The Student Room Group

S2 confusion

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?
Reply 1
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.
Reply 2
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
(edited 13 years ago)
Reply 3
Original post by ucasisannoying

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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