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Let $X_1, X_2, X_3$ be i.i.d. random variables such that $E[X_1] = 0$ and $E[X_1^2] = \displaystyle \frac{15}{4} $. Define $ \psi:(0, \infty) \rightarrow (0, \infty) $ by the conditional expectation $ \psi(t) = E[X_1^2 \ | \ X_1^2 + X_2^2 + X_3^2 = t] $. Find the value of $E[ \psi((X_1 + X_2)^2)] $.

So this is what I've tried (assuming continuous random variables):

Let $Y_i = X_i^2$ so that $$ \psi(z) = \int \int y_1 f_{Y_1}(y_1) f_{Y_2}(y_2) f_{Y_3}(z - y_1 - y_2) \,dy_2 \,dy_1 $$ (here, the $f_Y$'s are marginal distributions)

Now if $Z = (X_1 + X_2)^2$, then by the previous expression $$ E[ \psi(Z) ] = \int \int \int y_1 f_{Y_1}(y_1) f_{Y_2}(y_2) f_{Y_3}(z - y_1 - y_2) f_Z(z) \,dy_2 \,dy_1 \,dz $$ (Is this correct?) I then tried interchanging the order of integration and expressing $f_Z$ in terms of the $f_Y$'s, but that doesn't seem too helpful. The law of iterated expectation also doesn't seem to be applicable here.

This was one of the questions on a competitive exam and the value of the expectation was given as $ 2.5 $, but I have no idea how they got that. Any help on how to proceed would be appreciated, thanks!

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  • $\begingroup$ It is not that complicated - the key is you need to know that actually you can compute $\phi(t)$. Consider $E[X_i^2 | X_1^2 + X_2^2 + X_3^2 = t]$ for $i = 1, 2, 3$. $\endgroup$ – BGM Jan 24 '18 at 8:25
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Since $X_i$ are iid, we have $E[X_i^2|X_1^2 + X_2^2 + X_3^2] $ is the same for all $i = 1,2,3.$ By summing this, we get $\psi (t) = t/3,$ then $$ E[\psi((X_1+X_2)^2)] = E[(X_1+X_2)^2/3] = \frac{1}{3}E[X_1^2+X_2^2+ 2X_1X_2] = \frac{1}{3}(\frac{15}{4}+\frac{15}{4} + 2E[X_1]E[X_2]) = \frac{1}{3}(\frac{15}{4}+\frac{15}{4}) = 2.5. $$

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