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Math Help - Distribution of sample mean

  1. #1
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    Distribution of sample mean

    I am aware that for large sample sizes the sample mean tends towards a normal distribution but this is for either small sample sizes or for a more exact distribution than an approximate normal distribution. I think my idea works but part of the formulation of my conclusion is weak so I am posting it here for some feedback.

    Starting with the simple case of a sample size of 2.
    Suppose we are sampling from a continuous distribution with density function f(x) and range (-\infty,\infty). We take 2 samples which are random variables X1 and X2. The sample mean is \frac{x_1+x_2}{2}
    Given that the first sample takes the value a the second sample must take the value 2\bar{x}-a in order for the sample mean to equal \bar{x}
    The probability that X1 takes the value a and X2 takes the value b is f(a)f(b)
    P(x_1=a, x_2=b)= f(a)f(b)

    P(x_1=a, x_2=2\bar{x}-a)= f(a)f(2\bar{x}-a)

    This is the probability for the case where x1 took the value a, to account for all values of x1 the expression is integrated over the entire range.

    \int_{-\infty}^{\infty}f(a)f(2\bar{x}-a)da

    After evaluating the integral the resulting expression is the distribution of the sample mean.


    Now for the case where the sample size is 3

    The sample mean is \bar{x}=\frac{x_1+x_2+x_3}{3}

    Given that the first sample takes the value a and the second sample takes the value b the third sample must take the value 3\bar{x}-a-b in order for the sample mean to equal \bar{x}
    The probability that X1 takes the value a, X2 takes the value b and X3 takes the value c is f(a)f(b)f(c).
    P(x_1=a, x_2=b, x_3=c)= f(a)f(b)f(c)

    P(x_1=a, x_2=b, x_3=3\bar{x}-a-b)= f(a)f(b)f(3\bar{x}-a-b)

    This is the probability for the case where x1 took the value a, and x2 took the value c, to account for all values of x2 and x2 the expression is integrated over the entire range.

    \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f(a)f(b)f(3\bar{x}-a-b)da\cdot db

    After evaluating the integral the resulting expression is the distribution of the sample mean.

    You can see how this theory extends for a sample size n.
    Could I please have some feedback about this idea?
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  2. #2
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    Re: Distribution of sample mean

    Quote Originally Posted by Shakarri View Post
    I am aware that for large sample sizes the sample mean tends towards a normal distribution but this is for either small sample sizes or for a more exact distribution than an approximate normal distribution. I think my idea works but part of the formulation of my conclusion is weak so I am posting it here for some feedback.

    Starting with the simple case of a sample size of 2.
    Suppose we are sampling from a continuous distribution with density function f(x) and range (-\infty,\infty). We take 2 samples which are random variables X1 and X2. The sample mean is \frac{x_1+x_2}{2}
    Given that the first sample takes the value a the second sample must take the value 2\bar{x}-a in order for the sample mean to equal \bar{x}
    The probability that X1 takes the value a and X2 takes the value b is f(a)f(b)
    P(x_1=a, x_2=b)= f(a)f(b)

    P(x_1=a, x_2=2\bar{x}-a)= f(a)f(2\bar{x}-a)

    This is the probability for the case where x1 took the value a, to account for all values of x1 the expression is integrated over the entire range.

    \int_{-\infty}^{\infty}f(a)f(2\bar{x}-a)da

    After evaluating the integral the resulting expression is the distribution of the sample mean.


    Now for the case where the sample size is 3

    The sample mean is \bar{x}=\frac{x_1+x_2+x_3}{3}

    Given that the first sample takes the value a and the second sample takes the value b the third sample must take the value 3\bar{x}-a-b in order for the sample mean to equal \bar{x}
    The probability that X1 takes the value a, X2 takes the value b and X3 takes the value c is f(a)f(b)f(c).
    P(x_1=a, x_2=b, x_3=c)= f(a)f(b)f(c)

    P(x_1=a, x_2=b, x_3=3\bar{x}-a-b)= f(a)f(b)f(3\bar{x}-a-b)

    This is the probability for the case where x1 took the value a, and x2 took the value c, to account for all values of x2 and x2 the expression is integrated over the entire range.

    \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f(a)f(b)f(3\bar{x}-a-b)da\cdot db

    After evaluating the integral the resulting expression is the distribution of the sample mean.

    You can see how this theory extends for a sample size n.
    Could I please have some feedback about this idea?
    It is the same thing as the density of the sum of two independent RV being the convolution of their densities.

    .
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  3. #3
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    Re: Distribution of sample mean

    Alright thank you for the information, it's hard to find information on this stuff because most searches return something about the central limit theorem
    Last edited by Shakarri; July 18th 2013 at 05:36 AM.
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