4. Let X and Y be discrete random variables with joint...

4. Let X and Y be discrete random variables with joint probability mass function: | | Y = 1 | Y = 2 | Y = 3 | | :--- | :--- | :--- | :--- | | X = 0 | 0.01 | 0.10 | 0.05 | | X = 1 | 0.45 | 0.05 | 0.02 | | X = 2 | 0.08 | 0.20 | 0.04 | (i) Find P(X ? 1, Y = 2) (ii) By first finding the marginal pmf of X, compute E(X) and Var(X). (iii) By first finding the marginal pmf of Y, compute E(Y) and Var(Y). (iv) Using the results of (ii) and (iii), compute the covariance of X and Y i.e. Cov(X,Y).

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00:01 The question says let x and y be discrete random variable.

00:06 So given x and y discrete random variable.

00:24 So the first part of question is saying find the probability of probability of x less than equal to 1.

00:35 Y is equal to 2.

00:36 So this will be 0 .10 .05.

00:41 So after adding this would become 0 .15.

00:44 Coming to second part, probability, first of all, finding the marginal pmf of x, then compute e of x and variance of x.

00:56 So first of all, probability of x is equal to 0 will be probability of x is equal to 0, y is equal to 1 plus probability of x is equal to 0, comma y is equal to 2.

01:10 That means that x is equal to 0 what is the possible values of y so plus probability of x is equal to 0 y is equal to 3 this is equal to 0 .01 plus 0 .05 that is 0 .16 now probability of probability of of x is equal to 1 same method i have to use so this will become 0 .4 .9.

01:46 Plus 0 .05 plus 0 .05 plus 0 .02.

01:50 That means at x is equal to 1 what is the probability of y1, y2, y2, y 3.

01:56 So this is equal to 0 .52 probability of x is equal to 2 .08 plus 0 .2 plus 0 .04.

02:09 This is equal to 0 .32.

02:12 So from here, e of x will come as summation of x is equal to 0 .04.

02:19 To 2 probability of x to x...

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