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    calculating the E(X) of a pdf = integrate xf(x) with limits

    calculating E(X) of a cdf = just finding the midpoint of the cdf?

    is this correct?
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    Your first statement is correct.

    However for your second statement, if your cdf is F then what you're suggesting is to solve F(x)=1/2. This would give you the median value of your random variable, not the expectation. What you need to do instead is to differentiate F with respect to x to get the pdf f. Then integrate xf(x) with appropriate limits to get the expectation.
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    (Original post by ttoby)
    Your first statement is correct.

    However for your second statement, if your cdf is F then what you're suggesting is to solve F(x)=1/2. This would give you the median value of your random variable, not the expectation. What you need to do instead is to differentiate F with respect to x to get the pdf f. Then integrate xf(x) with appropriate limits to get the expectation.
    just to clear things up:

    pdf: integrate xf(x) with limits
    cdf: differentiate to get pdf, THEN integrate xf(x) with limits?

    if so, that makes perfect sense. thanks a lot
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    (Original post by ucasisannoying)
    just to clear things up:

    pdf: integrate xf(x) with limits
    cdf: differentiate to get pdf, THEN integrate xf(x) with limits?

    if so, that makes perfect sense. thanks a lot
    Yeah that's right.
 
 
 
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