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  • $X\sim N(\mu_1, \sigma_1^2), Y \sim N(\mu_2, \sigma_2^2)$ are independent, find $E[X|X+Y=s]$

  • I found the result in this question, but I don't know how to derive it

  • my approach

$$ E[X|X+Y=s] = \int x\; \Pr(X=x | X+Y=s) \; dx\\ = \int x\; \Pr(X=x) \Pr(Y=s-x) \; dx\\ = \int x\; {1\over \sqrt{2\pi} \sigma_1} e^{-{1\over 2} ({x-\mu_1\over \sigma_1})^2 } {1\over \sqrt{2\pi} \sigma_2} e^{-{1\over 2} ({s-x-\mu_2\over \sigma_2})^2 } \; dx\\ $$

  • i don't know how to continue from here. This is a quant interview question that is supposed to be calculated by hand, so I guess there is some trick I am missing?
em1971
  • 306
  • As explained in many threads here, independent Normal variables are bivariate Normal and linear combinations of multivariate Normals are also Normal. – whuber Aug 30 '22 at 12:32
  • @whuber I know the conclusion, I am asking about the derivation : ) – em1971 Aug 30 '22 at 12:56
  • I just gave you the derivation. If you need more details, you can find them in the duplicate. – whuber Aug 30 '22 at 13:17

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