Joint Probability Functions

[andrew2786]

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

[Rassis]

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

[andrew2786]

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

[Rassis]

Point 4 should say:

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

I attach a new file with the title correction.

Rui

[andrew2786]

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

[Rassis]

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