# Probability random variables

1. Oct 24, 2007

### Gott_ist_tot

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

A random variable has a distribution function F(z) given by
F(z) = 0 if z< -1
F(z) = 1/2 if -1 <= z < 2
F(z) = (1-z^{-3}) is 2 <= z

Find the associated mean and variance.

3. The attempt at a solution
I drew the distribution function. I started with the associated mean (if I can figure that out the variance should follow.) I have:

E[Z] = $$\sum$$ zp(z)

p(z) = P[X = z]

Therefore,
p[X = -1] = P[X= -1] - P[X<-1]
= F(-1) - lim F(1-1/n)
= 1/2 - (1-2)
= 3/2

Sorry, if I messed up badly somewhere. The class is taught without a book and I can't seem to get anything out of my notes for this homework. Thanks.

Last edited: Oct 24, 2007
2. Oct 24, 2007

### jhicks

Your notation is confusing to me, but for starters 1/2 - (1-2) isn't -1/2.

3. Oct 24, 2007

### Gott_ist_tot

Sorry, for the bad arithmetic. That is fixed. The problem statement is a piecewise function. As for the rest of the notation. I don't know how else to write it. My professor said this usually isn't presented in textbooks in this manner but he thinks it is a good way to do it. So the notation may be quite strange. If there is a specific part I might be able to explain what I am doing.

4. Oct 24, 2007

### HallsofIvy

Staff Emeritus
What are you adding over? The way you have written F it looks like a piecwise function and you should be integrating, not adding:
$$E[Z]= \int_{-1}^\infty zF(z)dz= \int_{-1}^2 \frac{z}{2}dz+ \int_2^\infty (z- z^{-2}dz$$
Unfortunately, it looks to me like that last integral won't converge. Are you sure it wasn't $F(z)= (1- z)^{-3}$ if z> 2?

If F is only defined for integer z (then you should have told us that),
$$E(Z)= \sum_{-1}^\infty nF(n)= \sum_{-1}^2 \frac{n}{2}+ \sum_2^\infty (n+ n^{-2})$$
Again, there is a problem with the final sum: it doesn't converge.