Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 2 / 9F (Probability)

The moment-generating function is $M_X(t)=\mathbb{E}[e^{tX}]$ at real $t$ for which that expectation is finite; $M_X(0)=1$. For independent random variables, the joint law factorizes, soThis is the transform form of independence. Its ordinary finite-valued interpretation applies where both factors are finite, with the transform of a zero multiple equal to $1$.

Finiteness of the moment-generating function on an open neighborhood of zero gives its differentiability there. Using the permitted interchange of differentiation and expectation,Hence $M_X(0)=1$, $M_X^{\prime }(0)=\mu$ and $M_X^{\prime \prime }(0)=s^2$. The second-order Taylor expansion with its Peano remainder givesHere $s^2$ is the raw second moment, not the variance: $\mathrm{Var}(X)=s^2-{\mu }^2$. This distinction matters before specializing to a centered variable.

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.