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Let $Y_n$ be a sequence of real random variables converging in distribution to a random variable $Y$ and let $M \in \mathbb{R}$ be a fixed real number. I would like to prove that $P(Y \geq M) \geq \liminf_{n\rightarrow \infty} P(Y_n \geq M)$. This seems like it should be very easy to prove since if $x_n \rightarrow x$ then $x \geq \liminf_{n\rightarrow \infty} x_n$ for any convergent sequence of real numbers.

But since we only have convergence in distribution and the limit is outside the measure $P$ I am having some trouble formalizing everything. How can one prove this?

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  • $\begingroup$ Fatou's Lemma?? $\endgroup$
    – Captuna
    Feb 12 at 3:21
  • $\begingroup$ @Captuna That was what I used in one of my attempts. I used Fatou's lemma and indicator functions but I couldn't get it to work since we only have convergence in distribution. $\endgroup$
    – CBBAM
    Feb 12 at 3:23
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    $\begingroup$ Which definition of convergence in distribution are you using? $\endgroup$ Feb 12 at 4:09
  • $\begingroup$ @BrianMoehring I am using the definition where $X_n \rightarrow X$ in distribution if $\mu_n \rightarrow \mu$ weakly where $\mu_n$ is the distribution of $X_n$ and $\mu$ the distribution of $X$. $\endgroup$
    – CBBAM
    Feb 12 at 6:12
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    $\begingroup$ @CBBAM You should add what definition of "weakly" are you using. That is what Brian is trying to ask. What you are asking follows trivially from one of the definitions of "weak" convergence of probability distributions on $\Bbb{R}$ (See my answer for more details). Since you are having trouble proving this, I would guess that you are using the definition that $\mu_{n}\to \mu$ if for all continuous bounded functions $f$, you have $\int_{\Bbb{R}}f\,d\mu_{n}\to\int_{\Bbb{R}} f\,d\mu$. $\endgroup$ Feb 12 at 10:25

2 Answers 2

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Let $0<\epsilon <M$ be such that $M-\epsilon$ is continuity point of $F_Y$. Then $\lim \inf P(Y_n \ge M)\le \lim \inf P(Y_n >M-\epsilon)=\lim P(Y_n >M-\epsilon)=P(Y>M-\epsilon)$.We can find a sequence $(\epsilon_i)$ decreasing to $0$ such that $M-\epsilon_i$ is continuity point of $F_Y$ for each $i$. We get $\lim \inf P(Y_n \ge M)\le P(Y>M-\epsilon_i)$ for each $i$ and letting $i \to \infty$ finishes the proof.

Second proof: Let $f(x)=1$ for $x \ge M, 0$ for $x \leq M-\epsilon$ and let $f(x)=\frac{x-M+\epsilon} {\epsilon}$ for $M-\epsilon \le x \le M$. Then $f$ is a bounded continuous function and $P(Y_n \geq M) \le \int fd\mu_n \to \int fd\mu \le P(Y >M-\epsilon)$ for every $\epsilon >0$. Can you finish the proof?

Note: The second proof shows that $\lim\sup P(Y_n \ge M) \le P(Y\ge M)$.

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  • $\begingroup$ How did you conclude the equality and inequality $$\lim \inf P(Y_n >M-\epsilon)=\lim P(Y_n >M-\epsilon)=P(Y>M-\epsilon)?$$ $\endgroup$
    – CBBAM
    Feb 12 at 6:17
  • $\begingroup$ Since $M-\epsilon $ is a continuity point of $F_Y$ and $Y_n \to Y$ in distribution it follows that $\lim P(Y_n>M-\epsilon)$ exists and equals $P(Y>M-\epsilon)$. @CBBAM $\endgroup$ Feb 12 at 6:19
  • $\begingroup$ Sorry I am still getting used to the concept of convergence in distribution, can you explain how this implies the equality? $\endgroup$
    – CBBAM
    Feb 12 at 6:28
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    $\begingroup$ Nice answer (+1). A stronger result also holds. i.e. we actually have $\lim\sup_{n\to\infty}P(Y_{n}\geq M)\leq P(Y\geq M)$ $\endgroup$ Feb 12 at 10:27
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    $\begingroup$ @Mr.GandalfSauron Thanks! My second proof also gives the inequality with $\lim \sup$. $\endgroup$ Feb 12 at 11:17
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Actually you have something stronger, in that $\lim\sup_{n\to\infty}P(Y_{n}\geq M)\leq P(Y\geq M)$. In fact, you have for all closed sets $K$ that $\lim\sup_{n\to\infty}P(Y_{n}\in K)\leq P(Y\in K)$ and this is both necessary and sufficient for convergence in distribution.

Here's an easy way to prove this by extending your idea that it would have been easier if we had almost sure convergence .

Use Skorokhod's Representation Theorem (Also here in Durrett Theorem $3.2.8$) to find a sequence $X_{n}$ and a random variable $X$ such that $X_{n}\xrightarrow{a.s.}X$ and $X_{n}=Y_{n}$ in distribuiton and $X=Y$ in distribution.

Now, notice that $\lim\inf\mathbf{1}_{\{X_{n}<M\}}\geq \mathbf{1}_{\{X< M\}}$

i.e. if for an $\omega\in\Omega$ (or more precisely, an $\omega$ from the almost sure set in which convergence holds pointwise) , if $X(\omega)<M$, then $X_{n}(\omega)<M$ for all large enough $n$ by basic properties of real sequences and hence $\lim\inf\mathbf{1}_{\{X_{n}<M\}}=1=\mathbf{1}_{\{X<M\}}$.

Now you have by the above and Fatou's lemma that

\begin{align}E(\mathbf{1}_{\{X<M\}})\leq E(\lim\inf\mathbf{1}_{\{X_{n}<M\}})\stackrel{Fatou}{\leq}& \lim\inf E(\mathbf{1}_{\{X_{n}<M\}})\\\\\implies P(X<M)\leq \lim\inf P(X_{n}<M)\end{align}

Now just use the fact that $P(X<M)=1-P(X\geq M)$ and $\lim\inf (-x_{n})=-\lim\sup x_{n}$ to get, $$\lim\sup P(X_{n}\geq M)\leq P(X\geq M)$$

and now, your result trivially follows as $\lim\inf P(X_{n}\geq M)\leq \lim\sup P(X_{n}\geq M)$ and you can just replace $X_{n}$ by $Y_{n}$ and $X$ by $Y$ as you have equality in distribution.

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    $\begingroup$ Thank you, this is very neat! I was not aware of the Skorokhod's representation theorem. $\endgroup$
    – CBBAM
    Feb 12 at 17:29

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