2
$\begingroup$

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 $E(A^2|A+B) = \frac{A^2+B^2}{2}$.

$\endgroup$
  • 3
    $\begingroup$ Note that you don't actually know $A$ and $B$, only their sum, which isn't sufficient to calculate $A^2 + B^2$ in your last line. $\endgroup$ – jbowman Dec 16 '13 at 19:03
  • 1
    $\begingroup$ Hint: $X=(A+B)$ and $Y=(A-B)$ are iid normals (each has variance $2$) and $A^2+B^2=(X^2+Y^2)/2.$ Can you find $\frac{1}{2}\mathbb{E}(X^2+Y^2|X)$? $\endgroup$ – whuber Dec 16 '13 at 22:00
1
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.