# Finding the expectation of functions of random variables with a bivariate normal distribution

X and Y have a bivariate normal distribution. I am given that $E[X] = 4$ and $E[Y] = 10$. I am asked to find $E[X^2 - Y^2]$ WITHOUT integration. I know how to solve for this using integration, but how can I find the solution without doing so?

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You must be given more information than that: you need the variances. Then use the definition of variance and the linearity of expected value. –  Robert Israel Nov 10 '13 at 4:19
You are right, I was also given that Var[X] = 9, Var[Y] = 61, and Cov[X,Y]=15. However, would this approach work? E[(X+Y)(X-Y)] = E[X+Y]*E[X-Y]. –  Luchia Nov 10 '13 at 4:30
@MichaelHardy Not in this case. –  Robert Israel Nov 10 '13 at 8:34
@Luchia : No. There's no reason to think that $X+Y$ and $X-Y$ are uncorrelated. –  Robert Israel Nov 10 '13 at 8:36

$$E[X^2-Y^2] = E[X^2] - E[Y^2] = (\operatorname{Var}(X)+E[X]^2)-(\operatorname{Var}(Y)+E[Y]^2)$$ $$= 9+16-[61+100]$$

The first equality follows from the fact that the expectation of a sum is the sum of an expectation. The second follows from the "computational formula" for variance: $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$.

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Why do you put "$=$", "$+$", "$-$", etc., OUTSIDE if MathJax? That gives you things like $A$+$B$-$C$ instead of $A+B-C$. You get a hyphen instead of a minus sign, and proper spacing and matching of fonts are neglected. –  Michael Hardy Nov 10 '13 at 17:52