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If you let $X_1$ and $X_2$ be the numbers obtainted on two rolls of a fair die. Let $Y_1 = max(X_1 , X_2), Y_2 = min(X_1 , X_2)$.

First of all how do I interpret $Y_1$ and $Y_2$, and then second, how do I form a joint probability table from this?

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2 Answers 2

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Firstly, all RVs are discrete, so we can look at their pmfs and their cdfs.

So, if you look at the pmf of $Y_1$, you have to look at $p(Y_1=a)$ for all $a$ in the support of $Y$ (which is all numbers 1 through 6).

$$p(Y_1 = 1) = p(\max(X_1,X_2) = 1)$$ There is only one pair of $X_1,X_2$ when this is true, (1,1). Thus, the answer for the problem is $\dfrac{1}{36}$. An interesting case would be for $Y_1 = 6$, which would mean that the maximum of both throws is 6. What do you think is the probability for that to happen?

Similarly, you can find out for the other cases and also extend the logic for $Y_2$.

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so if $Y_1$ = 6 then that includes all the throws of the die that have a 6 in them? And if $Y_1$ = 5 then it would be all the throws of the die that have a 5 maximum i.e. (5,3), (4,5), (5,5), (1,5), etc.? –  TheHopefulActuary Oct 15 '12 at 18:51
    
Yes. $Y_1 = 6$ would be the set of all pairs where the maximum of the 2 throws is 6. That would be (1,6) (6,1) (2,6) (6,2) (3,6) (6,3) (4,6) (6,4) (5,6) (6,5) (6,6). Similarly for 5. –  Inquest Oct 15 '12 at 18:56

Here $X_1$ is intended to the the number obtained on the first toss, and $X_2$ the result on the second. Let random variable $Y=\min(X_1,X_2)$ denote the minimum of $X_1$ and $X_2$, and $Z=\max(X_1,X_2)$ the maximum.

So for example if we had a $5$ on the first toss, and $2$ on the second, then $Y=2$ and $Z=5$. If we had a $3$ on each, then $Y=Z=3$.

To express it in game terms, the two dice are tossed, and Alicia gets, in dollars, the smaller of the two numbers showing. Then random variable $Y$ represents Alicia's gain.

We want to draw up a $6\times 6$ table of probabilities that tells us the probability that $Y=a$ and $Z=b$ for all pairs $a,b$. Kind of like a mileage chart.

From the definition, we have $Y\le Z$. So for any pair $(a,b)$ such that $a\gt b$, we have probability that $Y=a$ and $Z=b$ is $0$. So in your table, write $0$ for all such situations. The table is $6\times 6$, and we have already filled in $15$ places. Only $21$ to go.

For all pairs $(a,b)$ with $a\lt b$, we can get $Y=a$ and $Z=b$ in two ways: our first toss was $a$ and the second was $b$, or the other way around.

So if $a\lt b$, then with probability $\dfrac{2}{36}$, we have $Y=a$ and $Z=b$. This fills in $15$ more entries. You may prefer to enter them as $\dfrac{1}{18}$.

Finally, we want to find the entries on the main diagonal. What is, for example, the probability that $Y=4$ and $Z=4$? The $\max$ and $\min$ are both $4$ precisely if both tosses result in a $4$. The probability of this is $\dfrac{1}{36}$. All the diagonal entries, by the same reasoning, have value $\dfrac{1}{36}$.

As a check, let's see whether the table entries add up to $1$. There were $15$ entries of $\dfrac{2}{36}$, and $6$ of $\dfrac{1}{36}$, so the total is $15\cdot\dfrac{2}{36}+6\cdot\dfrac{1}{36}$. This sum is indeed $1$.

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