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Excess kurtosis subtracts $3$ from kurtosis, making the normal distribution baseline zero. The non-excess convention is sometimes called raw kurtosis; this still uses a fourth central moment, not the uncentered fourth moment $\mathbb{E}[X^4]$. Software may report either kurtosis or excess kurtosis, so the convention must be stated.^[ref]

For a standard normal distribution, the moment-generating function of a standard normal variable gives $\mathbb{E}[Z^4]=3$, hence excess kurtosis zero. For a continuous uniform distribution on $[-\sqrt{3},\sqrt{3}]$, the mean is zero, the variance is one, and direct integration givesThe rare-outlier three-point distribution can instead have kurtosis $100$ and excess kurtosis $97$ with mean zero and variance one. These are calculated properties of specified probability distributions, rather than a measured statistic.

Let $S=X+Y$ and $D=X-Y$. Since they are independent, applying the product rule for moment-generating functions to $S+D=2X$ givesAll factors are finite and strictly positive for every real $t$ by the stated hypothesis. Applying the same identity to $-t$ and dividing yields $\psi (2t)=\psi (t{)}^2$, orThe centered unit-variance Taylor expansion from part (ii) gives $M(h)=1+h^2/2+o(h^2)$ and the same expansion for $M(-h)$. Consequently $\psi (h)=1+o(h^2)$.

For the first step of dyadic rigidity of a moment-generating function, take logarithms, which are allowed because $\psi >0$. For each fixed $t$ and every integer $m$,Since $\log \psi (h)=o(h^2)$, the right side tends to zero: it is $2^{m}o(t^2/4^m)$. Thus $\psi (t)=1$ for every $t$, and the moment-generating function is even.

The original identity now becomes $M(2t)=M(t{)}^4$. Iterating this time with the correct fourth-power scaling givesNear zero, $\log M(h)=h^2/2+o(h^2)$, so the right side tends to $t^2/2$. ThereforeThis is the moment-generating function of a standard normal variable. By the uniqueness theorem for moment-generating functions, each of $X$ and $Y$ has the standard normal distribution. The proof establishes the Gaussian characterization by independent sum and difference directly from the two functional identities and the first two moments.

A useful method of moments in probability states that if real random variables $Z_N$ have all moments, $\mathbb{E}{Z}_N^j\to m_j$ for every integer $j\ge 1$, and $(m_j)$ is the moment sequence of a moment-determinate probability distribution, then $Z_N$ converges to that distribution in the sense of convergence in distribution. In particular, convergence to the standard normal distribution follows from limiting odd moments zero and limiting even moments $(2k-1)!!$. The moment-generating function of a standard normal variable, finite in a neighbourhood of zero, ensures moment determinacy.

For the classical Erdős-Kac theorem, put $L=\log \log N$ and choose, for example, $\phi (N)=L^{1/4}$, with a harmless modification for small $N$. Truncate the prime omega function to primes $p\le y=N^{1/\phi (N)}$. The omitted number of distinct prime factors of any $n\le N$ is at most $\phi (N)$, since a product of $r$ omitted primes exceeds $N^{r/\phi (N)}$. Thus this truncation changes the normalized variable by $o(1)$ uniformly. Also, the Mertens theorem for reciprocal primes gives

Part (b) transfers every fixed normalized central moment of the truncated prime omega function to the independent Bernoulli random variables model. Indeed, its unnormalized error is at most $O_j(N^{2j/\phi (N)-1})$, using ${\sum }_{p\le y}1\le y$, and this tends to zero for each fixed $j$. In the independent expansion, any singleton index has zero expectation; the leading contributions are pairings, giving exactly the standard normal distribution moments, while blocks of size at least three are negligible after normalization. The method of moments in probability, followed by the uniformly negligible truncation and centering errors, proves the Erdős-Kac theorem.

The same strategy proves the bounded-prime-weight form in part (a). Choose $\phi (N)\to \infty$ slowly enough that $\phi (N)=o(B(N))$. The omitted weighted contribution is $O(\phi (N))$; its mean is likewise $O(\phi (N))$, and its independent variance is $O(\log \phi (N)+1)=o(B(N{)}^2)$ by the Mertens theorem for reciprocal primes. Hence the truncated model has the same normalization, and part (b) again transfers each fixed central moment.

The standard normal density isIt is nonnegative, and the Gaussian integral gives ${\int }_{-\infty }^{\infty }\varphi (z)dz=1$, so it is a probability density function. Completing the square gives the moment-generating function of a standard normal variableThus $\mathbb{E}Z=M_Z^{\prime }(0)=0$ and

Complete the square in the integral against the standard normal density:The remaining integral is one because its integrand is the probability density function of the normal distribution $N(t,1)$. Hence the moment-generating function of a standard normal variable is finite on the whole real line.

For fixed real $t$, Taylor's theorem for the exponential function givesSubstituting into the previous moment-generating function,The continuity theorem for moment-generating functions states that, if the moment-generating functions of $Z_n$ and $Z$ are finite on a common open interval containing zero and $M_{Z_n}(t)\to M_Z(t)$ throughout that interval, then $Z_n\stackrel{\mathrm{d}}{\to }Z$. Here finiteness holds for all real $t$, and the limit is the moment-generating function of a standard normal variable. ThereforeThis proves the required instance of the Poisson central limit theorem directly from moment-generating functions; no appeal to a general central limit theorem is needed.

Integrate against the standard normal density by completing the square:Thus the moment-generating function of a standard normal variable is

Write $S_n=n^{-1/2}{\sum }_{k=1}^n{X}_k$. By independence of random variables, its moment-generating function isFor each fixed real $t$, the expansion in part (i) yieldsFor $t=0$ the equality is immediate. Exponentiation gives $M_{S_n}(t)\to e^{t^2/2}$, the moment-generating function of a standard normal variable. The functions are finite on a common neighborhood of zero, so the convergence theorem for moment-generating functions provesThis proves the central limit theorem under the exponential-moment hypothesis of this part.