Question

Q: Help in understanding and solving this example from Probability and Statistical (Conditional Distributions) with the steps of the solution to better understand, thanks.

**Please give the step by steps with details to completely see how the solution came about.

1) Let X and Y have the joint p.m.f:   f(x,y) = (x+2y)/33,   x = 1,2   y = 1,2,3.

a) Display the joint p.m.f and the marginal p.m.f.s on a graph.

b) Find g(x $\mid$ y) and draw a figure depicting the conditional p.m.f.s for y = 1,2,3.

c) Find h(x $\mid$ y) and draw a figure depicting the conditional p.m.f.s for x = 1 and 2.

d) Find P(1 $\leq$ Y $\leq$ 3 $\mid$ X = 1), P(Y $\leq$ 2 $\mid$ X = 2), and P(X = 2 $\mid$ Y = 3).

e) Find E(Y $\mid$ X = 1) and Var(Y $\mid$ X =1).

Answer 1

Plug the values of x and y in the given p.m.f. , then we get the following values of f(x, y)

for x = 1, y = 1

f( x, y) = (1 + 2*1 )/33 = 3/33

for x = 1, y = 2

f( x, y) = (1 + 2*2 )/33 = 5/33

for x = 1, y = 3

f( x, y) = (1 + 2*3 )/33 = 7/33

for x = 2, y = 1

f( x, y) = (2 + 2*1 )/33 = 4/33

for x = 2, y = 1

f( x, y) = (2 + 2*6 )/33 = 6/33

for x = 2, y = 3

f( x, y) = (2 + 2*3 )/33 = 8/33

Let's make table

Y|X	1	2	p(y)
1	3/33	4/33	7/33
2	5/33	6/33	11/33
3	7/33	8/33	15/33
p(x)	15/33	18/33	1

The column of p(y) is nothing but the marginal pmf of Y and the row of p(x) is nothing but the marginal pmf of X.

b) Let's find conditional pmf of x given y

for given y = 1

X|Y=1	1	2	Total
P(x|y=1)	3/7	4/7	1

Note that P(X =1|Y =1) = P(X = 1, Y = 1)/P(Y = 1) = (3/33)/(7/33) = 3/7

and P(X =2|Y =1) = P(X = 2, Y = 1)/P(Y = 1) = (4/33)/(7/33) = 4/7

for given y = 2

X|Y=2	1	2	Total
P(x|y=2)	5/11	6/11	1

for given y = 3

X|Y=3	1	2	Total
P(x|y=3)	7/15	8/15	1