# Distribution of a simple random sample

Homework Question

Hello, I have a simple homework question that asks:

We know that when we sample, with equal probability from a finite population {x1, x2,..., xN} without replacement, we obtain a simple random sample instead of an (iid) random sample. Let X1,...,Xn (n < N) be the sample obtained. It is claimed in class that X1,..., Xn, though no longer independent, are still identically distributed according to the pmf:

f(x) = 1/N, if x = xi, for some 1 <= i <= N; 0, for all other x.

The first choice X1 obviously follows this distribution. Show that the second choice X2 is distributed according to f(x) too.

My work: I did some research on exchangeable random variables. My thinking is that if I can show the sequence of random variables X1,..., Xn is exchangeable, then it follows that any of the Xi's can be the first random variable in the sequence. Therefore, each Xi will follow f(x).

My question: I am unsure if I properly understand what being exchangeable means. Also, I am not sure how you would properly prove that this sequence is exchangeable. I would really appreciate any tips on going about it.

Thank you.

• This has been asked and answered several times on this site. Search for "exchangabilty" and look through the answers. Here is one particularly highly voted question: stats.stackexchange.com/questions/3520/… – user31668 Mar 3 '14 at 3:48
• Also, given this is a homework question, you should add the self-study tag. – Patrick Coulombe Mar 3 '14 at 3:55
• An answer to the homework question (not the question about exchangeability proof) can be found here for the case $N = n = 2$. More generally, $$P(X_2=x_i)=\sum_{j=1}^NP(X_2=x_i\mid X_1=x_j)P(X_1=x_j)=\sum_{j=1}^{i-1}\frac{1}{N-1}\cdot\frac{1}{N}+0+\sum_{j=i+1} \frac{1}{N-1}\cdot\frac{1}{N} = \frac{1}{N}.$$ – Dilip Sarwate Mar 3 '14 at 3:57

$P(X_2)$ given some fixed $X_1$ is $1\over N-1$.
But there were $N-1$ choices for $X_1$, each with $P(X_1) = {1 \over N}$, for a given $X_2$.
Multiplying by all the possible $X_1$ we get
$P(X_2) = {1\over N-1} {N-1\over N}$