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 $\displaystyle (-\infty,\infty)$. We take 2 samples which are random variables X_{1} and X_{2}. The sample mean is $\displaystyle \frac{x_1+x_2}{2}$

Given that the first sample takes the value a the second sample must take the value $\displaystyle 2\bar{x}-a$ in order for the sample mean to equal $\displaystyle \bar{x}$

The probability that X_{1} takes the value a and X_{2} takes the value b is f(a)f(b)

$\displaystyle P(x_1=a, x_2=b)= f(a)f(b) $

$\displaystyle 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 x_{1} took the value a, to account for all values of x1 the expression is integrated over the entire range.

$\displaystyle \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 $\displaystyle \bar{x}=\frac{x_1+x_2+x_3}{3}$

Given that the first sample takes the value a and the second sample takes the value b the third sample must take the value $\displaystyle 3\bar{x}-a-b$ in order for the sample mean to equal $\displaystyle \bar{x}$

The probability that X1 takes the value a, X_{2} takes the value b and X3 takes the value c is f(a)f(b)f(c).

$\displaystyle P(x_1=a, x_2=b, x_3=c)= f(a)f(b)f(c) $

$\displaystyle 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 x_{1} took the value a, and x_{2} took the value c, to account for all values of x_{2} and x_{2} the expression is integrated over the entire range.

$\displaystyle \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?

Re: Distribution of sample mean

Quote:

Originally Posted by

**Shakarri** 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 $\displaystyle (-\infty,\infty)$. We take 2 samples which are random variables X_{1} and X_{2}. The sample mean is $\displaystyle \frac{x_1+x_2}{2}$

Given that the first sample takes the value a the second sample must take the value $\displaystyle 2\bar{x}-a$ in order for the sample mean to equal $\displaystyle \bar{x}$

The probability that X_{1} takes the value a and X_{2} takes the value b is f(a)f(b)

$\displaystyle P(x_1=a, x_2=b)= f(a)f(b) $

$\displaystyle 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 x_{1} took the value a, to account for all values of x1 the expression is integrated over the entire range.

$\displaystyle \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 $\displaystyle \bar{x}=\frac{x_1+x_2+x_3}{3}$

Given that the first sample takes the value a and the second sample takes the value b the third sample must take the value $\displaystyle 3\bar{x}-a-b$ in order for the sample mean to equal $\displaystyle \bar{x}$

The probability that X1 takes the value a, X_{2} takes the value b and X3 takes the value c is f(a)f(b)f(c).

$\displaystyle P(x_1=a, x_2=b, x_3=c)= f(a)f(b)f(c) $

$\displaystyle 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 x_{1} took the value a, and x_{2} took the value c, to account for all values of x_{2} and x_{2} the expression is integrated over the entire range.

$\displaystyle \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.

.

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