independence of random variables with random index

Question

Let $\{X_i\}$ be iid real random variables. And N be a random variable, taking values in $\mathbb{N}$.

My question is if $X_{N}$ has the same distribution as $X_{N+1}$, because of the iid assumption? If yes, have you any suggestions how to start this proof? Thank you verry much in advance!

Answer 1

We have:$$P\left(X_{N}\in A\right)=\sum_{n=0}^{\infty}P\left(N=n\right)P\left(X_{N}\in A\mid N=n\right)=$$$$\sum_{n=0}^{\infty}P\left(N=n\right)P\left(X_{n}\in A\mid N=n\right)$$ and:

$$P\left(X_{N+1}\in A\right)=\sum_{n=0}^{\infty}P\left(N=n\right)P\left(X_{N+1}\in A\mid N=n\right)=$$$$\sum_{n=0}^{\infty}P\left(N=n\right)P\left(X_{n+1}\in A\mid N=n\right)$$

So the answer is "yes" if: $$P\left(X_{n}\in A\mid N=n\right)=P\left(X_{n+1}\in A\mid N=n\right)\text{ for every }n\in\mathbb{N}$$

Sufficient conditions for this are:

• $N$ and $X_n$ are independent for every $n\in\mathbb N$.

• The rv's $X_{n}$ have identical distribution.

Answer 2

We have $X \overset{d}{=} Y$ if $\mathbb{E}(f(X)) = \mathbb{E}(f(Y))$ for all continuous bounded functions $f$. Now observe,

$$ \mathbb{E}(f(X_N)) = \mathbb{E}(\mathbb{E}(f(X_N)|N)) = \mathbb{E}(\mathbb{E}(f(X_n)|N) \vert_{n=N} ) = \mathbb{E}(\mathbb{E}(f(X_{n+1})|N) \vert_{n=N} ) = \mathbb{E}(f(X_{N+1})) $$

as long as $\mathbb{E}(f(X_{n+1})|N) = \mathbb{E}(f(X_{n})|N)$.

If we don't know that $X_k|N$ are identically distributed, then the proof doesn't go through. Here is a counter example: Let $X_k \overset{iid}{\sim} N(0, 1)$ and take $N = \mathbb{I}(X_1 > 0)$.