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I am having trouble proving this theorem.

I know that I am given $$\lim_{n \rightarrow \infty }Pr[X_n<x] = Pr[X<x]$$ and $$\lim_{n \rightarrow \infty }Pr[|Y_n| > \epsilon] = 0, \quad\forall\epsilon>0$$ where I have to show that $$\lim_{n \rightarrow \infty }Pr[X_n+Y_n<t]=Pr[X<t] \qquad(1)$$

Editted:

I know that I have to use the squeeze theorem, but I am not too strong in limit problems...

I cannot use Slutsky's Theorem because I am actually trying to prove it.

hyg17
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4 Answers4

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Hints: use the following: $\{X_n+Y_n \leq t\} \subset \{X_n \leq t+\epsilon\} \cup \{|Y_n| >\epsilon\}$ and $\{X_n \leq t -\epsilon\} \subset \{X_n+Y_n \leq t\} \cup \{|Y_n| >\epsilon\}$

Kavi Rama Murthy
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  • I am sorry I do not understand how the statements are true... – hyg17 Mar 18 '19 at 08:40
  • @hyg17 Suppose $X_n+Y_n \leq t$. Either $|Y_n| >\epsilon$ or $|Y_n| \leq \epsilon$. In the second case $X_n =(X_n+Y_n)-Y_n \leq t+\epsilon$ because $-Y_n \leq |Y_n| \leq \epsilon$. This proves the first part in my answer and the second part is similar. – Kavi Rama Murthy Mar 18 '19 at 08:48
  • Thank you, I will try to work on it. – hyg17 Mar 18 '19 at 08:54
  • I understand that $X_n+Y_n \le t$ then $X_n \le t+ \epsilon$ is true, but why would the set of ${X_n+Y_n \le t}$ be a subset of $X_n \le t + \epsilon \cup |Y_n| \le \epsilon$? I am sorry I am not good at sets. – hyg17 Mar 18 '19 at 20:37
  • To show that $A \subset B\cup C$ take $a \in A$. Consider the cases $a \in C$ and $a \notin C$. If the first case we already know that $a \in B \cup C$. In the second case you try to show that $a \in B$ so that $a \in B \cup C$. This is what I have done. – Kavi Rama Murthy Mar 18 '19 at 23:15
  • Hello, Mr. Murthy. Would it be okay if I ask you if I did some of it right? – hyg17 Mar 19 '19 at 03:02
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You can use Slutsky theorem, i.e., as $Y_n$ converges in probability to a constant $c=0$, then $$ X_n + Y_n \xrightarrow{D} X+c=X. $$

V. Vancak
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You should have the following for $lim_{n\rightarrow \infty}$

$X = Pr\{X_n\} \leq Pr\{X_n + Y_{n}\} \leq Pr\{X_{n}\} + Pr \{Y_{n}\} = X +0 = X $

so take the extreme and is valid

$X \leq Pr\{X_n + Y_{n}\} \leq X $

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Here's another way for the record (uses too much theory, because - as noted above - problem can be done directly):

$$E[e^{it (X_n + Y_n)}] = E[e^{it X_n}] + \underset{(*)}{\underbrace{E[e^{itX_n}(e^{itY_n} -1)]}}.$$

$|(*)|\le E[ |e^{itY_n} -1|] \to 0$ by dominated convergence (note that convergence in probability is sufficient for the conclusion of DCT to hold).

Fnacool
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