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Math Help - Exact distribution of the sample mean of a truncated continuous distribution

  1. #1
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    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_1the 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)=-7.8So where have I made an incorrect assumption? Please help.
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  2. #2
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    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.
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