how transforms prove the central limit theorem

The central limit theorem is one of the most surprising facts in mathematics: average enough independent copies of almost any distribution, rescale, and the result looks normal — regardless of the shape you started with. Stated baldly it sounds like magic. The transform method is what turns the magic into a short, honest proof, by working with characteristic functions instead of the unmanageable distribution of a giant sum.

Here is the whole argument in plain steps. Take independent identical variables, standardize each to have mean 0 and variance 1, and form the scaled average S_n* = (X_1 + ... + X_n)/sqrt(n). Because the pieces are independent, the characteristic function of the sum is a product, and standardizing makes it (phi(t/sqrt(n)))^n, where phi is the common characteristic function. Now Taylor-expand phi near 0: since the variables have mean 0 and variance 1, phi(s) = 1 - s^2/2 + (smaller terms). Plug in s = t/sqrt(n): phi(t/sqrt(n)) = 1 - t^2/(2n) + ... Raise to the n-th power. This is precisely the famous limit (1 - a/n)^n tending to e^(-a), so (1 - t^2/(2n) + ...)^n tends to e^(-t^2/2) as n grows. That limit, e^(-t^2/2), is exactly the characteristic function of the standard normal. Finally, Levy's continuity theorem converts this convergence of characteristic functions into convergence in distribution — so S_n* converges in distribution to Normal(0, 1).

Seen this way, the CLT is no longer mysterious: it is the consequence of one Taylor expansion plus the elementary limit (1 - a/n)^n to e^(-a). The proof also exposes exactly what is required and where it can fail. The Taylor step needs the second moment, so the theorem demands FINITE variance; for a heavy-tailed law like the Cauchy, which has no finite variance (indeed no mean), the expansion breaks and the CLT does not hold — averaging Cauchys gives back a Cauchy, never a normal. The transform proof makes the hypotheses visible rather than hiding them.