Register to reply

Probability random variables

by Gott_ist_tot
Tags: probability, random, variables
Share this thread:
Gott_ist_tot
#1
Oct24-07, 01:02 AM
P: 52
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] = [tex] \sum [/tex] 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.
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
jhicks
#2
Oct24-07, 02:07 AM
P: 336
Your notation is confusing to me, but for starters 1/2 - (1-2) isn't -1/2.
Gott_ist_tot
#3
Oct24-07, 02:29 AM
P: 52
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.

HallsofIvy
#4
Oct24-07, 06:55 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568
Probability random variables

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

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


Register to reply

Related Discussions
A Probability Problem Involving 6 Random Variables Calculus & Beyond Homework 1
Probability of continuous random variables Calculus & Beyond Homework 5
Probability inequality for the sum of independent normal random variables Set Theory, Logic, Probability, Statistics 3
Probability distribution of function of continuous random variables Set Theory, Logic, Probability, Statistics 1
Probability question: random variables Introductory Physics Homework 1