47 -138.67   -5.64  130.21  773.99 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -438.7059   140.2618  -3.128 0.001782 ** 
## x1            81.4724     1.3622  59.812  < 2e-16 ***
## x2            23.4856     6.9475   3.380 0.000735 ***
## x3             3.6309     4.5942   0.790 0.429417    
## x2:x3          4.9750     0.2275  21.869  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 199.4 on 2495 degrees of freedom
## Multiple R-squared:  0.9473, Adjusted R-squared:  0.9472 
## F-statistic: 1.121e+04 on 4 and 2495 DF,  p-value: < 2.2e-16
First we can see that the parameters have changed a bit.
In second place we can see that the standard errors are generally bigger in comparison to the parameter for the inference set and will generate a wider confidence interval.

ggplot(data = ratio_df) +
  geom_point(aes(x=parameter, y=ratio, col=set), size=3) +
  theme(legend.title = element_blank()) +
  theme_light() +
  xlab("") +
  ylab("Ratio") +
  ggtitle("Absolute ratio between SD and Estimate")

Image

My idea is to add a plot with the confidence intervals so the effect can be seen directly but I don't have the time today. Anyways, it is clear that the standad error to parameter ratio is bigger in the inference set, showing that the inference in the same dataset as model selection is invalid as it is overconfident in the results.

Remarks on R2

R2 depends on the variance on the variance of the predictors

Quoting from Shalizi[^1] Assuming a true linear model
$$ Y = aX + \epsilon$$
and assuming we know \(a\) exactly.
The variance of Y will be \(a^2\mathbb{V}[X] + \mathbb{V}[\epsilon]\).
So \(R^2 = \frac{a^2\mathbb{V}[X]}{a^2\mathbb{V}[X] + \mathbb{V}[\epsilon]}\)
This goes to 0 as \(\mathbb{V}[X] \rightarrow 0\) and it goes to 1 as \(\mathbb{V}[X] \rightarrow \infty\). "It thus has little to do with the quality of the fit, and a lot to do with how spread out the predictor variable is. Notice also how easy it is to get a high \(R^2\) even when the true model is not linear!"

Below a quick comparison between two linear relationships, one with much higher variance than the other in the predictor.
Added a different constant for better display in plot.

library(tidyverse)

x1 = rnorm(1000, mean=0, sd=1)
x2 = rnorm(1000, mean=0, sd=10)
error = rnorm(1000, mean=0, sd=0.5)

y1 = x1 + error
y2 = 10 + x2 +  error

df = data.frame(x1,x2,y1,y2)

model1 = lm("y1 ~ x1")
## Error in eval(predvars, data, env): object 'y1' not found
model2 =  lm("y2 ~ x2")
## Error in eval(predvars, data, env): object 'y2' not found

Linear Smoothers

Linear regression as smoothing

Let's assume the DGP (data generating process) is: $$ Y = \mu(x) + \epsilon$$ where \(\mu(x)\) is the mean Y value for that particular x and \(\epsilon\) is an error with mean 0.

When running OLS we are trying to approximate \(\mu(x)\) with a linear function of the form \(\alpha + \beta x\) and trying to retrieve the best \(\alpha\) and \(\beta\) minimizing the mean-squared error.

The conclusions don't change but the math gets easier if we assume both X and Y are centered (mean=0).
With that in mind we can write down the MSE and optimize to get the best parameters.

\[MSE(\alpha, \beta) = \mathbb{E}[(Y - \alpha - \beta X)^2] \\ = \mathbb{E}[\mathbb{E}[(Y - \alpha - \beta X)^2 | X]] \\ = \mathbb{E}[\mathbb{V}[Y|X]] + \mathbb{E}[Y- \alpha - \beta X | X])^2] \\ = \mathbb{E}[\mathbb{V}[Y|X]] + \mathbb{E}[(\mathbb{E}[Y- \alpha - \beta X | X])^2]\]

Deriving with respect to \(\alpha\) and \(\beta\) for optimization..
The first term can be dropped since doesn't include any parameter.

$$\frac{\partial MSE}{\partial \alpha} = \mathbb{E}[2(Y - \alpha - \beta X)(-1)] \ \mathbb{E}[Y - a - b X] = 0 \ a = \mathbb{E}[Y] - b \mathbb{E}[X] = 0 $$ when Y and X are centered..

and $$\frac{\partial MSE}{\partial \beta} = \mathbb{E}[2(Y - \alpha - \beta X)(-X)] \ \mathbb{E}[XY] - b\mathbb{E}[X^2] = 0 \ b = \frac{Cov[X,Y]}{\mathbb{V}[X]} $$

The optimal beta is a function of the covariance between Y and X, and the variance of X.

Putting together \(a\) and \(b\) we get \(\mu(x) = x \frac{Cov[X,Y]}{\mathbb{V}[X]}\)

Replacing with the values from the sampled data we get an estimation of \(a\) and \(b\).

Remember they are 0 centered so variance and covariance get simplified.

\[ \hat a = 0 \\ \hat b = \frac{\sum_i y_i x_i}{\sum_i x_i^2}\]

With all this we can see how OLS is a smoothing of the data.
Writing in terms of the data points:
$$\hat \mu(x) = \hat b x \ = x \frac{\sum_i y_i x_i}{\sum_i x_i^2} \ = \sum_i y_i \frac{x_i}{\sum_j x_j^2} x \ = \sum_i y_i \frac{x_i}{n \hat \sigma_x^2} x $$ where \(\hat \sigma_x^2\) is the sample variance of X.
In words, our prediction is a weighted average of the observed values \(y_i\) of the dependent variable, where the weights are proportional to how far \(x_i\) is from the center (relative to the variance), and proportional to the magnitude of \(x\). If \(x_i\) is on the same side of the center as \(x\), it gets a positive weight, and if it's on the opposite side it gets a negative weight. (Shalizi 2017)

If \(\mu(x)\) is really a straight line, this is fine, but when it's not, that the weights are proportional to how far they are to the center and not the point to predict can lead to awful predictions.

Alternative smoothers

For that, other methods smooth the data in another ways to help mitigate that.

As quick examples, we have KNN regression where the smoothing is done using only close observations to the one to predict (and getting quite noisy since depend a lot on the sample points around a small area).

Kernel smoothers are a variant where depending on the kernel selected we get different smoothing. The main idea is that we use a windowd of data with the idea of putting more weight to points close to the one to predict. Could be Gaussian weight around X for example, or uniform around a window. Note this is different than KNN regression since we do not take the average of those points, we get a regression for that area.
A nice thing about this smoothers (and KNN regression) is that if we want to predict points far from the training data we won't get a linear extrapolation as with OLS but it will be pushed towards the closest data points we had in training.

Bias Variance Tradeoff

Mean squared error (MSE) is a measure of how far our prediction is from the true values of the dependent variable. It's the expectation of the squared error.

The squared error being:

\[(Y - \hat \mu(x))^2\]

where Y is the true value and \(\hat \mu(x)\) is the prediction for a given x.

We can decompose it into:

\[(Y - \hat \mu(x))^2 \\ = (Y - \mu(x) + \mu(x) - \hat \mu(x)^2) \\ = (Y - \mu(x))^2 + 2(Y - \mu(x))(\mu(x) - \hat \mu(x)) + (\mu(x) - \hat \mu(x))^2\]

So, that's the squared error. The MSE is the expectation of that.

The expectation is a linear operator so we can apply it independently to different terms of a summation.
The expectation of the first term is the variance of the error intrinsic to the DGP.
The second term goes to 0 because involves \(E(Y-\mu(x))\) that is the expectation of the error and that's equal to 0.
The third term reamins as it is since doesn't involve random variables.

\[MSE(\hat \mu(x)) = \sigma^2_x + (\mu(x) - \hat \mu(x))^2\]

This is our first bias-variance decomposition. The first term is the intrinsic difficulty of the problem to model, is the variance of the error and can not be reduced, it is what it is.
The second term is how off our predictions are regarding the true expected value for that particular X.

This would be fine if we wouldn't need to consider \(\hat \mu(x)\) a random variable itself, since it is dependent on the specific dataset we are using. Given another dataset our estimation would be different despite using the same model methodology.
What we actually want is the MSE of the method used \(\hat M\) and not only the result of a particular realization.

\[MSE(\hat M_n(x)) = E[(Y - \hat M_n(X))^2 | X=x] \\ = ... \\ = \sigma^2_x + (\mu(x) - E[\hat M_n(x)])^2 - V[\hat M_n(x)] \]

This is our 2nd bias-variance decomposition.
The first term is still the irreducible error.
The second term is the bias of using \(\hat M_n\) to approximate \(\mu(x)\). Is the approximation bias/error.
The third term is the variance of the estimate of the regression function. If our estimates have high variance we can have large errors despite using an unbiased approximation.

Flexible methods will be able to approximate \(\mu(x)\) closely, however usually using more flexible methods involve increasing the variance of the estimate. That's the bias-variance tradeoff. We need to evaluate how to balance that, sometimes including some bias reduce much more the error by decreasing the variance.
Usually larger N decreases the MSE since it decreases bias and variance error.

Reference

Based on 1.4.1 from Advanced data analysis from a elementary point of view.

Spark and Pyspark

What's Spark?

prueba The definition says:

Spark is a fast and general processing engine compatible with Hadoop data. It can run in Hadoop clusters >through YARN or Spark's standalone mode, and it can process data in HDFS, HBase, Cassandra, Hive, and any >Hadoop InputFormat. It is designed to perform both batch processing (similar to MapReduce) and new >workloads like streaming, interactive queries, and machine learning.

Basically is a framework to work with big amounts of data stored in distributed systems instead of just one machine. This allows parallelization and hence much faster calculations.
It's biggest difference with plain Hadoop is that Spark uses RAM to process data while Hadoop doesn't.

Not being a data engineer myself I can tell you that you can use Spark to work with data stored in HDFS, S3 buckets or a data lake for example. All distributed systems.

Since those usually store huge big amount of data you can see how all this relate. The use case I have been exposed to, as a data scientist, is to query this distributed data and process it before using it for some purpose (modeling, reporting, etc).

How to use it?

I haven't deployed a distributed storage system myself but I think it's safe to assume that amount of data is gathered in big organizations and probably some data engineer has already done all the setup. You just want to access the data from an environment connected to the spark cluster.

There are several languages that can interact with Spark. Scala is the original one but you could use Java or Python. As data scientist we are probably more familiar with Python so I will show you Pyspark

Pyspark

Pyspark is an API to work with Spark using Python. In order to run you need also Java installed and Apache Spark. In our fictional organization a data engineer might have set up a server with Jupyter notebooks linked to the data lake and with all the dependencies.

There are probably ways to connect to the remote spark server from your local machine but I haven't done that.

So, Pyspark allows you to query the datalake/bigdata storage from a jupyter notebook and then convert that to a Pandas Dataframe and work as you are used to.

Spark/Pyspark has a particular syntax that is quite clear but has some particularities based on the parallelization notion. For example, many functions don't actually retrieve all the data, that only happens when you decide to. For example show() or collect() do retrieve the data (and can take a while if you are working with a lot of data) while filter() or withColumn() don't.

Another thing to notice is that you will need to create/initiate a sparkContext before actually being able to query data.

To understand this and have a good amount of examples regarding the functions and syntax I highly recommend THIS SITE.

How to practice?

You can practice Pyspark queries and scripts by installing Pyspark in your local machine despite not having a cluster running distributed data. With Pyspark installed you can create some data and use it as it was real.
You will be able to use all the functions and check them by yourself.

How to install it? You can check THIS GUIDE FOR WINDOWS.

I have struggled a bit to make it work so these are some things I learned during the way.

  • I have downloaded Java 8 since that's what the guide says and use that at my current organization.
  • To avoid creating an account in Oracle to download Java you can check THIS SOLUTION.
  • When creating Environment variables avoid blank spaces
  • If Pyspark doesn't run because can't find Java. Check the %JAVA_HOME% path.
  • If the error is related to missing Python3 , check the %PYTHONPATH% and create in the anaconda path a copy of python.exe but rename it python3.exe

Softmax vs sigmoid

When using Neural Nets for a multiclass classification problem it's standard to have a softmax layer at the end to normalize the probabilities for each class. This means that the output of our net is a vector of probabilities (one for each class) that sums to 1. If there isn't a softmax layer at the end, then the net will output a value in each of the last cells (one for each class) but without a delimited range.
Just a set of numbers where usually the highest is the one with the most probable class but it's not obvious how to value the differences between them.

So, you have a ordered set of numbers, you know which one is the most probable but you want to transform that into clear probabilities. You use the softmax layer.

You could use a sigmoid activation function in the last cell to have individual probabilities. For each class, it transforms the output of the net into a probability. However the sum of those probabilities is not guaranteed to sum 1, actually it won't in practice. It's a simple proxy but you can get better intuitions with softmax.

We will compare how these two approaches affect the last group of weights by inspecting the gradient after calculating the loss for an observation.

I'm using the reticulate package in R to include Python code in Rmarkdown. Pretty nice.

library(reticulate)

We import pytorch to handle tensors and neural net functions.

import numpy as np
import torch
import torch.nn.functional as F
torch.manual_seed(99)
## <torch._C.Generator object at 0x00000262714CF730>
  • 1 obs
  • 5 features (X)
  • 3 possible classes (index 1 = class 2)
  • W. 3 output cells, each one with 5 weights (one per feature)
  • W1 = W2 because we run it twice (two scenarios) and we can't re use the same weights because of the gradient calculated
X = torch.randn(5)
W1 = torch.randn(3,5)
W2 = W1.detach().clone() 
y = torch.tensor([1]) 

We transform everything to positives to make it cleaner and we add the requires_grad_() characteristic that tells pytorch that those tensors need the gradient backpropagated during training

X = X.abs()
W1 = W1.abs().requires_grad_()
W2 = W2.abs().requires_grad_()

We define both losses (softmax and sigmoid).

Softmax

  • Weights * input: cell value
  • we change dimension of output to use it as input of softmax
  • We calculate the softmax (probabilities of each class that sum 1)
  • Apply log because we will use the negative log likelihood
  • We calculate the loss (log of softmax probabilities vs actual class)

Sigmoid

  • Weights * input: cell value
  • we change dimension of output to use it as input of sigmoid
  • We calculate the sigmoid (probabilities of each class individually)
  • Apply log because we will use the negative log likelihood
  • We calculate the loss (log of sigmoid probabilities vs actual class)
# funcion con softmax al final
def softmax_loss(W):
    z = W @ X
    z = z.unsqueeze(0)
    z = torch.softmax(z, dim=1)
    z = torch.log(z)
    return F.nll_loss(z, y)

# funcion con una sigmoidea por activacion
def sigmoid_loss(W):
    z = W @ X
    z = z.unsqueeze(0)
    z = torch.sigmoid(z)
    z = torch.log(z)
    return F.nll_loss(z, y)

We run the forward pass and calculate the loss for the sigmoid first. Then we look for the gradient.
As we can see in the results, only the weights that go to the correct class' output cell are modified. Classes one and three rest untouched. This is because the sigmoid activation just has the individual weights (and cross entropy only look to the correct class)

out_sigmoid = sigmoid_loss(W1)
out_sigmoid.backward()
W1.grad

## tensor([[ 0.0000,  0.0000,  0.0000,  0.0000,  0.0000],
##         [-0.0452, -0.0867, -0.0564, -0.0492, -0.0549],
##         [ 0.0000,  0.0000,  0.0000,  0.0000,  0.0000]])
On the contrary, when running the same net but with softmax layer we see that all the weights are updated. The correct class has gradient with the same sign that for the sigmoid example but the other two classes have in this case opposite sign gradients (which makes sense since you want them to go in the other direction).
This happens because the softmax includes the other classes in each cell since they are needed to normalize and return probabilities that sum up to 1.

out_softmax = softmax_loss(W2)
out_softmax.backward()
W2.grad

## tensor([[ 0.5393,  1.0346,  0.6731,  0.5868,  0.6552],
##         [-0.5576, -1.0697, -0.6959, -0.6066, -0.6775],
##         [ 0.0183,  0.0351,  0.0228,  0.0199,  0.0222]])
This is a simple case with just one layer of weights so we can clearly see this. If you had a fully connected net with more layers, this is valid just for the last one because the gradient is backpropagated and the weights from "other paths" still affect the cell that corresponds to the second class.

Conclusion

The net should evolve during training in a similar way with both last layer activations but the way they do it is different and we try to show in here why. In the end, the sigmoid still reflects the preference for one of the classes and during each epoch it will go through the desired path but just updating some of the weights and not all at the same time.

Counter Strike: chance of winning

This CS GO Kaggle link has data about several competitive CS GO matches. In a few words:

  • those are 5 vs 5 matches where each team tries to kill the other or complete a task (planting or defusing the bomb depending the role you are playing) before the time expires.
  • The goal is to win 16 rounds before the other team.
  • After 15 rounds both teams switch sides/role.

The data has mostly information about each time a player damages another one (and eventually kills it), some grenades usage and some general data of each round as the amount of money spent by each team and the result of that round.

In here I have followed Namita Nandakumar hockey example to obtain and model some basic winning probability based on the lead and how many rounds have been played so far.

This is how probability of winning looks as the game progresses, grouped by how much the current winner is leading. (Averaging leads greater than 4 to keep it clean ).
The thin line is the empirical probability, based solely on segmenting the data.
The thick line is a local regression with its standard deviation. Image

So, as we see there is some noise around the trend and the approximation wiggles a bit as you go through X. We would like to have a model where winning by some amount is always better if you are closer to 16. Let's say it is not crazy to assume that if you are winning by 3, 15 to 12, you should always have higher chances to win than if you are leading 6-3.

Namita shows that xgboost is a nice tool to impose that kind of constraint to a simple model using the monotone constraint parameter.

params = list(objective = "binary:logistic",
              eval_metric = "logloss",
              max_depth = 2,
              eta = 0.1,
              monotone_constraints = c(1,1)) 

What we get is a model that follows the constraints, although has some bias for the lower leading categories. Nevertheless is a quick approach to approximate the probabilities in a credible way.
You could use the dataset to explore other stuff since it has some rich information about locations.

Image