two fair dice are rolled let the random variable x be dened as the minimum of the two numbers sh

Step 1: The support of X is the set of all possible values that X can take. In this case, X repr

Two fair dice are rolled let the random variable x be dened...

Two fair dice are rolled. Let the random variable X be defined as the minimum of the two numbers showing on the dice. (a) Find the support of X. Is X discrete or continuous? Justify. (b) Find the probability mass function (pmf) of X. Display the final answer neatly in table form. (c) Calculate the cumulative distribution function (CDF) of X. Display the final answer as a step function. Then, graph the CDF. (d) Find the following: p(3), F(3), p(5.6), F(5.6), p(20), F(20), p(-20), F(-20).

Transcript

00:01 So for this problem to begin, we can think of x1 and x2 as being the results of the individual dice rolls.

00:09 So they would be uniform, discrete random variables from one up to six.

00:14 And we have that x is equal to the minimum of x1 and x2, whichever is smaller.

00:24 Now, for part a, now we would have that the support of x would just be the set 1, 2, 3, 4, 5, and 6.

00:42 We note that this is a discrete distribution because we have that the minimum of two discrete values will be discrete itself.

01:12 Effectively, we can think of it as the minimum of two values would have to be a value that is valid.

01:18 If we have x1 and x2, the minimum of the two would have to be something in the range of x1 and x2.

01:27 Therefore, if x1 and x2 are discrete, then the minimum must be discrete as well.

01:33 Now, for part b, finding the pmf, the first thing that we'll want to do here is create a little bit of a table of the sample space.

01:43 So we know that we can quickly enumerate, or not necessarily quickly, but we can thoroughly enumerate the sample space.

01:50 By setting the first result to a particular value, 1, for instance, and then iterating the second result from 1 up to 6.

01:58 And then we iterate the first value up by 1, and again, iterate the second value up to 6.

02:05 Now, just to save myself some wrist pain and save you as the viewer a little bit of time, i'm just going to use a little bit of a trick here to quickly generate the table of the different possible pairs.

02:18 So, i'm just going to adjust the view.

02:24 All right, so now we can start about our probability mass function for our x, where x is, of course, the minimum of the two values.

02:33 So we can see that for everything across this first row, one will be the minimum.

02:38 So that occurs for six out of 36 different possibilities, so the probability would be one in six.

02:44 Then for two to be the minimum, it's only these five outcomes.

02:55 Or, pardon me, not those five outcomes.

02:57 Pardon me.

03:00 And actually, i also spoke too soon on the first portion, excuse me.

03:03 So let me fix this up here.

03:05 So the mistake that i made, so we have one is the minimum of everything across the first row.

03:13 One is also the minimum of everything across the first column.

03:17 So we would have a total of six one way, six the other way, but one would be overcounted.

03:23 So we can say that it's for 11 out of 36 different possible outcomes.

03:31 Then if two is the minimum value, then we can think of it as everything along this row and this column.

03:43 So we can see we have one, two, three, four.

03:46 We have four in the sort of legs, and we have two legs, plus one, the sort of diagonal element.

03:55 So 4 plus 4, 8 plus 1 gives us 9 over 36...

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