Week19: Central Limit Theorem

Considering X to be the sum of random variables of samples of the population (X1 + X2 + … + Xn), with mean n*mu and std (sqrt(n)*sigma).
The transformed variable Y = (X - mu)/sigma converges to a normal distribution with mean 0 and std (sqrt(n)) instead of 1 as is being suggested. I tried the problem both empirically as well as from mathematical theorems.

Please clarify.
Thank You

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Hi @saag,
Can you please share your proof that you tried?

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mean(X) = n*mu
std(X) = sqrt(n)*sigma
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I suppose sir considered mu and sigma as mean and std of random variable X for the explanation and established a relationship between X and Z instead of the population from which the random variables X1, X2, Xn are being taken.

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I haven’t seen the Week 19 lectures yet, but here is my understanding for this interesting discussion.

If X is defined as the sum, then for Y even the mean will not be 0 (and variance not 1 as you pointed out).

As shown in your response above, mean of X is n * nu and variance is ' n * sigma^2.
So to get a R.V from X with mean 0 and std 1, Y probably need to be constructed as

Y = \frac{X - mean\ of\ X}{std(X)}

ie.

Y = \frac{X - n\mu}{\sigma\sqrt{n}}

Calculation below for Expectation and Var for the two cases(left and right) based on how Y is defined:

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