Thread: Geometric random variable

Let X be a random variable with distribution function px(x) defined by:

px(0) = a and px(x) = Px(-x) = ((1-a)/2) * p * (1-p)^(x-1) x = 1,2...

where a and p are two constants between 0 and 1.

a) What is the mean of X?
b) Use the variance of a geometric random variable to compute the variance of X.

Okay so for a) I just used the geometric random variable formula, but made ((1-a)/2) = b. Since the mean of a geometric random variable with probability of success p is E(X) = 1/p, I just multiplied it by b, giving me -(a-1)/2p. Is this correct? If it is correct then I'm pretty sure I know how to do part b), so if anyone could help me with this I'd really appreciate it.

Thanks in advance.

what does px(0) = a and px(x) = Px(-x) mean?

Is px(0) the probability that X=0?
and is this symmetric about 0, with the px(x) = Px(-x)?

Originally Posted by matheagle

what does px(0) = a and px(x) = Px(-x) mean?

Is px(0) the probability that X=0?
and is this symmetric about 0, with the px(x) = Px(-x)?

Yup, px(0) is meant to be that the probability that X=0, and I *think* it is meant to be symmetric about 0 (with px(x) = Px(-x)). Do they play some part in the mean of X? (I didn't really take them into account when I tried to work out the mean because I didn't think they were important...).