Problem 1 Consider two random variables X and Y with joint...

Problem 1 Consider two random variables X and Y with joint PMF given in Table 5.3. a. Find P(X ? 2, Y ? 4). b. Find the marginal PMFs of X and Y. c. Find P(Y = 2|X = 1). d. Are X and Y independent? | | Y = 2 | Y = 4 | Y = 5 | | :--- | :---: | :---: | :---: | | X = 1 | 1/12 | 1/24 | 1/24 | | X = 2 | 1/6 | 1/12 | 1/8 | | X = 3 | 1/4 | 1/8 | 1/12 | Problem 2 Let X and Y be as defined in Problem 1. I define a new random variable Z = X ? 2Y. a. Find the PMF of Z. b. Find P(X = 2|Z = 0).

Transcript

00:01 So for part a, we're looking for probability that x is less than or equal to 2, and y is less than or equal to 4.

00:11 So the way that we'll do that is we'll sum over the corresponding events.

00:16 So what i'll do first is indicate those events.

00:20 So we have 1 over, we have x equals 2, y equals 2, x equals 1, y equals 2, x equals 2, and x equals 1 y equals 4 so we would simply sum up those corresponding events so that's 1 over 12 plus 1 over 6 plus 1 over 24 plus 1 over 12 for a result of 0 .375 for part b to find the marginal pmfs of x and y we'll start with finding the different probabilities for x so actually one second here i'll do this as a little table so we have x x has different values of 1, 2, and 3, then we can find the corresponding probabilities simply by, first of all, summing over each row.

01:11 So, to find probability that x equals 1, that's 1 over 12, plus 1 over 24, plus 1 over 24, for a result of 1 over 6, and i'll leave it as 1 over 6 there.

01:27 For probability x equals 2, that's 1 over 6 plus 1 over 12, plus 1 over 8.

01:32 So that's 3 over 8.

01:35 And then for probability x equals 3, we get 1 over 4 plus 1 over 8 plus 1 over 12.

01:42 So that's 11 over 24.

01:46 Then for the pmf for y, we can see y can take on values of 2, 4, or 5.

01:53 Then we have the probabilities would be found by summing over the columns.

01:59 So 1 over 12 plus 1 over 6 plus 1 over 4 from the first column...

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