# Showing two random variables independent despite seemingly looking dependent

I just met this in probability and it got me completely stumped:

We define an i.i.d sequence of normally distributed random variables $\{ X_n \}_{n=1}^{\infty}$ such that $X_n \sim \mathcal{N} (0,\sigma ^2)$ and we define a random variable $N \sim Pois(\Lambda)$ independent of the $X_n$'s

We are asked to show that $X_N$ and N are independent random variables

I cannot seem to get past the simple fact that these seem clearly dependent as N clearly appears as a subscript for the X random variable so I do not really know what is wrong or how to prove that these are independent random variables so this is where I need help. Thanks all.

## 2 Answers

Independence follows from the fact that for every fixed $k\in\mathbb N$:$$P(X_N\in A)=\sum_{n=0}^{\infty} P(X_N\in A\mid N=n)P(N=n)=$$$$\sum_{n=0}^{\infty} P(X_n\in A)P(N=n)=\sum_{n=0}^{\infty} P(X_k\in A)P(N=n)=$$$$P(X_k\in A)=P(X_N\in A\mid N=k)$$

This is not different from rolling a green and a red die ans using a coin flip to decide which of the dice you use and which you ignore. You still have a coin flip and a die.

All that matters is that the $X_n$ are identically distributed (they need not be normally distributed and not independent of one anaother) and that $N$ is independent of all $(X_1,X_2,\ldots)$.

For any measurable $A$ and $n\in\Bbb N$, we have (letting $Y=X_N$ for clarity) \begin{align} P(Y\in A\mid N=n)&=P(X_n\in A\mid N=n)&\text{def of Y}\\&=\frac{P(X_n\in A\wedge N=n)}{P(N=n)}&\text{def of conditional prob.}\\&=P(X_n\in A)&\text{independent}\\&=P(X_1\in A)&\text{identically distributed}\end{align} so that (as one might have guessed) \begin{align}P(Y\in A)&=\sum_n P(Y\in A\mid N=n)P(N=n)\\&=\sum_nP(X_1\in A)P(N=n)\\&=P(X_1\in A)\sum_nP(N=n)\\&=P(X_1\in A)\end{align} and hence $$P(Y\in A\mid N=n)=P(Y\in A).$$

• Something seems wrong here. First computations a conditional probability is equal to a product of multiplications maybe you mean AND instead of conditional probability? – kroner Apr 19 '16 at 8:36
• @Hagen Something seems wrong with your answer – Don John Prep Apr 19 '16 at 8:46
• Your first sentence makes things clear on a nice way. If you repair (e.g. by deleting everything from: "for any measurable $A$...") then I will turn my downvote into an upvote. $P(Y\in A| N=n)$ and $P(Y\in A\wedge N=n)$ are essentially different. – drhab Apr 20 '16 at 14:04
• I decided to do it myself. My downvote stays a downvote. – drhab Apr 21 '16 at 8:35