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So I'm trying to understand how to derive the central limit theorem but I'm confused about step 3 in Peter Young's Derivation

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In step 3, rightmost part of the equation, why are we taking the sum of all $x_i$, then subtracting them by $X$, which is essentially equal to zero. Won't it turn to $\delta(0)$ ? Are we really saying $\delta(1+2+3-6)$

Iancovici
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  • Does he mean "density" by "distribution"? Anyway, the point of the delta function is just to limit the range of integration to those $x_i$ that satisfy the constraint. I think maybe you are confused because I would think equation (1) should have "$X_i$" for "$x_i$" and also the random variable mentioned in the first sentence should be "$X$" not "$x$". Does it make sense then? – user39080 Nov 13 '13 at 04:32
  • I know the point of a delta function, but i don't think someone would make it that redundant for no reason. No it doesn't make sense. – Iancovici Nov 13 '13 at 12:02
  • I'm not saying the delta function is redundant. It isn't! But yet another bad thing about the notation in this handout is that if appears as if the $x_i$'s in the delta function expression are outside the scope of the binding $dx_i$'s. It would be better to write all the integral signs, then all the $P(x_i)$ expressions, then the delta function, then all the $dx_i$'s. Does it make sense to you written like that? You can view $X$ as fixed and the $x_i$ as varying. – user39080 Nov 13 '13 at 13:04
  • I know this is an old question and I am a beginner. But I don't think it makes sense to say that as $N \rightarrow \infty$ you get a normal with mean $N\mu$ and standard deviation $\sqrt{N}\sigma$ because (a) the mean and variance go to infinity as $N$ goes to infinity and (b) you shouldn't have the sample size $N$ in the limiting distribution. I think it makes sense to say that for large finite $N$ the sum is very close to but not actually $Normal(N\mu,N\sigma^2)$ – HJ_beginner Oct 11 '18 at 00:29

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