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Suppose $X$ and $Y$ are independent random variables. The mgf for $X$ is $M_{X}(t) = (1-t)^{-1}, \ t<1$. The mgf for $Y$ is $M_{Y}(t) = (1-2t)^{-1}, \ t< 0.5$. Two other random variables $U$ and $V$ are defined by $U = \frac{1}{2}(X+Y)$ and $V = \frac{1}{2}(X-Y)$. Calculate $\text{Cov}(U,V)$.

So $\text{Cov}(U,V) = E(UV)-E(U)E(V)$. This is the same thing as $\text{Cov}(\frac{1}{2}(X+Y), \frac{1}{2}(X-Y))$. Expanding out, we get

$\text{Cov}(\frac{1}{2}(X+Y), \frac{1}{2}(X-Y)) = \frac{1}{4} \text{Var}(X) -\frac{1}{4} \text{Cov}(X,Y)+\frac{1}{4} \text{Cov}(Y,X)-\frac{1}{4} \text{Var}(Y)$ which equals $\frac{1}{4}(\text{Var}(X)-\text{Var}(Y))$.

Is there an easier way to obtain this without going through all this work? Could one use the fact $U+V = X$ and $U-V = Y$?

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What you did looks fine to me, but another way would be to expand the first equation to $\text{Cov}(U,V) = E(UV) - E(U)E(V) = ¼E(X²-Y²) - ¼\left(E(X)² - E(Y)²\right)$, which leads to the same thing. –  Rahul Nov 4 '10 at 19:54
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