I have to find the conditional pmf and show it sums to one. Using the formula for conditional probability, I have written down $$\mathbb{P}(X_1=a_1,...,X_n=a_n \mid S_n=k) = \frac{\mathbb{P}(X_1=a_1 ,..., X_n=a_n, S_n=k)}{\mathbb{P}(S_n=k)}$$

I know $S_n$ follows a $Binomial(n,p)$ distribution. What I don't understand is how to deal with the top. If i just multiply each Bernoulli pmf I get $\frac{k!(n-k)!}{n!}$, which I'm pretty sure is not the right answer. Then I have to deduce whether $X_1,...,X_n$ are conditionally independent given $S_n=k$. This confuses me for the same reason, I'm not sure how to deal with $\mathbb{P}(X_1=a_1,...,X_n=a_n \mid S_n=k)$ or how to get $\mathbb{P}(X_1=a_1\mid S_n=k)$. Thanks in advance.


You have the right conditional pmf provided $a=(a_1,\ldots,a_n)$ is in the set $$ S:=\Big\{a:a_i\in\{0,1\},|\{i:a_i=1\}|=k\Big\},$$ and the pmf is zero otherwise. To show that the pmf sums to $1$ you just have to count the number of elements in $S$, which you should be able to do.

To test for independence, try a checking events of the form $A=\{X_1=X_2=\ldots=X_k=1\}$ and $B=\{X_{k+1}=\ldots=X_n=0\}$. If $(X_1,\ldots,X_n)$ were independent conditioned on $\{S_n=k\}$ then you would have $\mathbb P(A\cap B\,|\,S_n=k)=\mathbb P(A\,|\,S_n=k)\mathbb P(B\,|\,S_n=k)$. Is this the case here?

  • $\begingroup$ k of the n $a_i$'s have to be 1, so should it seems like there are n choose k elements. However, I don't see how this connects to the pmf summing to 1. $\endgroup$ – N Dizzle Sep 12 '18 at 2:29
  • $\begingroup$ Since $\mathbb P(X=a\,|\,S_n=k)$ does not depend on $a$, one simply has that the sum is equal to the number of elements in $S$ times the probability of one of them, that is, ${{n}\choose{k}}\frac{k!(n-k)!}{n!}$. Now expand ${{n}\choose{k}}$ and simplify. $\endgroup$ – Jason Sep 12 '18 at 13:26
  • $\begingroup$ A gentleman and a scholar. Thank you kind sir. $\endgroup$ – N Dizzle Sep 13 '18 at 0:02

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