
#1
Oct2407, 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) = (1z^{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(11/n) = 1/2  (12) = 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. 



#2
Oct2407, 02:07 AM

P: 337

Your notation is confusing to me, but for starters 1/2  (12) isn't 1/2.




#3
Oct2407, 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.




#4
Oct2407, 06:55 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877

Probability random variables[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. 


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