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

  1. Dec 12, 2011 #1
    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. jcsd
  3. Dec 12, 2011 #2

    Stephen Tashi

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    Science Advisor

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

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