Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 28 / 6 / c / Solution

Write the constant limit as $c$. First, convergence in distribution to a constant implies convergence in probability: the Portmanteau theorem applied to the closed set $\{y:|y-c|\ge \epsilon \}$ givesNow compare characteristic functions without imposing independence. For fixed $t\in \mathbb{R}$,The second bound uses $|e^{iu}-e^{iv}|\le \min \{2,|u-v|\}$. By convergence in probability, the limiting upper bound is $|t|\epsilon$; then let $\epsilon \downarrow 0$. Meanwhile $X_n$ has weak convergence of random variables to $X$, so $\mathbb{E}{e}^{it{X}_n}\to \mathbb{E}{e}^{itX}$. Thus the characteristic function of the sum tends to $e^{itc}\mathbb{E}{e}^{itX}$, the characteristic function of $X+c$, which is continuous at zero. The Lévy continuity theorem provesThis is the additive case of the Slutsky theorem, with its proof supplied. The constant limiting marginal removes the dependence obstruction in part (b).