# Some Help on Probability

1. Nov 27, 2006

### JaysFan31

I'm in a university-level probability course and I'm stuck on a homework problem. Currently we are working on multivarate probability distributions.

I have this problem:
In an earlier exercise, we considered two individuals who each tossed a coin until the first head appeared. Let Y1 and Y2 denote the number of times that persons A and B toss the coin, respectively. If heads occurs with probability p and tails occurs with probability q=1-p, it is reasonable to conclude that Y1 and Y2 are independent and that each has a geometric distribution with parameter p. Consider Y1-Y2, the difference in the number of tosses required by the two individuals.

I need to find a lot of information about the experiment, like E(Y1), E(Y2), E(Y1-Y2), etc. I think I can determine all of the other information if I can just find what E(Y1) and E(Y2) are. This problem is different from all the others we've done since it does not give the joint probability density function. Thus, I have no idea where to start. What is the joint probability density function and what are E(Y1) and E(Y2)?
I would guess that E(Y1) and E(Y2) simply equal (1/p) [from geometric distribution], but these seem horribly wrong.

Any help would be greatly appreciated.

2. Nov 27, 2006

### quasar987

Since Y1 and Y2 are geometric of parameter p, their expectation value is 1/p, there's no question about that; it just follows from algebra. Why do you say it looks wrong?

And haven't you seen that if two random variables are independant, then their joint probability density is just the product of their respective densities.

3. Nov 27, 2006

### JaysFan31

Thanks for the response.

If this is true, then does E(Y1-Y2)=0?
I'm using the notion that E(Y1-Y2)=E(Y1)-E(Y2).

4. Nov 27, 2006

### JaysFan31

Ok. I think I figured everything out.
I just need help with one thing. I have found a lot of information,
E(Y1), E(Y2), E(Y1-Y2), E[(Y1)^2], E[(Y2)^2], and E(Y1Y2).

However I need to find E((Y1-Y2)^2) and V(Y1-Y2).
I would use V(Y1-Y2)= E((Y1-Y2)^2)-(E(Y1-Y2))^2, but I don't know the first two.

5. Nov 27, 2006

### quasar987

Maybe this is not the easiest way but,

E[(Y1-Y2)²] = E[Y1²-2Y1Y2+Y2²]=E[Y1²]-2E[Y1Y2]+E[Y2²]

You can easily find the laws of these 3 new variables. For instance, the density of Y1² is the derivative of the repartition function of Z=Y1²:

$$F_Z(z)=P[Y1^2\leq z]=P[Y_1\leq \sqrt{z}]\mathbb{I}_{[0,\infty)}(z)=\int_0^{\sqrt{z}}f_{Y_1}(y)dy$$

etc.

Last edited: Nov 27, 2006