*Dr.Rigol's Statistics*

**CENTRAL LIMIT THEOREM: A DEMONSTRATION**

Now I will give a simple demonstration of the Central Limit Theorem based on the results we got when we discussed the problem about adding Standard Deviations or adding Variances (see :"here!"). There, we concluded that when adding N independent variables randomly selected, the mean value of the sum Y defined as:

Y = X_{1} + X_{2} + .....+ X_{N}.

will be equal to: **μ _{Y} = N
μ_{X},** and the variance for Y will be equal to N times
the variance for X:

σ^{2}_{Y} = N σ^{2}_{X}

In the Central Limit Theorem we are
interested in a sample of dimension N. The mean value for this sample <X>
will be equal to <X> = Y/N. As you can see, we have a linear relationship
between <X> and Y. As we have shown in other place ( see here!)
the mean value for <X> will be equal to the mean value of Y divided
by N, and the standard deviation for <X> will be equal to the
standard deviation for Y divided by N, or what is equivalent, the variance for
<X> will be equal to the variance for Y divided by N^{2}.
So,

μ_{<X>}
= μ_{Y}/N = Nμ_{X}/N = μ_{X},

that means that the mean value of the sampling distribution will be equal to the mean value of the original distribution for X, and

σ^{2}_{<X>} = σ^{2}_{Y}
/ N^{2}.

As we saw before, σ^{2}_{Y} = N σ^{2}_{X},
so we can conclude that :

σ^{2}_{<X>} = σ^{2}_{Y}
/ N^{2}.= σ^{2}_{X}/N,

or, what is equivalent, the standard deviation for <X> will be equal to the standard deviation for X divided by the square root of N. So, the standard deviation for the sampling distribution will be equal to the standard deviation of the original distribution divided by the square root of N, the dimension of the sample.