I just wonder if anybody can help me to prove the following identity:
Given a series of i.i.d. non-negative random variables $X_1, X_2, ..., X_n$, then $$E(X_1+X_2+ \cdots +X_k \mid X_1+X_2+ \cdots +X_n=b)=b \cdot \frac{k}{n} .$$
I just wonder if anybody can help me to prove the following identity:
Given a series of i.i.d. non-negative random variables $X_1, X_2, ..., X_n$, then $$E(X_1+X_2+ \cdots +X_k \mid X_1+X_2+ \cdots +X_n=b)=b \cdot \frac{k}{n} .$$
You can reduce yourself to the case where $k = 1$ because the expectation is a linear operator.
Since $X_i$'s are i.i.d., $$ \mathbb E(X_i \, | \, X_1 + \dots + X_n = b ) $$ does not depend on $i$ (as long as $1 \le i \le n$). Thus $$ n \, \mathbb E \left( X_i \, \left| \sum_{i=1}^n X_i = b \right. \right) = \sum_{i=1}^n \, \mathbb E \left(X_i \, \left| \, \sum_{i=1}^n X_i = b \right. \right) = \mathbb E \left( \sum_{i=1}^n X_i \, \left| \, \sum_{i=1}^n X_i = b \right. \right) = b $$ so that $$ \mathbb E \left( X_i \, \left| \sum_{i=1}^n X_i = b \right. \right) = \frac bn. $$ Your case can then be solved by linearity of expectation.
Hope that helps,