Joint Probability distribution

Question

Let X and Y be discrete random variables with joint probability mass function P(X = x,Y = y) = (x + 1)(y + 2) / 54 , for x = 0,1,2

and y = 0,1,2

0, elsewhere

Calculate E(Y|X = 1).

Answer 1

Our goal is to calculate the conditional expectation of Y given X = 1. The definition of expectation of a random variable for a discrete distribution is...

E[X] = ∑ⁱ x_i • f(x_i)

or (the probability of getting a value) x (the value itself). Expanding this to a conditional expectation...

E[Y| X=1] = ∑ⁱ y_i • f(y_i | x = x)

Note that the only thing that changed was that we use the conditional probability of getting a y_i given x = 1. We'll use Bayes' theorem to calculate that probability.

Bayes Theorem:

P(Y = y| X = x) = P(X =x | Y = y)P(Y = y)/ P(X = x)

To put this into words, the probability of seeing y given x, is equal to the probability of seeing y and x divided by the probability of seeing x. Replace "x" with "1" and we have our given problem. Solving the right hand side...

P(X = 1 and Y = y) = (1 + 1)(y + 2)/54 <- note this is for an individual y_i so we don't need to sum it

P(X = 1) = ∑_{{i = 0,1,2}} (1 + 1)(y_i + 2)/54 <- different from above as it's P(x=1) for all y_i which is why we sum

P(X = 1) = (1+1)(0+2)/54 + (1+1)(1+2)/54 + (1+1)(2+2)/54

P(X = 1) = 4/54 + 6/54 + 8/54 = 18/54

P(X = 1 and Y = y | X = 1) = [(1 + 1)(y + 2)/54] / [18/54] = (y + 2)/9

We're all set to find the expected value now

E[Y| X=1] = ∑_{i=0,1,2} y_i • f(y_i | x = x)

E[Y| X=1] = ∑_{i=0,1,2}y_i • (y_i + 2)/9

E[Y| X=1] = 0 • (0+2)/9 + 1 • (1+2)/9 + 2 • (2+2)/9

E[Y| X=1] = 0 + 3/9 + 8/9 = 12/9 or 1.333

Hope this all helps :) feel free to ask any followups