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The random variables X and Y have joint probability function $p(x,y)$ for $x = 0,1$ and $y = 0,1,2$. Suppose $3p(1,1) = p(1,2)$, and $p(1,1)$ maximizes the variance of $XY$. Calculate the probability that $X$ or $Y$ is $0$.

Solution: Let $Z = XY$. Let $a, b$, and $c$ be the probabilities that $Z$ takes on the values $0, 1$, and $2$, respectively. We have $b = p(1,1)$ and $c = p(1,2)$ and thus $3b = c$. And because the probabilities sum to $1, a = 1 – b – c = 1 – 4b$.

Then, $$E(Z) = b + 2c = 7b,$$$$ E(Z\cdot Z) = b + 4c = 13b.$$ Then, $$Var (Z) = 13b-19b^2.$$$$ \frac{dVar(Z)}{db}=13-98b=0 \implies b =\frac{ 13}{98}.$$ The probability that either $X$ or $Y$ is zero is the same as the probability that $Z$ is $0$ which is as $a = 1 – 4b = \frac{46}{98} = \frac{23}{49}$.

I am not sure how they got: $E(Z) = b + 2c = 7b$. Can someone explain this step where $b$ and $2c$ come from?

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    $\begingroup$ I don't think it's a good idea to delete the questions. Even if they no longer interest you, they should still exist for the purpose of the community. I've flagged a number of these "deleted" questions for attention. Thanks. $\endgroup$ Mar 15, 2018 at 0:17

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\begin{align}\mathbb{E}(Z)&=0\cdot P(Z=0)+1\cdot P(Z=1)+2\cdot P(Z=2)\\ &=1\cdot P(Z=1)+2\cdot P(Z=2)\\ &=1\cdot P(X=1,Y=1)+2\cdot P(X=1,Y=2)\\ &=b+2c \end{align}

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