Question 5 - Chapter 9 (Resnick, 2005)
Suppose \(X_{n}\) and \(Y_{n}\) are independent for each \(n\) and \[\begin{equation*} \begin{array}{lclllllll} X_{n} & \Rightarrow & X_{0}, \hspace{2ex} Y_{n} & \Rightarrow & Y_{0}, \end{array} \end{equation*}\]
Prove using characteristic functions that \[\begin{equation*} \begin{array}{lclllllll} X_{n} + Y_{n} & \Rightarrow & X_{0} + Y_{o}. \end{array} \end{equation*}\]
Solution
Assuming that \(X_{n}\) and \(Y_{n}\) are independent for each \(n\) and \[\begin{equation*} \begin{array}{lclllllll} X_{n} & \Rightarrow & X_{0} \hspace{2ex} \mbox{and} \hspace{2ex} Y_{n} & \Rightarrow & Y_{0}, \end{array} \end{equation*}\]
Then, by the continuous mapping theorem (Corollary 8.3.1) we have that
\[\begin{equation*} \begin{array}{lclllllll} e^{itX_{n}} & \Rightarrow & e^{itX_{0}} \hspace{2ex} \mbox{and} \hspace{2ex} e^{itY_{n}} & \Rightarrow & e^{itY_{0}} \end{array} \end{equation*}\] since for each \(t\), the functions \(e^{itX_{n}}\) and \(e^{itY_{n}}\) are bounded continuous functions, that is, we have to
\[\begin{equation*} \begin{array}{lclllllll} |e^{itX_{n}}| & \leq & 1 \hspace{2ex} \mbox{and} \hspace{2ex} |e^{itY_{n}}| & \leq & 1 \end{array} \end{equation*}\]Thus, by the dominated convergence theorem, we obtain
\[\begin{equation*} \begin{array}{lclllllll} \mathbb{E}(e^{itX_{n}}) & \xrightarrow[n \to \infty]{} & \mathbb{E}(e^{itX_{0}}) \hspace{2ex} \mbox{and} \hspace{2ex} \mathbb{E}(e^{itY_{n}}) & \xrightarrow[n \to \infty]{} & \mathbb{E}(e^{itY_{0}}) \end{array} \end{equation*}\] that is,
\[\begin{equation*} \begin{array}{lclllllll} \phi_{X_{n}}(t) & \xrightarrow[n \to \infty]{} & \phi_{X_{0}}(t) \hspace{2ex} \mbox{and} \hspace{2ex} \phi_{Y_{n}}(t) & \xrightarrow[n \to \infty]{} & \phi_{Y_{0}}(t) \end{array} \end{equation*}\] for all \(t\in\mathbb{R}\)
Where such an implication is part of the Continuity Theorem (Theorem 9.5.2, part (i)).
Now, if \(X_{n}\) is independent of \(Y_{n}\), for each \(n\), so by property 7 of characteristic function (page 297), we have that,
\[\begin{equation*} \begin{array}{lclllllll} \phi_{X_{n} + Y_{n}}(t) & = & \mathbb{E}(e^{it(X_{n}+Y_{n})}) \\ & = & \mathbb{E}(e^{itX_{n}} \cdot e^{itY_{n}}) \\ & \stackrel{ind.}{=} & \mathbb{E}(e^{itX_{n}}) \cdot \mathbb{E}(e^{itY_{n}}) \\ & = & \phi_{X_{n}}(t) \cdot \phi_{Y_{n}}(t) \\ \end{array} \end{equation*}\] for each \(t\in \mathbb{R}\).
In which, \(\phi_{X_{n} + Y_{n}}(t)\) is the characteristic function of \(X_{n} + Y_{n}\).
Thus, \[\begin{equation*} \begin{array}{lclllllll} \lim\limits_{x \to \infty} \left[\phi_{X_{n}+Y_{n}}(t) \right] & = & \lim\limits_{x \to \infty} \left[\phi_{X_{n}}(t)\cdot\phi_{Y_{n}}(t) \right] \\ \end{array} \end{equation*}\]
and as seen previously, the limits of \(\phi_{X_{n}}\) and \(\phi_{Y_{n}}\) exist, so
\[\begin{equation*} \begin{array}{lclllllll} \lim\limits_{x \to \infty} \left[\phi_{X_{n}}(t) \cdot\phi_{Y_{n}}(t) \right] & = & \lim\limits_{x \to \infty} \phi_{X_{n}}(t)\cdot\lim\limits_{x \to \infty}\phi_{Y_{n}}(t) \\ & = & \phi_{X_{0}}(t)\cdot\phi_{Y_{0}}(t), \\ \end{array} \end{equation*}\] for all \(t\in\mathbb{R}\).
So, if \(X_{0}\) and \(Y_{0}\) are independent, then we will have that by property 7 of the characteristic function \[\begin{equation*} \begin{array}{lclllllll} \phi_{X_{0}+Y_{0}}(t) & = & \phi_{X_{0}}(t)\cdot\phi_{Y_{0}}(t), \\ \end{array} \end{equation*}\] for all \(t\in\mathbb{R}\).
And then we have \[\begin{equation*} \begin{array}{lclllllll} \lim\limits_{x \to \infty} \left[\phi_{X_{n}+Y_{n}}(t) \right] & = & \lim\limits_{x \to \infty} \left[\phi_{X_{n}}(t) \cdot\phi_{Y_{n}}(t) \right] \\ & = & \phi_{X_{0}}(t)\cdot\phi_{Y_{0}}(t) \\ & = & \phi_{X_{0}+Y_{0}}(t), \end{array} \end{equation*}\] for all \(t\in\mathbb{R}\).
Thus, let \(\phi_{X_{0}}(t)\) and \(\phi_{Y_{0}}(t)\) continuous functions at \(t=0\), then \(\phi_{X_{n}}(t) \cdot\phi_{Y_{n}}(t) = \phi_{X_{0}+Y_{0}}(t)\) is continuous at \(t=0\).
In which, \(\phi_{X_{0}+Y_{0}}(t)\) is the characteristic function of \(X_{0}+Y_{0}\). Therefore, by the Continuity Theorem (Theorem 9.5.2, part (ii)), we will have that \[\begin{equation*} \begin{array}{lclllllll} X_{n} + Y_{n} & \Rightarrow & X_{0} + Y_{0}, \end{array} \end{equation*}\] as we wanted to demonstrate.
Note: “\(\Rightarrow\)” means convergence in distribution.