If a random variable converge in distribution then scalar multiplication also converges in distribution.

Question

Let c be a constant and suppose that $X_n\xrightarrow{d}X$, prove that

i) $cX_n\xrightarrow{d}cX$
ii) $X_n+c\xrightarrow{d}X+c$

I've tried to use that:

$lim_{n\rightarrow\infty} F_{cX_n}(x) = lim_{n\rightarrow\infty} P(cX_n\leq x) = lim_{n\rightarrow\infty} P\left(X_n\leq \dfrac{x}{c}\right) = lim_{n\rightarrow\infty} F_{X_n}\left(\dfrac{x}{c}\right) $

But that's not exactly what I was looking for.

Answer 1

Remember that $X_n \xrightarrow{d} X$ it's by definition: $\lim\limits_{n \rightarrow \infty} F_{X_n} (x) = F_X(x)$

Hence:
a) \begin{equation*} \begin{split} \lim\limits_{n \rightarrow \infty} F_{cX_n} (x) & = \lim\limits_{n \rightarrow \infty} \mathbb{P} (cX_n \leq x) \\ & = \lim\limits_{n \rightarrow \infty} \mathbb{P} \left( X_n \leq \dfrac{x}{c} \right) \\ & = \lim\limits_{n \rightarrow \infty} F_{X_n} \left(\dfrac{x}{c}\right) \\ & = F_{X} \left(\dfrac{x}{c}\right) \\ & = \mathbb{P} \left( X \leq \dfrac{x}{c} \right) \\ & = \mathbb{P} \left( cX \leq x \right) \\ & = F_{cX} \left({x}\right) \\ \end{split} \end{equation*}

\begin{equation*} \therefore \lim\limits_{n \rightarrow \infty} F_{cX_n} (x) = F_{cX} \left({x}\right) \end{equation*}

\begin{equation*} \therefore cX_n \xrightarrow{d} cX \end{equation*}

Analogously:
b) \begin{equation*} \begin{split} \lim\limits_{n \rightarrow \infty} F_{X_{n}+c} (x) & = \lim\limits_{n \rightarrow \infty} \mathbb{P} (X_n + c \leq x) \\ & = \lim\limits_{n \rightarrow \infty} \mathbb{P} \left( X_n \leq x-c \right) \\ & = \lim\limits_{n \rightarrow \infty} F_{X_n} \left( x-c \right) \\ & = F_{X} \left( x-c \right) \\ & = \mathbb{P} \left( X \leq x-c \right) \\ & = \mathbb{P} \left( X+c \leq x \right) \\ & = F_{X+c} \left( x \right) \\ \end{split} \end{equation*}

\begin{equation*} \therefore \lim\limits_{n \rightarrow \infty} F_{X_{n}+c} (x) = F_{X+c} \left({x}\right) \end{equation*}

\begin{equation*} \therefore X_n+c \xrightarrow{d} X+c \end{equation*}

Answer 2

The sequence $X_n$ converges in distribution to $X$ as $n \rightarrow \infty$ (denoted $X_n \Rightarrow X$) if and only if, for any bounded continuous function $f$ (denoted $f \in C_b$), we have $$\lim_{n\rightarrow\infty} E[f(X_n)] = E[f(X)].$$ Let $c \in \mathbb{R}$ be given. The maps $g_c^\times(x) = cx$ and $g_c^+(x) = x + c$ are continuous, so for any $f \in C_b$, $f\circ g_c^\times$ and $f\circ g_c^+$ are also in $C_b$. Therefore, because $X_n \Rightarrow X$ we have $$\lim_{n\rightarrow\infty} E[f(X_n^{c,\times})] = \lim_{n\rightarrow\infty} E[(f\circ g_c^\times)(X_n)] = E[(f\circ g_c^\times)(X)] = E[f(X_c^\times)],$$ and ditto for $X_n^{c,+}$.