Two fair and independent dice (each with six faces) are thrown. Let $X_1$ be the score on the first die and $X_2$ the score on the second. Let $X = X_1 + X_2$ , $Y = X_1 X_2$ and $Z = \min(X_1; X_2)$.
How would you calculate the variance of $Z$?
Two fair and independent dice (each with six faces) are thrown. Let $X_1$ be the score on the first die and $X_2$ the score on the second. Let $X = X_1 + X_2$ , $Y = X_1 X_2$ and $Z = \min(X_1; X_2)$.
How would you calculate the variance of $Z$?
$Z$ can assume one of six possible values: $ \mathbb{P}(Z=k) = \mathbb{P}(X_1 = k) \mathbb{P}(X_2 \geqslant k) + \mathbb{P}(X_2=k) \mathbb{P}(X_1 >k) = \frac{1}{6} \frac{6-k+1}{6} + \frac{1}{6} \frac{6-k}{6} = \frac{13-2k}{36} $ The variance is computed as $\mathbb{Var}(Z) = \mathbb{E}(Z^2) - \mathbb{E}(Z)^2$. $ \mathbb{E}(Z) = \sum_{k=1}^6 k \cdot \frac{13-2k}{36} = \frac{13}{36} \frac{6\cdot(6+1)}{2} - \frac{1}{18} \frac{6 \cdot (6+1) \cdot (2 \cdot 6 +1) }{6} = \frac{91}{36} $ Similarly $ \mathbb{E}(Z^2) =\sum_{k=1}^6 k^2 \cdot \frac{13-2k}{36} = \frac{301}{36} $
Enumeration combined with standard formulas
Edit to make it clear, since apparently it wasn't -
Enumeration: You go through the sample space for $(X_1, X_2)$ and find the minimum for each point (the corresponding $Z$), adding up the probabilities that go with the points in the sample space contributing to each value of $Z$, yielding the probabilities of each value for $Z$. Having obtained the p.f. for $Z$, you evaluate its variance using standard formulas.