Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 3 / 24J (Probability and Measure)

The characteristic function is ${\varphi }_X(u)=\mathbb{E}(e^{iuX})$. It satisfies ${\varphi }_{-X}(u)={\varphi }_X(-u)=\overline{{\varphi }_X(u)}$. Thus a real-valued characteristic function gives ${\varphi }_X={\varphi }_{-X}$, and the uniqueness theorem for characteristic functions implies equality of the two distributions. Conversely, equality in distribution gives equality with the complex conjugate, making the characteristic function real. Reality for all $u$ is equivalent to symmetry of the distribution about zero.

Choose a bounded strictly increasing function, for example $h(x)=\arctan x$. Independence and identical distribution imply $\mathrm{Var}(h(X)-h(Y))=2\mathrm{Var}(h(X))$. If $X=Y$ almost surely, the left side is zero. Therefore $h(X)$ is almost surely constant; injectivity of $h$ makes $X$ almost surely constant. Using a bounded function avoids any unassumed second moment of $X$.

Independence gives ${\varphi }_{X-Y}(u)={\varphi }_X(u){\varphi }_X(-u)=|{\varphi }_X(u){|}^2=1$ for $|u|<\epsilon$. Taking real parts shows $\mathbb{E}[1-\cos (u(X-Y))]=0$. The integrand is nonnegative, so it vanishes almost surely for each such $u$.

Choose a countable sequence of positive $u_j\to 0$ in that interval, and intersect the corresponding probability-one events. On their intersection, a nonzero difference $D=X-Y$ would give $0<|u_{j}D|<2\pi$ for large $j$, contradicting $\cos (u_{j}D)=1$. Thus $X=Y$ almost surely, and part (a) proves $X$ is almost surely constant.

Condition on $X$. The normal characteristic function gives $\mathbb{E}(e^{iuXY}\mid X)=e^{-u^2{X}^2/2}$, whose Gaussian integral is $(1+u^2{)}^{-1/2}$. A second Gaussian integral gives $\mathbb{E}(e^{-iu{Z}^2})=(1+2iu{)}^{-1/2}$, taking the square-root branch continuous from one at $u=0$. Independence of $(X,Y)$ and $Z$ therefore gives