Conditional expectation of i.i.d random variables

Question

consider the sum of random variables $Q_k=\sum_k R_k $, $R_k$ are i.i.d. Now I want to calculate:

$$E(Q_j| Y_{k+j}=n) =j \frac{k}{k+j}$$

Answer 1

Since $X_i$'s are i.i.d. it follows that $c=E(X_i|Y_{n+m}=n)$ is independent of $i$. Summing over $i \leq n+m$ we get $(m+n) c=n$ so $c =\frac n {n+m}$. Finally the given expectation is $mc=\frac {mn} {m+n}$.

Answer 2

Note that all permutations of the $X_j$ are equally likely. Thus the desired expectation is invariant if you average over them. If the sum of the first $m+n$ of the $X_j$ is $n$ (or in fact any other value, there’s no reason why it should be linked to the indices), then since each of these $m+n$ occurs in the first $m$ places in a fraction $\frac m{m+n}$ of the permutations, the expectation of the first $m$ is $\frac m{m+n}$ times that value.

Answer 3

Just another solution:

$$E(Y_m| Y_{m+n}=n)=E(\sum_{k=1}^{m} X_k| \sum_{i=1}^{m+n} X_i=n)$$ $$=\sum_{k=1}^{m}E( X_k| \sum_{i=1}^{m+n} X_i=n)=\sum_{i=1}^{m} \frac{n}{n+m}= m \frac{n}{n+m} $$

It is enough to show $E( X_k| \sum_{i=1}^{m+n} X_i=n)=\frac{n}{n+m}$

$$E( \sum_{k=1}^{m+n} X_k| \sum_{i=1}^{m+n} X_i=n)=n$$ $$\Leftrightarrow$$ $$\sum_{k=1}^{m+n}E( X_k| \sum_{i=1}^{m+n} X_i=n)=n$$

$$\Leftrightarrow$$ $$(m+n)E( X_k| \sum_{i=1}^{m+n} X_i=n)=n$$

Answer 4

Just another solution

lets $E(X_i)=\mu$. In non-parametric family, $\bar{X}$ is sufficient and complete estimator for $\mu$(that coincide with this example).

$$E(X_k|\sum_{i=1}^{m+n} X_i)=E(X_k|\frac{\sum_{i=1}^{m+n} X_i}{m+n})=E(X_k|\bar{X}_{(m+n)})=g(\bar{X}_{(m+n)})$$

I want to show $g(\bar{X}_{(m+n)})=\bar{X}_{(m+n)}$ almost surely.

$$E(g(\bar{X}_{(m+n)})-\bar{X}_{(m+n)})=E(E(X_k|\bar{X}_{(m+n)}))-\mu=\mu-\mu=0$$ since $\bar{X}_{(m+n)}$ is complete and sufficient and
$g(\bar{X}_{(m+n)})-\bar{X}_{(m+n)}$ is a function of $\bar{X}_{(m+n)}$ so $$P(g(\bar{X}_{(m+n)})-\bar{X}_{(m+n)}=0)=1$$

so $g(\bar{X}_{(m+n)})=\bar{X}_{(m+n)}$ almost surely. so

$$E(X_k|\sum_{i=1}^{m+n} X_i)=\bar{X}_{(m+n)}$$ and

$$E(X_k|\sum_{i=1}^{m+n} X_i=n)=E(X_k|\frac{\sum_{i=1}^{m+n} X_i}{m+n}=\frac{n}{m+n})=\frac{n}{m+n}$$. Finally

$$E(\sum_{k=1}^{m} X_k| \sum_{i=1}^{m+n} X_i=n)=m\frac{n}{m+n}$$