# Cumulative distribution function to obtain moments?

• Feb 18th 2013, 10:00 PM
Reefer
Cumulative distribution function to obtain moments?
I am given a continuous random variable F which is the sum of 3 exponential distributed random variables X, Y, and Z. The parameter of X is λ and the parameter for Y and Z is μ.

X, Y and X, Z are independent, and the correlation between Y and Z is B. I am told to obtain the first 2 moments of F. How would I go about solving this problem?

I know to find the moments for a random variable is E[x^n]=n!/
λ^n.

• Feb 19th 2013, 01:59 AM
chiro
Re: Cumulative distribution function to obtain moments?
Hey Reefer.

Hint: For this problem you will need to consider the co-variance term and how it relates to correlation.

Covariance and correlation - Wikipedia, the free encyclopedia

Once you link correlation to covariance use the fact that Var[X+Y] = Var[X] + Var[Y] + 2*Cov(X,Y)
• Feb 19th 2013, 01:25 PM
Reefer
Re: Cumulative distribution function to obtain moments?
Isn't correlation just the covariance divided by the standard deviation of both variables? How does this help me find the first 2 moments?
• Feb 19th 2013, 03:04 PM
chiro
Re: Cumulative distribution function to obtain moments?
The first moment is the mean and the second moment is E[X^2] which is Var[X] - E[X]^2, and for the correlated variable, this involves co-variance when you consider E[X+Y+Z] and E[(X+Y+Z)^2] (which are the first two standard moments).

Also you may want to point out if they are central moments with respect to the mean or not.
• Feb 19th 2013, 11:01 PM
Reefer
Re: Cumulative distribution function to obtain moments?
I'm not sure if I understand it fully.. but since Y and Z are of the same parameter, can't we say that Y+Z= 2Y?

Then from there we can find the moments to be E[X+2Y] and E[(X+2Y)^2]??
• Feb 20th 2013, 03:10 PM
chiro
Re: Cumulative distribution function to obtain moments?
Y equals Z only when they have a perfect correlation (i.e. correlation = 1) which is a very special case and not a general one.