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Question:

Given $X\sim N(\mu_X, \sigma_X^2)$ and $Y\sim N(\mu_Y, \sigma_Y^2)$ are independent, and you know $X+Y=s$. What is the expected value of $X$?

I encountered this during an interview. My thoughts were to use conditional expectation $E(X|Z=s)$ where $Z\sim N(\mu_X+\mu_Y, \sigma_X^2+\sigma_Y^2)$. However, it involves a lot of calculations, which I don't think would be an interview question. Could anyone suggest?

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    $\begingroup$ (1) Find the joint distribution of $(X, X+Y).$ (2) Use your understanding of univariate ordinary least square regression. $\endgroup$
    – whuber
    Aug 17, 2022 at 22:58
  • $\begingroup$ related: stats.stackexchange.com/questions/9071/… $\endgroup$
    – Henry
    Aug 18, 2022 at 0:42
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    $\begingroup$ I took an approach similar to the one suggested by @whuber in my answer where I replaced (2) with your understanding/knowledge of a property, the conditional expectation, of bivariate normal distributions. Arguably, for your question this is equivalent to knowing how to express the so-called population regression function in the simple linear regression model (i.e., the conditional expectation of the dependent variable $X$ given the intercept and regressor $\left(X+Y\right)$) in terms of $\mathrm{Cov}(X,X+Y)$, $\mathbb E(X)$, $\mathbb E(X+Y)$, and $\mathbb V(X+Y)$. $\endgroup$
    – statmerkur
    Aug 18, 2022 at 9:44

1 Answer 1

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Since $X$ and $Y$ are both univariate normal and independent, we know that all linear combinations of $X$ and $X+Y$ are univariate normal. Thus, $\left(X,X+Y\right)^\top$ is bivariate normal and we have $$ X+Y \sim \mathcal N\left(\mu_X+\mu_Y, \sigma_X^2+\sigma_Y^2\right), \\ \mathrm{Cov}\left(X,X+Y\right)=\mathrm{Cov}\left(X,X\right)+\mathrm{Cov}\left(X,Y\right)=\sigma_X^2, \\ \begin{pmatrix} X\\ X+Y\\ \end{pmatrix} \sim \mathcal N\left(\begin{pmatrix} \mu_X\\ \mu_X+\mu_Y \end{pmatrix}, \begin{pmatrix} \sigma_X^2 & \sigma_X^2\\ \sigma_X^2 & \sigma_X^2+\sigma_Y^2 \end{pmatrix}\right). $$ The well-known formula for the conditional expectation in the bivariate normal case then yields $$ \mathbb E\left(X|X+Y=s\right) = \mu_X+\frac{\sigma_X^2}{\sigma_X^2+\sigma_Y^2}\left(s-\left(\mu_X+\mu_Y\right)\right). $$

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