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How to find the expected value of a continuous variable with pdf fy= y^(-2)?

  • #1

Homework Statement



Find the expected value of a continuous variable y with pdf fy= alpha*y^-2, 0<y<infinity.

I know it is the integral from zero to infinity of y*fy, but I don't know where to go from there. I'm then supposed to use the expected value to find the method of moments estimator.

Does this pdf fall under a certain type of distribution?
Any help is appreciated.
 

Answers and Replies

  • #2
Simon Bridge
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What is f(y) if y≤0?

The next step is to write out the integral and carry it out.
But I'm thinking there is information not included?
 
  • #3
@Simon Bridge.

No. That's all I'm given.

f(y;alpha)= alpha*y^(-2) 0<alpha<= y< infinity.

It says to determine E(Y) and then find the method of moments estimator of alpha.


The integral of y*fy gives me [alpha* lny] evaluated from zero to infinity, but isn't it infinity?
 
  • #4
Simon Bridge
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f(y;alpha)= alpha*y^(-2) 0<alpha<= y< infinity.
$$f(y;\alpha) = \frac{\alpha}{y^2}: 0 < \alpha \leq y < \infty$$
No. That's all I'm given.
Well then it cannot be done because the function in not determined for part of the range... though the above is already different from what you gave in post #1. Perhaps the extra information is implicit in the context - in your course notes for eg?

Perhaps you are supposed to assume the pdf is zero outside the given range?
The integral of y*fy gives me [alpha* lny] evaluated from zero to infinity, but isn't it infinity?
But maybe you are supposed to evaluate from ##\alpha## to ##\infty##?
Still not much help -

And what about normalization? Isn't $$\int_{-\infty}^\infty f(y)dy=1$$ ... unless ##\alpha## is a function of y?

I'd go ask the person who set the problem. (After checking with other students.)

http://en.wikipedia.org/wiki/Method_of_moments_(statistics [Broken])
 
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  • #5
HallsofIvy
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It is standard to define a probability density on a restricted range- it does not have to be from negative infinity to infinity. However, the first given problem, from 0 to infinity has that the problem that the integral goes to infinity. The revised problem, from [itex]\alpha[/itex] to infinity, is good:
[tex]\int_\alpha^\infty \frac{\alpha}{y^2}dy= \alpha\int_\alpha^\infty y^{-2}dy= \alpha\left[-\frac{1}{y}\right]_\alpha^\infty= 1[/tex]

And the expected value is just [itex]\int yf(y)dy[/itex] which, for this problem, is
[tex]\alpha\int_{\alpha}^\infty \frac{1}{y}dy[/tex]
Now that has a problem at infinity!
 
  • #6
Ray Vickson
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Homework Statement



Find the expected value of a continuous variable y with pdf fy= alpha*y^-2, 0<y<infinity.

I know it is the integral from zero to infinity of y*fy, but I don't know where to go from there. I'm then supposed to use the expected value to find the method of moments estimator.

Does this pdf fall under a certain type of distribution?
Any help is appreciated.
There IS NO SUCH probability density function! What you have integrates to +∞ for any α > 0, so cannot integrate to 1. Are you really sure you have written the question correctly?

RGV
 
  • #7
HallsofIvy
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No, it doesn't. The integral of [itex]\alpha y^{-2}[itex], from [itex]\alpha[/itex] to infinity is 1 so it is a pdf. It just does not have finite expected value.
 
  • #8
Ray Vickson
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No, it doesn't. The integral of [itex]\alpha y^{-2}[itex], from [itex]\alpha[/itex] to infinity is 1 so it is a pdf. It just does not have finite expected value.
OK, I missed the revised statement that had y ε [α,∞) instead of [0,∞).

However, there is still a problem with EY, so I still wonder if the question has been stated correctly.

RGV
 
  • #9
Simon Bridge
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... are the other moments finite?
 

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