$X_i \sim^{iid}\operatorname{Ber}(p)$ and $Y_m = \sum_{i=1}^{m}X_i$. find $E[Y_m|Y_n]$

Question

I have a math problem regarding condition expectancy. Let there be $$X_i \sim^{iid}\operatorname{Ber}(p), Y_m = \sum_{i=1}^{m}X_i$$.

Now we know that $$Y_m\sim \operatorname{Bin}(m,p), m \leq n$$

Im tring to find $E[Y_m\mid Y_n]$ but I dont know how to approach it to begin with, since Im not sure whether to keep it with $\Sigma$ oreither change to two binomial conditioned.

Suppose $m \leq n$ so the smaller sum can definately give us information about the greater one, so I know Bayes‏ is needed. So far what came to mind was:

Create a variable $Y_k = Y_n-Y_m$ so that I can divide the bigger sum into the dependent stuff and what's not. but I'm not so sure about it. I'd like a guide for how to proceed, not even a full solution but a hint.

Answer 1

Since $X_i$'s are i.i.d $E(X_i|X_1+X_2+...+X_n)$ is the same for each $i \leq n$. The sum of these conditional expectations is $$E(X_1+X_2+...+X_n|X_1+X_2+...+X_n)$$ $$=X_1+X_2+...+X_n=Y_n.$$ Hence $$E(X_i|X_1+X_2+...+X_n)=Y_n/n$$ for each $i\leq n$. Adding then first $m$ of these we get $E(Y_m|Y_n)= \frac m n Y_n$.

Answer 2

Hint

Let $m<n$. $$\mathbb E[Y_n\mid Y_m]=Y_m+\mathbb E[Y_n-Y_m\mid Y_m].$$

Moreover, you have that $Y_n-Y_m$ and $Y_m$ are independent.