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A fair die is tossed twice. Let $d_1=\text{value of die on roll 1}$ and $d_2=\text{value of die on roll 2}$ Let $X=d_1+d_2$, the sum of the faces; $Y=\max\left\{d_1,d_2\right\}$, the maximum of the two faces; and $Z = |d_1-d_2|$.

What is the value of $X, Y, Z,$ and $ W = XZ$ for each outcome $\omega\in\Omega$? What is $\mathrm{Range}(X)$, $\mathrm{Range}(Y)$, $\mathrm{Range}(Z)$ and $\mathrm{Range}(W)$? Describe the partitions $A_X$ and $A_Z$ induced by these random variables.

From my understanding $\left(d_1,d_2\right)=\omega\in\Omega=\left\{1,2,3,4,5,6\right\}^2=\left\{\left(1,1\right),\left(1,2\right),\dotsc,\left(6,5\right),\left(6,6\right)\right\}$

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  • $\begingroup$ This is about dice not coins right? $\endgroup$
    – user142299
    Apr 14, 2014 at 14:56
  • $\begingroup$ Yes it is about dice. $\endgroup$
    – petrov
    Apr 14, 2014 at 15:17

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With the sample space $\Omega=\left\{1,2,3,4,5,6\right\}^2=\left\{\left(1,1\right),\left(1,2\right),\dotsc,\left(6,5\right),\left(6,6\right)\right\}$, we can easily find what is being asked.

The range of a Random Variable is the set of values it can assume, so:

$\mathrm{Range}\left(X\right)=\left\{2,3,\dotsc,11,12\right\}$ as these are the possible sums of the sides of a fair die rolled twice.

$\mathrm{Range}\left(Y\right)=\left\{1,\dotsc,6\right\}$ as whatever is rolled, any of these values could be the maximum.

$\mathrm{Range}\left(Z\right)=\left\{0,1,\dotsc,5\right\}$ as these are the possible differences between the values of the two rolls, note that we have $0$ when $d_1=d_2$ and 6 isn't possible as the largest difference will occur when we get $\left(6,1\right)$ or $\left(1,6\right)$.

$\mathrm{Range}\left(W\right)$ is a little more difficult as we need to calculate $W\left(\omega\right)$, $\forall\omega\in\Omega$. When done, we find that $\mathrm{Range}\left(W\right)=\left\{0,3,5,7,8,9,11,12,15,16,20,21,24,27,32,35\right\}$

A partition of a random variable is the set of points $\omega\in\Omega$ that give rise to each possible value that the random variable can take.

The partitions $A_X$ and $A_Z$ can now be found \begin{align*} A_{X_1=2} &= \left\{\left(1,1\right)\right\} \\ A_{X_2=3} &= \left\{\left(1,2\right),\left(2,1\right)\right\} \\ A_{X_2=4} &= \left\{\left(1,3\right),\left(2,2\right),\left(3,1\right)\right\} \\ &\vdots \\ A_{X_{10}=11} &= \left\{\left(5,6\right),\left(6,5\right)\right\} \\ A_{X_{11}=12} &= \left\{\left(6,6\right)\right\} \\ \end{align*} and \begin{align*} A_{Z_1=0} &= \left\{\left(1,1\right),\left(2,2\right),\dotsc,\left(5,5\right),\left(6,6\right)\right\} \\ A_{Z_2=1} &= \left\{\left(1,2\right),\left(2,3\right),\dotsc,\left(3,2\right),\left(2,1\right)\right\} \\ &\vdots \\ A_{Z_6=5} &= \left\{\left(1,6\right),\left(6,1\right)\right\} \\ \end{align*}


In response to the comment about why the value $5\times12\not\in\mathrm{Range}\left(W\right)$, we see that $W\left(\omega\right)=X\left(\omega\right)\times Z\left(\omega\right)$, so therefore it isn't possible to obtain $60=12\times5$ because the only value $\omega\in\Omega$ such that $X\left(\omega\right)=12$ is $\omega=\left(6,6\right)$ but clearly $Z\left(\omega\right)=0$ if $\omega=\left(6,6\right)$. Similarly for $Z\left(\omega\right)=5$, we need $\omega=\left(1,6\right)$ or $\omega=\left(6,1\right)$ and then clearly $X\left(\omega\right)=7$ for both of these $\omega$.

In order to compute $\mathrm{Range}\left(W\right)$, it is necessary to check the value of $W\left(\omega\right)=X\left(\omega\right)\times Z\left(\omega\right)$ for all $\omega\in\Omega$ where $\Omega$ is defined as above.

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  • $\begingroup$ How did you get the range of W? Why is 60 not included (12*5) and other values if you multiply Range(X) and Range(Z)? $\endgroup$
    – o.o
    Nov 16, 2015 at 23:52
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    $\begingroup$ @CeeC I will address your query in an addition to my answer. $\endgroup$
    – wilsnunn
    Nov 18, 2015 at 9:45
  • $\begingroup$ Thank you very much! I have one more question, would the partition of W be this: i.imgur.com/7oRQTEE.jpg? $\endgroup$
    – o.o
    Nov 20, 2015 at 22:58
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    $\begingroup$ @CeeC That looks good to me. $\endgroup$
    – wilsnunn
    Nov 21, 2015 at 0:57

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