0
$\begingroup$

Suppose two four-sided fair dice. Let $X = min(X_{1}, X_{2})$ and $Y = max(X_{1}, X_{2})$ where $X_{1}$ is the result of the first dice and $X_{2}$ the result of the second dice. What is the PMF of the joint probability distribution?

How do I solve this problem? So there are $4\cdot 4 = 16$ different outcomes. What I need to do is to find the probability of each $min, max$ combination. How do I do it?

$\endgroup$
  • 1
    $\begingroup$ Well, you know that if $\min(X_1, X_2) = \max(X_1, X_2)$ then $X_1 = X_2 = \min(X_1, X_2)$, and if $\min(X_1, X_2) < \max(X_1, X_2)$ then you know the values of both dice, only not their order. This should help you solve the problem. Just enumerate all combinations for the min and max and add up probabilities. $\endgroup$ – dsaxton Dec 8 '15 at 20:31
  • $\begingroup$ As an aside, I think I've seen this exact problem somewhere before. Is it published in a textbook perhaps? I'd be interested in the reference. $\endgroup$ – Sycorax Dec 8 '15 at 20:36
  • $\begingroup$ Probably, it sounds like a lot of other probability exercises. $\endgroup$ – dsaxton Dec 8 '15 at 21:42
  • $\begingroup$ Welcome to Cross Validated! Please add the self-study tag, read its tag-wiki and modify your question to follow the guidelines on asking such questions. In particular, you'll need to clearly identify what you've done to solve the problem yourself, and indicate the specific help you need at the point you struck difficulty. $\endgroup$ – Glen_b Dec 9 '15 at 1:03
3
$\begingroup$

Here is the brute force way to solve this problem. Simply enumerate all possible outcomes of rolling the two die and then calculate each min and max for the possible outcomes. From there we can just calculate the probability of each event occurring by counting how often we see it out of the total.

So first step, make the list of outcomes

\begin{array}{c |c| c| c} X_1 & X_2 & \min(X_1,X_2) & \max(X_1,X)\\ \hline 1 & 1 & 1 & 1\\ 1 & 2 & 1 & 2\\ 1 & 3 & 1 & 3\\ 1 & 4 & 1 & 4\\ 2 & 1 & 1 & 2\\ 2 & 2 & 2 & 2\\ 2 & 3 & 2 & 3\\ 2 & 4 & 2 & 4\\ 3 & 1 & 1 & 3\\ 3 & 2 & 2 & 3\\ 3 & 3 & 3 & 3\\ 3 & 4 & 3 & 4\\ 4 & 1 & 1 & 4\\ 4 & 2 & 2 & 4\\ 4 & 3 & 3 & 4\\ 4 & 4 & 4 & 4\\ \hline \end{array}

Now, we would like to obtain the joint probability mass function of $(X,Y)$. we can do this by simply counting the number of ways each combination of $x$ and $Y$ can occur and then dividing by the total number of ways which is 16.

\begin{array}{c |c| c| c} X & Y & f_{X,Y}(x,y)\\ \hline 1 & 1 & 1/16 \\ 1 & 2 & 2/16 \\ 1 & 3 & 2/16 \\ 1 & 4 & 2/16 \\ 2 & 1 & 0/16 \\ 2 & 2 & 1/16 \\ 2 & 3 & 2/16 \\ 2 & 4 & 2/16 \\ 3 & 1 & 0/16 \\ 3 & 2 & 0/16 \\ 3 & 3 & 1/16 \\ 3 & 4 & 2/16 \\ 4 & 1 & 0/16 \\ 4 & 2 & 0/16 \\ 4 & 3 & 0/16 \\ 4 & 4 & 1/16 \\ \hline \end{array}

If we add up the third column of probabilities we see that it sums to 1 and so we have not made any mistakes here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.