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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.

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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*}

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  • $\begingroup$ Brilliant! @passerby51 $\endgroup$
    – Ruth
    Jan 4, 2021 at 7:13
  • $\begingroup$ When I revisit this, for the general case, I understand how Y is independent of each of the V_i. But does that imply that Y is independent of max(V_i)? Anyone could help. That link is not clear to me. $\endgroup$
    – Ruth
    Dec 2, 2021 at 4:43
  • $\begingroup$ @Ruth It is not only that Y is independent of each V_i. It is more: Y is independent of $\mathbf{V} = (V_2,...V_n)$ as a whole. Although in the Gaussian case, these two statements are equivalent since both statements reduce to a certain part of the covariance matrix being zero. Once you have that $Y$ is independent of $\textbf{V}$, then any (measurable) function of $Y$ will be independent of any (meas.) function of $\mathbf{V}$. This is a fact of probability theory. $\endgroup$
    – passerby51
    Dec 2, 2021 at 17:01

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