# summation and joint probability function

#### eizzn

##### New member
I can't remember how to do this kind of summation:

(summation over X)(summation over Y) (constant c)(x^2 + y^2)

this is a prabability problem, so the summation should add up to 1 and the goal is to find the value of the constant c.
Am I right to just seperate the summations and do the X's and Y's seperatly?

The reason I am confused is because the following question asks for the probability of Y < X, given that X can only be {1,2,4} and Y can only be {1, 3}.
my probability comes out to be greater then 1.

#### pka

##### Elite Member
Please post the exact wording of the problem.
From what you wrote, for that sum to be 1 then c=1/72.
But we do not know the probability function.

#### eizzn

##### New member
P(X,Y) = c(x^2 + y^2) where x = {1,2,4} and y = {1,3}
and P(X,Y) = 0 otherwise

a) find c

b) find P(Y < X) and P( Y > X)

#### pka

##### Elite Member
Here is the way it works. I used a computer algebra system.

The factor Y<SUB>j</SUB><X<SUB>k</SUB> is 1 or 0, for true or false.

#### eizzn

##### New member
thanks a lot. I got the answer to c by drawing a matrix and putting in all the individual probabilities and I also came up with 72. from there the rest was easy.

#### eizzn

##### New member
I am still having difficulty with the summation.
Now I need to find the marginal probability mass functions P(x) and P(y)

I know P(x) = (summation over y) P(x,y)
P(x,y) = 1/72(x^2 + y^2)

I just need to know how to take summations of multiple variables. In this case do I treat x as a constant?

#### pka

##### Elite Member
Keeping with my notation above:
$$\displaystyle \L \begin{array}{l} P(X) = \sum\limits_{k = 1}^2 {c\left( {X^2 + \left[ {Y_k } \right]^2 } \right)} \\ P(Y) = \sum\limits_{k = 1}^3 {c\left( {Y^2 + \left[ {X_k } \right]^2 } \right)} \\ \end{array}$$