# Joint PMF of p x,y

1. Aug 15, 2011

### orangesun

1. The problem statement, all variables and given/known data

Hi, if you could offer help to any of these questions, it would be great, i was unable to attend my lectures this week and I had no idea how to do these.

The joint pmf of x and y is p x,y = c(x^2 + y^2) x,y = 1,2,3 Find:

a. c and the marginal pmfs of x and y
b. e(x)
c. the pmf of 3x-2y
d. e(3x-2y) in two ways

2. Relevant equations

3. The attempt at a solution
From first investigation, I think that I have to create a table with
1,1 1,2 1,3
2,1 2,2, 2,3
3,1 3,2 3,2
and x^2 + y^2 = 1 and the p = 1/9 since theres only 9 options.

i have no idea how to continue.

2. Aug 15, 2011

### HallsofIvy

Staff Emeritus
" p = 1/9 since theres only 9 options." I have no idea what you mean by this. "p" is a function, not a number. Did you mean "c= 1/9"?

No, $x^2+ y^2$ is NOT 1. $x^2+ y^2= 1+ 1= 2$ when x= y= 1, $x^2+ y^2= 4+ 1= 5$ when x= 2, y= 1, etc.

Start by actually filling in your table: What is p(1, 1), p(1, 2), etc.? Those will have a c in them. Then find c by using the fact that the sum of all the probabilities must be 1.

3. Aug 15, 2011

### orangesun

I'll give it a shot now.

x,y 1 2 3
1 2c 5c 10c
2 5c 8c 13c
3 10c 13c 18c
is this right so far?
if so, then is it just 83c = 1? c=1/84

i hope it is...