# X, Y are independent normal distributions, find E[X | X+Y = s] [duplicate]

• $$X\sim N(\mu_1, \sigma_1^2), Y \sim N(\mu_2, \sigma_2^2)$$ 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?
• As explained in many threads here, independent Normal variables are bivariate Normal and linear combinations of multivariate Normals are also Normal.
– whuber
Commented Aug 30, 2022 at 12:32
• @whuber I know the conclusion, I am asking about the derivation : ) Commented Aug 30, 2022 at 12:56
• I just gave you the derivation. If you need more details, you can find them in the duplicate.
– whuber
Commented Aug 30, 2022 at 13:17