The inverse of uniform random variable

[giglamesh]

Hi all
I'm looking for solving this problem to find the closed form solution if it is possible:

Y=\frac{1}{X}

Where X is uniform random variable > 0
I know the expected value for X which is \overline{X}

is there a method to find the expected value of Y which is \overline{Y} in term of \overline{X} as closed form solution?

I know how to calculate it easily using numerical solution, but I need it for modeling problem and I need the analytical solution.

Thanks

 

[chiro]

Hi all
I'm looking for solving this problem to find the closed form solution if it is possible:

Y=\frac{1}{X}

Where X is uniform random variable > 0
I know the expected value for X which is \overline{X}

is there a method to find the expected value of Y which is \overline{Y} in term of \overline{X} as closed form solution?

I know how to calculate it easily using numerical solution, but I need it for modeling problem and I need the analytical solution.

Thanks

Hey giglamesh and welcome to the forums.

What is your statistics and probability background like?

A standard intro year course in university will give you the tools to solve this problem. Do you know about transformation theorems in statistics?

http://www.ncur20.ws/presentations/2/216/presentation.pdf

 

[giglamesh]

hello chiro
Thanks for replying

I have a background with Random Variables and stochastic processes
I've read about MLE once but never use it in my applications, I remember that it is used to estimate the random variable from sample data vectors.

which is not what I'm looking for.

Maybe I didn't explain my problem well:

X is random variable I know only it's expected variable
Y=1/X is a random variable I need to know it's expected value using only E[X]

I'll check the transformation methods you mentioned, I know there are Laplace and Z-transform, I've used Z-transform but it didn't give the required result.

I'll try to search more.
Thanks

 

[micromass]

Aah, that changes it. Given a continuous random variable X with pdf f_X and a function g, we can always calculate

E[g(X)]=\int_{-\infty}^{+\infty}{g(x)f_X(x)dx}

So in your case, you need to calculate

E[1/X]=\int_{-\infty}^{+\infty}{\frac{f_X(x)}{x}dx}

So if X is uniform(1,2) for example, then

E[1/X]=\int_1^2{\frac{1}{x}dx}=\log{2}

 

[giglamesh]

hi micromass
Actually X is discrete I need to say: X is not uniform but Y=1/X is constructed as a uniform distribution from X, that means gives that X=3 then Y=1/3
P(y)=E[1/X]
I know only X then I need to get E[1/X] using only E[X] which is known but the distribution of X is not known.

Thanks for replying

 

[giglamesh]

I think what chiro said makes sense for me right now

X 0 1 2 3 ...H
P(Y|X=i)=1/i 1 0.5 1/3 ...1/H

From the second line I'll try to estimate the PMF of Y using MLE, I'll try it

 

[giglamesh]

I just closed this thread, I will open new one and try to make it more clear.