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

1. Oct 11, 2012

1. The problem statement, all variables and given/known data

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.

2. Oct 11, 2012

### Simon Bridge

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. Oct 11, 2012

@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. Oct 11, 2012

### Simon Bridge

$$f(y;\alpha) = \frac{\alpha}{y^2}: 0 < \alpha \leq y < \infty$$
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?
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])

Last edited by a moderator: May 6, 2017
5. Oct 11, 2012

### HallsofIvy

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 $\alpha$ to infinity, is good:
$$\int_\alpha^\infty \frac{\alpha}{y^2}dy= \alpha\int_\alpha^\infty y^{-2}dy= \alpha\left[-\frac{1}{y}\right]_\alpha^\infty= 1$$

And the expected value is just $\int yf(y)dy$ which, for this problem, is
$$\alpha\int_{\alpha}^\infty \frac{1}{y}dy$$
Now that has a problem at infinity!

6. Oct 11, 2012

### Ray Vickson

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. Oct 11, 2012

### HallsofIvy

No, it doesn't. The integral of $\alpha y^{-2}[itex], from [itex]\alpha$ to infinity is 1 so it is a pdf. It just does not have finite expected value.

8. Oct 11, 2012

### Ray Vickson

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. Oct 11, 2012

### Simon Bridge

... are the other moments finite?