# Transformation of Probabilistic Distributions

Break Yang April 14, 2023 #math #reinforcement learning #machine learning# Motivation

To use

`rsample`

or not to use`rsample`

, that is a question.

If you ever come across the above when implementing a deep learning algorithm, for example, a policy gradient algorithm for reinforcement learning, this post is about that.

**Disclaimer**: Please note that I am more interested in making the math intuitive rather than strict here.

# The concepts

## Transformation of Random Variables

Our exploration starts with a math problem that you might find in the assignments from an introductory level statistics course. Suppose we have an 1D random variable $\mathbf{X}$ following Gaussian distribution,

$$ \mathbf{X} \sim \mathcal{N}(0, 1) $$

the probabilistic density function (pdf) of this distribtion is

$$ p(x) = \frac{1}{\sqrt{2\pi}} \exp \left( -\frac{x^2}{2} \right) $$

Now, assume $\mathbf{Y}$ is another random variable that has a close relationship with $\mathbf{X}$.

$$ \mathbf{Y} = \sigma \mathbf{X} + \mu =: f(\mathbf{X}) $$

**Question**: What is the pdf of $\mathbf{Y}$? What kind of distribution does $\mathbf{Y}$ follow?

To answer this question, we can start from simple principles. Let's denote the pdf of $\mathbf{Y}$ as $q(\cdot)$. The integral of both pdf $p(\cdot)$ and $q(\cdot)$ over $\mathbb{R}$ must be equal to 1.

$$ 1 = \int_{-\infty}^{+\infty}p(x) \mathrm{d}x = \int_{-\infty}^{+\infty}q(y) \mathrm{d}y $$

It is not hard to deduce that, based on intuition, the above equation that constraints $p(\cdot)$ and $q(\cdot)$ is too **loose**. The probability of $\mathbf{X}$ taking any value $x$ must be equal to the probability of $\mathbf{Y}$ taking the corresponding value $y = f(x)$. This means that

$$ \forall x \in \mathbb{R}, \mathbb{P} \{ X = x \} = \mathbb{P} \{ Y = f(x) \} $$

The above equation can be expressed with derivative form (well intuively but inaccurately you can understand this as the form you usually write to the right of $\int$) equivalence as below

$$ \forall x \in \mathbb{R} \text{ and } y = f(x), p(x) \mathrm{d}x = q(y) \mathrm{d}y $$

Note that since $y = f(x)$, and fortunately $f$ is a linear function so that it can be inversed, we can now simplify the above as

$$ \begin{darray}{rcl} &&p(x) \mathrm{d}x && = q(y) \mathrm{d}y \\ \Rightarrow && p(f^{-1}(y)) \mathrm{d}(f^{-1}(y)) && = q(y) \mathrm{d}y \end{darray} $$

We are now blocked by $\mathrm{d}(f^{-1}(y))$, which is a derivative form. Without a formal definition, we can simplify it with very intuitive rules. For example, in this case since $f(x) = \sigma x + \mu$,

$$ f^{-1}(y) = \frac{y - \mu}{\sigma} $$

The form $\mathrm{d}(f^{-1}(y))$ is pretty much in plain word just the area under $f^{-1}(\cdot)$ within an infinitesimal neighborhood around a specific $y$. By just applying a symbolic trick (dividing $\mathrm{d}y$ and multiply it back), we can derive:

$$ \begin{darray}{rcl} \mathrm{d}(f^{-1}(y)) && = && \frac{\mathrm{d}(f^{-1}(y))}{\mathrm{d}y} \cdot \mathrm{d} y \\ && = && \frac{\mathrm{d}((y - \mu) / \sigma)}{\mathrm{d}y} \cdot \mathrm{d}y \\ && = && \frac{1}{\sigma} \cdot \mathrm{d} y \\ && = && \frac{\mathrm{d}y}{\sigma} \end{darray} $$

Please note that $\mathrm{d}(\cdot)/\mathrm{d}y$ is just what we usually call **the derivative w.r.t $y$a**. More generally, if the transformation function $f$ is invertible, the above equation is reduced to

$$ \begin{equation} \mathrm{d}(f^{-1}(y)) = \frac{\mathrm{d} y}{f'(f^{-1}(y))} \end{equation} $$

Now, we have

$$ \begin{darray}{rcl} &&p(x) \mathrm{d}x && = q(y) \mathrm{d}y && \\ \Rightarrow && p(f^{-1}(y)) \mathrm{d}(f^{-1}(y)) && = q(y) \mathrm{d}y && \\ \Rightarrow && p(f^{-1}(y)) \frac{\mathrm{d} y}{f'(f^{-1}(y))} && = q(y) \mathrm{d}y && \\ \Rightarrow && p(\frac{y - \mu}{\sigma}) \cdot \frac{\mathrm{d}y}{\sigma} && = q(y) \mathrm{d}y && \\ \Rightarrow && \frac{1}{\sigma} p(\frac{y - \mu}{\sigma}) && = q(y) && (\text{ eliminate } \mathrm{d}y)\\ \Rightarrow && q(y) = \frac{1}{\sqrt{2 \pi} \sigma} \exp -\frac{(y - \mu)^2}{\sigma^2} && \end{darray} $$

Now, we have proved (sort of intuitively and symblically via derivative form arithmetics) that such linear transformation of the random variable $\mathrm{X}$ still follows Gaussian distribution. In fact,

$$ \mathrm{Y} \sim \mathcal{N}(\mu, \sigma^2) $$

Pretty straight forward, right?