Joint Probability Functions

Hi.
Here's the problem im not quite sure how to do:

The joint probability function f(x,y) of the discrete random variables X and Y are shown below.

X
______0____1_____2______3___
__1|(0.1)_(0.15)__(0.2)___(p)
Y_2|(p)___(0.15)_(0.15)_(0.05)
__3|(0.05)_(0)____(p)____(0)


Find the value of p
[ i do this and work it out as p=0.05]
Determine whether or not X and Y are independent
[ i do this by finding E(x)=1.4, E(Y)=1.6, E(XY)=2.2, then the Cov(X,Y)=-0.04 and as the covariance equals 0 when X and Y are independent, the variables show some degree of dependency]

Here's where my problem starts:
Evaluate P(X+Y=Z) for each possible values z of the random variable X+Y.

I don't know how to do this and would appreciate any help!

Thanks

Andy

Hi Andrew,

I hope this is still on time and helpful.

The answer is Z = X + Y = [1(0.1); 2(0.23); 3(0.34); 4(0.24); 5(0.08); 6(0.01)]

1. List all possible pair combinations of X and Y – it gives 12;
2. Sum each of these 12 combinations (X + Y);
3. List the probabilities of each value of X and Y – P(X) and P(Y);
4. Multiply P(X) and P(Y);
5. List all the unique numbers that resulted from point 2 – it gives 1, 2, 3, 4, 5, and 6;
6. Sum all the probabilities of each value obtained in point 5 and you get the results above P(Z = X + Y).

If you sample at random values from the distribution Z, you will get number 1 with probability 0.1, number 2 with probability 0.23, number 3 with probability 0.34, number 4 with probability 0.24, number 5 with probability 0.08 and number 6 with probability 0.01.

To make it clearer I attach an EXCEL file with the solution.

Rui

Hi Rui,

Thank you very much for that, im very greatful indeed. I've looked over it all and it seems so simple now I can see it like it is (along with the answers!).

Many thanks indeed

Andrew

Point 4 should say:

4. Multiply P(X) and P(Y);

I attach a new file with the title correction.

Rui

One other thing, in your excel file, you have E(XY)=-0.04.

I've worked this out to be the covariance.

E(XY) = (1x0x0.1)+(1x1x0.15)+(1x2x0.2)+(1x3x0.05)+(2x0x0.05)+(2x1x0.15)+(2x2x0.15)+(2x3x0.05)+(3x0.05x0)+(3x 1x0)+(3x2x0.05)+(3x3x0)= 2.2

Cov(X,Y) = E(XY) - E(X) E(Y)

=2.2-(1.4x1.6)
=-0.04

You are right. I haven´t used the most adequated symbology. Please see the new version (3) of the Excel file that I replaced in my two previous posts.

Rui