# Why am I getting the wrong result for the expected value of an F-distribution?

1. Dec 12, 2011

### thisguy12

Let W_1 and W_2 be independent Chi-Squared distributed random variables with v_1 and v_2 degrees of freedom, respectively. Then F = (W_1/v_1)/(W_2/v_2) = (v_2/v_1)(W_1/W_2) is said to have an F-distribution with v_1 numerator degrees of freedom and v_2 denominator degrees of freedom.

I want to find E(F).

Here is where I get confused:
E(F) = E((v_2/v_1)(W_1/W_2))
= (v_2/v_1) E(W_1/W_2) since v_1, v_2 are constants
= (v_2/v_1) E(W_1)/E(W_2) because W_1 and W_2 are independent random variables
= (v_2/v_1) (v_1/v_2) because if X is chi-squared r.v. with v degrees of freedom, E(X) = v
= 1

However, the correct result for the expected value of an F-distributed r.v. is v_2/(v_2 - 2). Where is the hiccup in my logic?

2. Dec 12, 2011

### Stephen Tashi

You are assuming E( 1/W2) = 1/ E(W2).

If W is a random variable with density f(w), this would say
$\int \frac{1}{w} f(w) dw$ = $\frac{1}{\int f(w) dw}$