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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 $Z$?

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hint: $X_1$ and $X_2$ are discrete uniform random variables –  jay-sun Oct 31 '12 at 23:29
    
so would i just take the expected value of the two and sum them? –  tamefoxes Oct 31 '12 at 23:33
    
Nope. Just calculate the distribution for $Z$ and then it's variance. –  jay-sun Oct 31 '12 at 23:36
    
I'm not too sure how to get that... I'm very confused when it involves the minimum and maximum –  tamefoxes Oct 31 '12 at 23:44
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Is this homework? If so, please add the homework tag. Note that HW questions receive special treatment on CV (you may want to read our FAQ). Lastly, what role are $X$ & $Y$ playing in this question? Are they there just to throw students off? –  gung Nov 1 '12 at 1:06

2 Answers 2

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.

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But how would I take into the account that the event Z=min(X1,X2)? –  Jesse Nov 1 '12 at 0:50
    
Exactly as it says in my answer - enumeration. You go through the sample space and find the minimum for each point, adding up the corresponding probabilities contributing to each value of $Z$, yielding the probabilities of each value for $Z$. –  Glen_b Nov 1 '12 at 0:53
    
@Jesse To be specific, this is the sample space for two dice (ignore the shading in that image). Find $Z$ in each cell and the corresponding probability. Add probabilities where the $Z$'s are the same. Is this homework? Actually, I'll tag it homework, but feel free to explain how it doesn't count as homework. –  Glen_b Nov 1 '12 at 0:59
    
I got the variance to being approximately 2.92. and yes it is homework –  Jesse Nov 1 '12 at 1:06
    
You may want to check your calculations. What did you get for the mean? –  Glen_b Nov 1 '12 at 1:23

$Z$ 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 $$ \mathbb{E}(Z^2) =\sum_{k=1}^6 k^2 \cdot \frac{13-2k}{36} = \frac{301}{36} $$

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(+1) Here is just another way to compute $\mathbb{P}(Z=k)=\mathbb{P}(X_j\ge k)^2-\mathbb{P}(X_j\ge k+1)^2=\left(\frac{7-k}{6}\right)^2-\left(\frac{6-k}{6}\right)^2=\frac{13-2k}{36‌​}$ –  robjohn Nov 1 '12 at 7:04

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