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Suppose $X_1\sim N(\mu,\sigma^2)$ and $X_2,X_3,...,X_n\sim N(0,\sigma^2)$, is the following identity correct? $$E(X_1+X_2+...+X_n|X_1>\max(X_2,X_3,...,X_n))=E(X_1+X_2+...+X_n)$$
It seems that it is correct based on simulation results. But I couldn't figure out how to prove it. Could anyone kindly help? Thanks.
I am going to assume the random variables are independent, otherwise the problem is not well-defined. Without loss of generality, I am also going to assume $\sigma^2 = 1$.
Consider $n = 2$ and let $X_1 = \mu + Z_1$ and $X_2 = Z_2$ where $Z_1, Z_2$ i.i.d. $\sim N(0,1)$. Let $U = Z_1 + Z_2$ and $V = Z_1 - Z_2$. One can verify that $U$ and $V$ are distributed as $N(0,2)$. Moreover their covariance is zero, hence being jointly normal, they are also independent. Then, \begin{align*} \mathbb E(X_1 + X_2 \mid X_1 > X_2) &= \mathbb E (\mu + Z_1 + Z_2 \mid \mu + Z_1 > Z_2) \\ &= \mu + \mathbb E (U \mid V > -\mu) \\ &= \mu + \mathbb E(U) \quad \text{by indept.} \\ &= \mu. \end{align*}
You can stop reading here and try to argue the general case yourself.
How about the general case? Let $X_1 = \mu + Z_1$ as before. Let $V_i = X_i - Z_1$ for $i \ge 2$. Let $$ Y = Z_1 + X_2 +\cdots +X_n. $$ Then, argue that $Y$ is independent of $(V_2,V_3,\dots,V_n)$. (Hint: compute the covariance of $Y$ and each of $V_i, i \ge 2$.) We have \begin{align*} \mathbb E(X_1 + \dots +X_n \mid X_1 > \max\{X_2,\dots,X_n\}) &= \mathbb E (\mu + Y \mid \max_{i \ge 2}V_i > -\mu) \\ &= \mu + \mathbb E (Y \mid \max_{i \ge 2}V_i > -\mu) \\ &= \mu + \mathbb E(Y) \quad \text{by indept.} \\ &= \mu. \end{align*}
