Exact distribution of the sample mean of a truncated continuous distribution

Please first note that I do know about the central limit theorem but I wish to derive an exact expression for the sample mean for any continuous distribution with probability density function f(x). I thought I derived it correctly here but there is some mistake I can't spot. Can anyone tell me where I have made a mistake?

I take a simple case toderive the sample mean for with only 2 samples. The probabilitydensity function f(x) is defined as

The maximum value of x is 10 and the minimum value of x is 0. Note that

The samples are random variables and

If the sample mean is equal to then given that the first sample is the second sample must equal . But if is too large or too small then a value of may not exist in the range[0,10].

We know

And

The probability that takes the value and takes the value is To account for all cases of that could result in we integrate with respect to

I did this integration and it ended up as

The sample mean mustlie inside [0,10] so the integral of all the eventualities of the sample mean over the range [0,10] should equal 1. But

So where have I made an incorrect assumption? Please help.

Re: Exact distribution of the sample mean of a truncated continuous distribution

Hey Shakarri.e

I would work it out by getting the joint distribution of all your random variables and then using that to calculate 1/n*E[X1 + X2 + .. + XN].

As an example for two random variables you would get 1/2 * Integral (over space of all possiblities) (x1+x2)*f(x1,x2)dx1dx2. This works in all cases including truncated distributions.

You can extend this to as many variables as you want: the difference will be the number of integrals and the joint density function will be of n variables.