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# Continuous Random Variable Problem watch

1. 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.
2. The cdf of X is x.
3. (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?
4. (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.
5. (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
6. (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.
7. (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.
8. (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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