# Thread: Expectation of a geometric random varable

1. ## Expectation of a geometric random varable

Let X be a geometric random variable with parameter p. Find E[min(X, 3)]

My work:

min(X, 3) = X if X < 3 and 3 if X ≥ 3 so

E[min(X, 3)] = 0* P(X = 0) + 1* P(X = 1) + 2*P(X = 2) + 3* P(X ≥ 3)

= 1*p + 2*p*(1-p) + 3* (1-p)^2

as
P(X ≥ 3) = P(X > 2) = (1-p)^2

Is my answer of
= 1*p + 2*p*(1-p) + 3* (1-p)^2 correct?

2. ## Re: Expectation of a geometric random varable

Originally Posted by Capital
Let X be a geometric random variable with parameter p. Find E[min(X, 3)]
min(X, 3) = X if X < 3 and 3 if X ≥ 3 so

E[min(X, 3)] = 0* P(X = 0) + 1* P(X = 1) + 2*P(X = 2) + 3* P(X ≥ 3)

= 1*p + 2*p*(1-p) + 3* (1-p)^2

as
P(X ≥ 3) = P(X > 2) = (1-p)^2

Is my answer of
= 1*p + 2*p*(1-p) + 3* (1-p)^2 correct?
That is correct.