# Independence in Variables

## Discrete

Two discrete random variables $X$ and $Y$ are called independent if: \begin{align*} \P(X=x,Y=y) = \P(X=x)\P(Y=y) \text{ for all } x,y \end{align*} Intuitively: knowing the value of $X$ tells us nothing about the distribution of $Y$. If two variables are not independent, they are called dependent. This is a similar conceptually to independent events, but we are dealing with multiple*variables*. Make sure to keep your events and variables distinct.

## Continuous

Two continuous random variables $X$ and $Y$ are called independent if: \begin{align*} \P(X\leq a, Y \leq b) = \P(X \leq a)\P(Y \leq b) \text{ for all } a,b \end{align*} This can be stated equivalently using either the CDF or the PDF: \begin{align*} F_{X,Y}(a,b) &= F_{X}(a)F_{Y}(b) \text{ for all } a,b \\ f(X=x,Y=y) &= f(X=x)f(Y=y) \text{ for all } x,y \end{align*} More generally, if you can factor the joint density function then your random variable are independent (or the joint probability function for discrete random variables): \begin{align*} &f(X=x,Y=y) = h(x)g(y) \\ &\P(X=x,Y=y) = h(x)g(y) \end{align*}## Example: Showing Independence

Let $N$ be the # of requests to a web server/day and that $N \sim \Poi(\lambda)$. Each request comes from a human with probability = $p$ or from a "bot" with probability = $(1 – p)$. Define $X$ to be the # of requests from humans/day and $Y$ to be the # of requests from bots/day. Show that the number of requests from humans, $X$, is independent of the number of requests from bots, $Y$.

Since requests come in independently, the probability of $X$ conditioned on knowing the number of requests is a Binomial. Specifically: \begin{align*} (X|N) &\sim \Bin(N,p)\\ (Y|N) &\sim \Bin(N, 1-p) \end{align*} To get started we need to first write an expression for the joint probability of $X$ and $Y$. To do so, we use the chain rule: \begin{align*} \P(X=x,Y=y) = \P(X = x, Y=y|N = x+y)\P(N = x+y) \end{align*} We can calculate each term in this expression. The first term is the PMF of the binomial $X|N$ having $x$ "successes". The second term is the probability that the Poisson $N$ takes on the value $x+y$ : \begin{align*} &\P(X = x, Y=y|N = x+y) = { {x + y} \choose x}p^x(1-p)^y \\ &\P(N = x + y) = e^{-\lambda}\frac{\lambda^{x+y}}{(x+y)!} \end{align*} Now we can put those together we have an expression for the joint: \begin{align*} &\P(X = x, Y=y) = { {x + y} \choose x}p^x(1-p)^y e^{-\lambda}\frac{\lambda^{x+y}}{(x+y)!} \end{align*} At this point we have derived the joint distribution over $X$ and $Y$. In order to show that these two are independent, we need to be able to factor the joint: \begin{align*} \P&(X = x, Y=y) \\ &= { {x + y} \choose x}p^x(1-p)^y e^{-\lambda}\frac{\lambda^{x+y}}{(x+y)!} \\ &= \frac{(x+y)!}{x! \cdot y!} p^x(1-p)^y e^{-\lambda}\frac{\lambda^{x+y}}{(x+y)!} \\ &= \frac{1}{x! \cdot y!} p^x(1-p)^y e^{-\lambda}\lambda^{x+y} && \text{Cancel (x+y)!} \\ &= \frac{p^x \cdot \lambda^x}{x!} \cdot \frac{(1-p)^y \cdot \lambda ^{y}}{y!} \cdot e^{-\lambda} && \text{Rearrange} \\ \end{align*} Because the joint can be factored into a term that only has $x$ and a term that only has $y$, the random variables are independent.

## Symmetry of Independence

Independence is symmetric. That means that if random variables $X$ and $Y$ are independent, $X$ is independent of $Y$ and $Y$ is independent of $X$. This claim may seem meaningless but it can be very useful. Imagine a sequence of events $X_1,X_2, \dots$. Let $A_i$ be the event that $X_i$ is a "record value" (eg it is larger than all previous values). Is $A_{n+1}$ independent of $A_n$? It is easier to answer that $A_n$ is independent of $A_{n+1}$. By symmetry of independence both claims must be true.## Expectation of Products

**:**

*Lemma: Product of Expectation for Independent Random Variables*If two random variables $X$ and $Y$ are independent, the expectation of their product is the product of the individual expectations. \begin{align*} &E[X \cdot Y] = E[X] \cdot E[Y] && \text{ if and only if $X$ and $Y$ are independent}\\ &E[g(X)h(Y)] = E[g(X)]E[h(Y)] && \text{ where $g$ and $h$ are functions} \end{align*} Note that this assumes that $X$ and $Y$ are independent. Contrast this to the sum version of this rule (expectation of sum of random variables, is the sum of individual expectations) which does

**not**require the random variables to be independent.