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    Let X be uniformly distributed on [0,1], find the expectation of X^p for -1<p<infinity

    I got the c.d.f. of X to be 1 and calculated the expectation of X^p to be the integral of x^p between 0 and 1 which came to 1/p+1

    Is this right so far?

    The question then said to do the same for p<=-1

    I think this should be the same formula, but for p=-1 the expectation doesn't exist.

    THe problem here is that X is distributed on [0,1], yet for p<-1 the expectation of X is negative according to my formula, which intuitivly makes no sense.

    Where have I gone wrong.
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    The cdf of X is x.
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    (Original post by Post121)
    The cdf of X is x.
    Really? But integral of x between 0 and 1 is 1/2=/=1. I thought the total area under the curve had to be 1?
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    (Original post by james22)
    Really? But integral of x between 0 and 1 is 1/2=/=1. I thought the total area under the curve had to be 1?
    You are confusing pdf with its cdf.
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    (Original post by Post121)
    You are confusing pdf with its cdf.
    I have in my notes that the c.d.f. is a function, f(x), which staisfies f(x)>=0 and integral of f(x) between -infinity and infinity is 1
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    (Original post by james22)
    I have in my notes that the c.d.f. is a function, f(x), which staisfies f(x)>=0 and integral of f(x) between -infinity and infinity is 1
    You sure it's not integral f'(x) between -infinity and infinity is 1?

    Edit: Your first expectation is correct. However, when p<-1, you are dividing by zero to achieve that formulae.
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    (Original post by Post121)
    You sure it's not integral f'(x) between -infinity and infinity is 1?
    I've double checked and you are correct, but I think I was still using the finction correctly, even if I named it wrong.
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    (Original post by james22)
    I've double checked and you are correct, but I think I was still using the finction correctly, even if I named it wrong.
    See my edit in previous post.
 
 
 
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Updated: December 31, 2012
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