# Exact distribution of the sample mean of a truncated continuous distribution

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• Sep 2nd 2013, 08:47 AM
Shakarri
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
$f(x)=\frac{x}{50}$ The maximum value of x is 10 and the minimum value of x is 0. Note that
$\int_0^{10} f(x) dx=1$
The samples are random variables $X_1$ and $X_2$
$\bar{x}=\frac{x_1+x_2}{2}$
If the sample mean is equal to $\bar{x}$ then given that the first sample is $x_1$the second sample $x_2$ must equal $2\bar{x}-x_1$. But if $x_1$ is too large or too small then a value of $x_2$ may not exist in the range[0,10].
We know $x_2 \leq 10 \therefore 2\bar{x}-x_1 \leq 10$

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

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

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

The probability that $X_1$ takes the value $x_1$ and $X_2$ takes the value $2\bar{x}-x_1$ is $f(x_1)f(2\bar{x}-x_1)$ To account for all cases of $X_1$ that could result in $\bar{X}=\bar{x}$ we integrate with respect to $x_1$ $\int_{2\bar{x}-10}^{2\bar{x}}f(x_1)f(2\bar{x}-x_1)dx_1$
I did this integration and it ended up as
$\frac{1}{7500}(1000-570\bar{x}-120\bar{x}^2)$

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
$\int_0^{10}\frac{1}{7500}(1000-570\bar{x}-120\bar{x}^2)=-7.8$So where have I made an incorrect assumption? Please help.
• Sep 2nd 2013, 06:31 PM
chiro
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.