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Let $A$, $B$, $\epsilon_a$ and $\epsilon_b$ be independent normal random variables:

$A \sim N(\mu_A, \sigma^{2}_A)$

$B \sim N(\mu_B, \sigma^{2}_B)$

$\epsilon_a \sim N(0, \sigma^{2}_{\epsilon_A})$

$\epsilon_b \sim N(0, \sigma^{2}_{\epsilon_B})$

Let Z be defined as $Z \equiv A + B + \epsilon_a + \epsilon_b$.

I am interested in computing the expectation: $E(AB|Z)$.

From questions like this I realize that I (most likely) cannot simply write $E(AB|Z) = E(A|Z)E(B|Z)$, which would be easy to solve using the bivariate normal distribution.

How can I go about computing $E(AB|Z)$ in terms of means and variances?

Bob
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1 Answers1

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One can deduce the desired identity from the following pair of results.

Let $X$, $Y$ and $Z$ denote independent centered normal random variables with respective variances $x^2$, $y^2$, and $z^2$.

Linear case: Let $S=X+Y$. There exists some real numbers $(a,u)$ such that $$X=aS+uU$$ where $U$ is standard normal and independent of $S$. Thus,

$$E(X\mid S)=aS$$ The parameter $a$ solves $$x^2=a(x^2+y^2)$$

Quadratic case: Let $T=X+Y+Z$. There exists some real numbers $(a,b,u,v,w)$ such that $$X=aT+uU\qquad Y=bT+vU+wW$$ where $U$ and $W$ are standard normal independent random variables, independent of $T$. Thus, $$XY=abT^2+((av+b)uU+awW)T+uvU^2+uwUW$$ which implies that

$$E(XY\mid T)=abT^2+uv$$ The parameters $(a,b,uv)$ solve $$x^2=a(x^2+y^2+z^2)\qquad y^2=b(x^2+y^2+z^2)\qquad uv=-ab(x^2+y^2+z^2)$$

Note that $$E(XY\mid T)=ab(T^2-E(T^2))$$

Did
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  • First of all, thank you very much! I was able to find a and b by solving the conditional expectation as usual, but I feel this is wrong as I should use the method to find the expectation. Hence, I couldn't find uv. When trying to find a in the linear case through Var(x), I end up with $x^2 = a^2(x^2 + y^2) + u^2$, which is not a solution. Could you please elaborate on finding the parameters? Further, could you please tell me how the results are influenced by taking variable with non-zero mean instead? – Bob May 03 '17 at 08:30
  • The extension to nonzero means is direct and I omitted it on purpose. Please describe what you tried and where you failed to perform it. // Yes $x^2=a(x^2+y^2)$ and $x^2=a^2(x^2+y^2)+u^2$, what is the problem with these identities? – Did May 03 '17 at 08:42
  • There is no problem. As I said, I computed $E(x|s)=(x^2 + y^2)S$, from which i found a. However, I believe this is not the right approach, as had I been able to compute the conditional expectation already, I would not have needed to use these results and find a in the first place. As I cannot find $E(XY|T)$ in the same way (my OP question), I can't find uv this way. The other approach I tired was to take Var(X)=Var (aS + uU) and similarly for the 2nd case, which also did not lead me to find a,uv. These are the two approaches I tried, which didn't work for me. – Bob May 03 '17 at 09:36
  • Further, for non-zero means, is it still true that $X=aS + uU$, where U is standard normal? I suppose that U would be normal, but not standard normal from my calculations. Thank you again. – Bob May 03 '17 at 09:38
  • To identify the missing parameters, you might want to use identities such as $$E(XS)=E(E(X\mid S)S)=aE(S^2)\qquad E(XY)=E(E(XY\mid T))=abE(T^2)+uv$$ – Did May 03 '17 at 11:14
  • Thank you! With these I was able to find a,b,uv (for zero central rvs) as well as solve E(X|S) for non-zero mean. However, when attempting $E(XY|T)$ for non-zero mean variables, I seem to lack the creativity to come up with enough identities to identify all the relevant parameters. I have less equations than unknowns. Using identities for $E(XT)$, $E(XY)$, $E(YT)$, $E(X)$, $E(Y)$ I can find $a,b, uE(u)$ and $[vE(U) + wE(W)]$. I cannot identify $v$ itself nor $E(U^2)$ (or $Var(U)$ or $u^2$ from which I could deduce $E(U^2)$). Could you please propose some other useful identities for this case? – Bob May 03 '17 at 16:37
  • For the quadratic case: Is it also true that $X=aT + uU$ and $Y=bT + vU$ here? If so, then I can solve the quadratic case with non-zero mean and would be done. Thank you for your support so far! @Did – Bob May 04 '17 at 10:12
  • "For the quadratic case: Is it also true that X=aT+uU and Y=bT+vU here?" Of course not (why do you think I felt the need to introduce W?). But the general quadratic case is direct from the centered quadratic case I explained... What is your trouble here, exactly? – Did May 04 '17 at 11:31
  • I thought not, but did not know what else to do. For non-zero means, I arrive at $E(XY|T)=abT^2 + T[(av+b)uE(U) + awE(W)] + uvE(U^2) + uwE(U)E(W)$. With the identities you mentioned, I can identify $a,b,uE(U)$ and $[vE(U)+wE(W)]$. My trouble is that I cannot identify the other parameters. I suppose I would need to use some other identities, which I cannot come up with. As it stands, I do not have enough equations to identify all the parameters. Thank you again! – Bob May 04 '17 at 11:50
  • I can identify one more parameter from using $E(XY)=E(E(XY|T))=\mu_x \mu_y$. However, I would still have too many unkowns. – Bob May 04 '17 at 11:58
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    Rather use the decompositions $X=X_0+\mu_X$, $Y=Y_0+\mu_Y$, and $T=T_0+\mu_X+\mu_Y+\mu_Z$, hence $$E(XY\mid T)=E(X_0Y_0\mid T_0)+\mu_YE(X_0\mid T_0)+\mu_XE(Y_0\mid T_0)+\mu_X\mu_Y$$ and you know how to compute each term on the RHS since $E(X_0Y_0\mid T_0)$ fits the centered quadratic case while $E(X_0\mid T_0)$ and $E(Y_0\mid T_0)$ both fit the centered linear case. – Did May 04 '17 at 15:58