Find expected value and variance from pmf.

Question

I have a question about pmf:

The n candidates for a job have been ranked 1, 2, ..., n. Let X be the rank of a randomly selected candidate, so that X has pdf has pmf

p(x) = 1/n if x=1,...,n & 0 if otherwise.

Compute E(X) and V(X).

I know the E(X) of pmf is sum of x*p(x) and V(x) is E(X^2) - [E(X)]^2. But it's a bit hard to get start it. Any comment would be very appreciated.

score 1 · Accepted Answer · answered Sep 07 '14 at 21:31

1

You have the formulas. Now all you need to do is plug stuff in and simplify the result:

$E[X] = \displaystyle\sum_{x}x \cdot p(x) = \sum_{x = 1}^{n}x \cdot \dfrac{1}{n} = \cdots$

$E[X^2] = \displaystyle\sum_{x}x^2 \cdot p(x) = \sum_{x = 1}^{n}x^2 \cdot \dfrac{1}{n} = \cdots$

$\text{Var}[X] = E[X^2]-E[X]^2 = \cdots$

The identities $\displaystyle\sum_{x=1}^{n}x = \dfrac{n(n+1)}{2}$ and $\displaystyle\sum_{x=1}^{n}x^2 = \dfrac{n(n+1)(2n+1)}{6}$ will be particularly useful.

answered Sep 07 '14 at 21:31

JimmyK4542

54,331

E(X)=(1+2+...+n)/n and V(X)=[(1^2+2^2+...+n^2)]/2 - [(1+2+3+...+n)/n]^2? – Keris Sep 07 '14 at 21:38
Those are correct equalities, but you should simplify those expressions. – JimmyK4542 Sep 07 '14 at 21:41
From what you given, E(x)=(n+1)/2, E(x^2)=[(n+1)(2n+1)]/6 and V(x)=(n^2-1)/12. I hope this is correct. – Keris Sep 07 '14 at 21:51

Find expected value and variance from pmf.

1 Answers1