# Linear combination of a sequence converging in distribution (weakly)

I've been trying to prove that if $X_n \rightarrow X$ in distribution, that is for $F_n, F$ - distribution functions of $X_n, \ X$ resp we have:

$$\forall x: \ F \ \text{is continuous in} \ x : \ \Rightarrow F_n(x) \rightarrow F(x) \ \ \ (n \rightarrow \infty) \ \ \ \ \ \ \ (*)$$

then also $aX_n + b \rightarrow aX+b$ in distribution.

I've tried using:

$1)$ A sequence of distributions converges weakly $P_n \rightarrow P$ by definition, if the sequence of their distribution functions satisfies $(*)$

And it is equivalent to: for any bounded continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ we have $\int_{\mathbb{R}} f dP_n \rightarrow \int_{\mathbb{R}} f dP \ \ \ (n \rightarrow \infty)$

2) By Portmanteau lemma we have $E(f(X_n)) \rightarrow E(f(X)) \ \ \ \ (n \rightarrow \infty)$

But neither has lead me anywhere so far.

Could you help me out a bit?

Thank you!

By Portmanteau's lemma, it suffices to show

$$\mathbb{E}g(a X_n+b) \to \mathbb{E}g(aX+b) \tag{1}$$

for any continuous bounded function $g$. Since

$$f(x) := g(ax+b)$$

is also a continuous bounded function, this is equivalent to

$$\mathbb{E}f(X_n) \to \mathbb{E}f(X). \tag{2}$$

Can you take it from here?

• If $a, b \in \mathbb{R}$, then $E(aX_n+b) = aE(X_n)+b$. And $X_n \rightarrow X$ in distr., $\iff E(X_n) \rightarrow E(X)$ by Portmanteau lemma. – Bilbo Mar 8 '15 at 13:01
• $X_n \rightarrow X$ in distribution $\iff$ for any $f, E(f(X_n)) ...$ (in particular for $f(x)= g(ax+b)$) $\ \ \Rightarrow$ $E(g(aX_n +b)) \rightarrow E(g(aX+b)) \ \iff aX_n +b \rightarrow aX+b$, by the lemma – Bilbo Mar 8 '15 at 13:11
• @Bilbo Yes, it is still true, but the proof is more difficult. See this question: math.stackexchange.com/q/317706 – saz Mar 8 '15 at 13:21
• @Bilbo Yes, I was referring to "coinciding". "$c_n X_n'$ coincides wich $c_n X_n$ in distribution" means simply that for each fixed $n \in \mathbb{N}$, we have $c_n X_n = c_n X_n'$ in distribution. – saz Mar 8 '15 at 14:06
• @Bilbo Well, I guess you mean the right thing: $$P_n(c_n X_n \in B) = P_n (c_n X_n' \in B)$$ for any Borel set $B$. (It does not suffice to have equality for $B$ of the form $B=\{x\}$.) – saz Mar 8 '15 at 14:13