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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 variables2012-10-31
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    so would i just take the expected value of the two and sum them?2012-10-31
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    Nope. Just calculate the distribution for $Z$ and then it's variance.2012-10-31
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    I'm not too sure how to get that... I'm very confused when it involves the minimum and maximum2012-10-31
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    How would I calculate the distribution for Z?2012-10-31
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    look [here](http://stats.stackexchange.com/questions/220/how-is-the-minimum-of-a-set-of-random-variables-distributed)2012-10-31
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    What I get from this is that I would find the cdf, differentiate to get the pdf and find variance from there?2012-11-01
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    Also in that link, it says that to calculate the cdf, we would evaluate the function at the certain value that we want to be smaller than.. In this case, would I use the case where x1x22012-11-01
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    I was asked to find the variance for the X and Y but I already calculated those2012-11-01
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    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

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$Z$ 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}$2012-11-01
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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)?2012-11-01
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    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 $Z$.2012-11-01
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    @Jesse To be specific, [this](http://www.thewinternet.com/mcknight/BIMR6/MR_Prob_6%28Tb2%29.gif) 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.2012-11-01
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    I got the variance to being approximately 2.92. and yes it is homework2012-11-01
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    You may want to check your calculations. What did you get for the mean?2012-11-01
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    for the mean i got 3.52012-11-01
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    That's not the mean of $Z$. You can't get the variance right if you don't get the mean right. What was the probability you got for $Z=6$? (For that matter, how did you work out the variance of Y?)2012-11-01
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    z = 1/36. I used the mean for the roll of a die, so I recalculated and got 2.53 for the mean2012-11-01
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    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