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Suppose $a_1 = b + c_1$ and $a_2 = 2b + c_2$ where $b, c_1, c_2$ are all $N(0,1)$

Find $E[b|a_1,a_2]$

My attempt: As $E[b] = 0$, I assume $E[b|a_1, a_2] = 0$. Is this a logical assumption?

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    $\begingroup$ Hint: when you know $a_2$ and $a_1,$ you also know $a_2-a_1$ via subtraction. What information does that give you about $b$? $\endgroup$
    – whuber ♦
    Oct 25, 2013 at 19:43
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    $\begingroup$ Is this a logical assumption? No, a conditional mean such as $E[b\mid a_1, a_2]$ is not necessarily equal to the unconditional mean $E[b]$. $\endgroup$
    – Dilip Sarwate
    Oct 25, 2013 at 20:50
  • $\begingroup$ I think a better way to do this is to apply law of total expectation in reverse and work out the joint distribution of a1 and a2? $\endgroup$
    – honeychip
    Oct 26, 2013 at 1:15
  • $\begingroup$ This is the second homework problem you've posted today... Neither had much of an "attempt". $\endgroup$
    – nomen
    Oct 26, 2013 at 4:11

1 Answer 1

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Let $B,C_1,C_2$ be independent $\mathrm{N}(0,1)$ random variables. Define $A_1=B+C_1$ and $A_2=2B+C_2$. Since we are conditioning on the same information, and $C_1$ and $C_2$ have the same distribution, by symmetry we have $$ \mathrm{E}[C_1\mid A_1,A_2] = \mathrm{E}[C_2\mid A_1,A_2] $$ almost surely (we haven't used the independence assumption yet). Hence, $$ \mathrm{E}[B\mid A_1,A_2] = \mathrm{E}[B\mid A_1,A_2] + \mathrm{E}[C_2\mid A_1,A_2] - \mathrm{E}[C_1\mid A_1,A_2] $$ $$ = \mathrm{E}[B+C_2-C_1\mid A_1,A_2] = \mathrm{E}[A_2-A_1\mid A_1,A_2] $$ $$ = \mathrm{E}[A_2\mid A_1,A_2] - \mathrm{E}[A_1\mid A_1,A_2] = A_2 - A_1 = B+C_2-C_1 $$ almost surely. Therefore (Why? Remember the independence assumption and use this. What is the distribution of $-C_1$?), $$ \mathrm{E}[B\mid A_1,A_2]\sim \mathrm{N}(0,3) \, . $$

(If you have any doubts about $\mathrm{E}[B\mid A_1,A_2]$ being a random variable, check this answer.)

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