0
$\begingroup$

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 $Z$?

  • 0
    It would have been better not to include information that relates to things you didn't need a response on; it worried me at first as well.2012-11-01

2 Answers 2

1

$Z$ 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} $

  • 0
    (+1) Here is just another way to compute $\mathbb{P}(Z=k)=\mathbb{P}(X_j\ge k)^2-\mathbb{P}(X_j\ge k+1)^2=\left(\frac{7-k}{6}\right)^2-\left(\frac{6-k}{6}\right)^2=\frac{13-2k}{36}$2012-11-01
1

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.

  • 0
    You can't use the mean of one random variable (like $X_1$) for the mean of another random variable ($Z$). You look like you got the mean okay (but don't round it off like that when using it in the variance). You should now be able to do the variance using the formula I pointed to.2012-11-01