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Here is the setup I am trying to solve:

Alice and Bob are playing a game. Alice turns a spinner and wins an amount with mean $\mu$ and variance $\sigma^2$. Bob then flips a coin, if heads he takes Alice's winnings from her spin, if tails he wins nothing. What are the mean and variance for Bob's winnings?

I think it's trivial that Bob's winnings have mean $\frac{\mu}{2}$, and I would think similarly it has variance $\frac{\sigma^2}{2}$. From a perspective of symmetry we know Alice and Bob both will have the same distribution of winnings, and the sum of their distributions will be the distribution of the spinner. And by additivity of mean and variance then the aforementioned result must hold.

The answer above is not correct, where am I going wrong?

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    $\begingroup$ Well, Bob's reward is $0$ half the time, which induces an extra variability beyond the variability of the spinner value, which should have some contribution to the variance (for instance, even if $\sigma^2 = 0$, Bob's reward would still have a variance if $\mu \neq 0$.). You can just go at this formulaically, though. Let $X$ be the value drawn from the spinner, and let $Y$ be $1$ when the coin is heads, and $0$ when the coin is tails. Bob's reward is $XY$. The variance of Bob's reward, then is $\mathbb{E}[X^2 Y^2] - \mathbb{E}[XY]^2$. Work this out in terms of $\mu,\sigma^2$. $\endgroup$ Commented Nov 14, 2023 at 21:10

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Let $W_A$ denote a random variable modeling Alice's winnings and let $X$ denote a random variable modeling the outcome of the coin flip. Assume that the coin is fair and note that $W_A$ and $X$ are independent.

Bob's winnings is the random variable $W_B = W_A 1_{X=H}$, thus $$E[W_B]=E[W_A]E[1_{X=H}]=\frac \mu 2.$$

Furthermore, $V[W_B] = E[W_B^2] - E[W_B]^2= E[W_A^2 1_{X=H}] - \frac {\mu^2} 4=\frac 12 (V[W_A]+E[W_A]^2)- \frac {\mu^2} 4 = \frac {\sigma^2}2 + \frac {\mu^2}4.$

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