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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!

Bilbo
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1 Answers1

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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?

saz
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  • 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 08 '15 at 13:01
  • @Bilbo Well, yes, but we are interested in $\mathbb{E}g(aX_n+b)$, not in $\mathbb{E}(aX_n+b)$. – saz Mar 08 '15 at 13:03
  • Oh, right. Won't it help to notice that composition of two continuous bounded functions is continuous and bounded? – Bilbo Mar 08 '15 at 13:07
  • And apply the lemma to $h(x)=ax+b, f(x)=g(h(x))$? – Bilbo Mar 08 '15 at 13:08
  • @Bilbo Yes, this tells us that $f(x) := g(ax+b)$ is bounded+continuous. So how can we deduce $(1)$ from $(2)$ and the fact that $X_n \to X$ in distribution? – saz Mar 08 '15 at 13:08
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    $X_n \rightarrow X$ 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 08 '15 at 13:11
  • @Bilbo Yes, exactly. – saz Mar 08 '15 at 13:13
  • Great! Thank you! Could you tell me if this result would still be true if instead of constant $a, b$ we took $a_n \rightarrow a, \ \ b_n \ \rightarrow b$? – Bilbo Mar 08 '15 at 13:14
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    @Bilbo Yes, it is still true, but the proof is more difficult. See this question: http://math.stackexchange.com/q/317706/ – saz Mar 08 '15 at 13:21
  • So thanks to Skorokhod's theorem we get new random variables $X_n', X'$ such that $X_n(\omega) \rightarrow X(\omega) a.s.$ and $X_n, X_n'$ have the same distribution $P_n$, and $X, X'$ have distr. $P$. $$|c_nX_n' - cX'| \le |c_nX_n' - cX_n'+cX_n'-cX'|$$ $$\le |c_n - c||X_n'| + |c||X_n'-X'|$$ and as $n \rightarrow \infty$, $$P_n(\omega : |c_n - c||X_n'| + |c||X_n'-X'| > \varepsilon ) \rightarrow 0$$ – Bilbo Mar 08 '15 at 13:34
  • Could you explain to me what it means that two sequences coincide in distribution? Does it mean that the measure of the set on which the disagree tends to zero as $n \rightarrow \infty$? If so, why do the coincide? – Bilbo Mar 08 '15 at 13:38
  • @Bilbo I don't know, this is not commonly used in probability theory ... where did you read this? – saz Mar 08 '15 at 13:52
  • Ok, I see. Could you tell me how to justify the fact that if $c_n \rightarrow c$, then $ c_nX'_n\to cX'$ – Bilbo Mar 08 '15 at 13:57
  • Or maybe you're referrimg to coinciding. I read this in the answer to your question: "But each $c_nX'_n$ coincides with $c_nX_n$ in distribution and $cX'$ coincides with $cX$ in distribution, hence $c_nX_n\to cX$ in distribution." – Bilbo Mar 08 '15 at 13:58
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    @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 08 '15 at 14:06
  • So $P_n(c_nX_n = x) = P_n(c_nX_n' = x)$? – Bilbo Mar 08 '15 at 14:11
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    @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 08 '15 at 14:13
  • Yes, of course. Thank you very, very much! – Bilbo Mar 08 '15 at 14:16