# Solved – conditional expectation of squared standard normal

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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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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