Let $A,B$ independent standard normals. What is $E(A^2|A+B)$?

Is the following ok?

$A,B$ iid and hence $(A^2,A+B),(B^2,A+B)$ iid.

Therefore we have $int_M A^2 dP = int_M B^2 dP$ for every $A+B$-measurable set $M$ and hence $E(A^2|A+B) = E(B^2|A+B)$.

We obtain $2 cdot E(A^2|A+B) = E(A^2|A+B) + E(B^2|A+B) = E(A^2+B^2|A+B) = A^2+B^2$ where the last equation holds since $A^2+B^2$ is $A+B$-measurable.

Finally we have $E(A^2|A+B) = frac{A^2+B^2}{2}$.

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#### Best Answer

There are similar questions on CV, but I haven't seen anyone give the details on the distribution of $A,$ so here goes.

Let $Z = A+B.$ Following the logic given by Xi'an here: Simulation involving conditioning on sum of random variables , the pdf for $A| Z$ is

$$ f_{A|Z}(a|z)=frac{f_B(z-a) f_A(a)}{f_Z(z)}=frac{frac{1}{sqrt{2 pi}}e^{-left(z-a right)^2 over 2 } frac{1}{sqrt{2 pi}}e^{-a^2 over 2}}{frac{1}{sqrt {4 pi}} e^{{-z^2 over 4 }}} $$

This simplifies to $$ f_{A|Z}(a|z) = frac{1}{sqrt{pi}} e^{- left( a-frac{z}{2} right)^2} $$

If we let $w=frac{1}{sqrt{2}}$ then we can write it as

$$ f_{A|Z}(a|z) = frac{1}{sqrt{2 pi w^2}} e^{- left( a-frac{z}{2} right)^2 over 2w^2} $$

Now we can see the conditional distribution is normal with mean $z over 2$ and variance $frac{1}{2}.$

So to answer the question, since for any random variable $X$ with finite variance we know that $E[X^2]=mu^2+sigma^2,$ we have

$$E[A^2|A+B=z]=left( frac{z}{2} right)^2 + frac{1}{2}= frac{z^2+2}{4} = frac{left( A + B right)^2+2}{4} $$

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