Getting wrong expected value integrating over survivor function

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peddy

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I'm working on a problem where I'm given the cumulative probability density function (F(x)=1-x-5, x>1). To get the expected value, I integrated 1-F(x) over 1 to infinity. When I do this, I get 0.25.

When I take the derivative to get the pdf, I get f(x)=5x-4. When I integrate xf(x) from 1 to infinity, I get 1.25.

Why am I getting different values? The answer is definitely 1.25. I thought that to get the expected value, you could integrate either xf(x) or 1-F(x) over the range of possible values. Is this not universally true? Or am I missing something?

Any help would be greatly appreciated. Thanks in advance.

Peter
 
peddy said:
I'm working on a problem where I'm given the cumulative probability density function (F(x)=1-x-5, x>1). To get the expected value, I integrated 1-F(x) over 1 to infinity. When I do this, I get 0.25.

When I take the derivative to get the pdf, I get f(x)=5x-4. Incorrect

\(\displaystyle \frac{d}{df}(x^{-n}) \ = \ (-n) * x^{-(n+1)}\)

When I integrate xf(x) from 1 to infinity, I get 1.25.

Why am I getting different values? The answer is definitely 1.25. I thought that to get the expected value, you could integrate either xf(x) or 1-F(x) over the range of possible values. Is this not universally true? Or am I missing something?

Any help would be greatly appreciated. Thanks in advance.

Peter
Click to expand...

,
 
Sorry for the typo. The pdf is f(x) = 5x-6. That gives E[X] = 1.25. Can you explain why integrating the survivor function isn't working?
 
Could you explain why you think integrating F(x) should give the expected value?
 
Hi HallsofIvy,

If you use integration by parts, you'll see that the integral of xf(x) from 0 to infinity equals the integral of 1-F(x) from 0 to infinity, if F(x) is the integral of f(t) over 0 to x.

Anyway, your question made me look at this from another angle. Since this cpdf is not fixed by x=0 (i.e., the range starts at x=1), the above identity isn't applicable. That's why it's not working.

Thanks everyone for considering this. Is there a way to mark this thread as resolved?

Peter
 
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