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

$\displaystyle f(x)=\frac{x}{50}$ The maximum value of x is 10 and the minimum value of x is 0. Note that

$\displaystyle \int_0^{10} f(x) dx=1$

The samples are random variables $\displaystyle X_1$ and $\displaystyle X_2$

$\displaystyle \bar{x}=\frac{x_1+x_2}{2}$

If the sample mean is equal to $\displaystyle \bar{x}$ then given that the first sample is $\displaystyle x_1$the second sample $\displaystyle x_2$ must equal $\displaystyle 2\bar{x}-x_1$. But if $\displaystyle x_1$ is too large or too small then a value of $\displaystyle x_2$ may not exist in the range[0,10].

We know $\displaystyle x_2 \leq 10 \therefore 2\bar{x}-x_1 \leq 10$

$\displaystyle x_1 \geq 2\bar{x}-10$

And $\displaystyle x_2 \geq 0 \therefore x_1 \leq 2\bar{x}$

$\displaystyle 2\bar{x}-10 \leq x_1\leq 2\bar{x}$

The probability that $\displaystyle X_1$ takes the value $\displaystyle x_1$ and $\displaystyle X_2$ takes the value $\displaystyle 2\bar{x}-x_1$ is $\displaystyle f(x_1)f(2\bar{x}-x_1)$ To account for all cases of $\displaystyle X_1 $ that could result in $\displaystyle \bar{X}=\bar{x} $ we integrate with respect to $\displaystyle x_1$$\displaystyle \int_{2\bar{x}-10}^{2\bar{x}}f(x_1)f(2\bar{x}-x_1)dx_1$

I did this integration and it ended up as

$\displaystyle \frac{1}{7500}(1000-570\bar{x}-120\bar{x}^2)$

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

$\displaystyle \int_0^{10}\frac{1}{7500}(1000-570\bar{x}-120\bar{x}^2)=-7.8$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.