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Let $x_n$ an increasing convergent sequence (that converges to $x$), and $y_n$ a decreasing sequence that converges to $y$, how can I prove that: $$ \lim_{n -> \infty} P([x_n,y_n])=P([x,y]) $$

and this make me wonder if it is necessary that the sequences $x_n$ and $y_n$ converges to $x$ and $y$ respectively ?

I am trying to solve this statement of sequences in probability but I can solve it, if someone could help me please.

Thanks for your time and help.

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  • $\begingroup$ What do you mean by "increasing" and "decreasing" here? That e.g. $X_n\leqslant X_{n+1}$ almost surely (and similarly $Y_n\geqslant Y_{n+1}$ almost surely)? Moreover, in what sense do these sequences of random variables converge? In probability? In distribution? Almost surely? $\endgroup$ – Math1000 Feb 20 '17 at 6:32
  • $\begingroup$ @Math1000 Real numbers, not random variables. Yes, bad title... $\endgroup$ – Did Feb 20 '17 at 7:59
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First solution:

Let $I_n:=[x_n,y_n]$. Since $x_n$ is increasing and $y_n$ is decreasing we have: $I_1\subseteq I_2\subseteq\cdots$, that is, $(I_n)$ is a decreasing sequence of events, so it satisfies $$\lim_{n\rightarrow \infty}(I_n)=\cap_{n=1}^\infty I_n=\left[\sup_{n\in \mathbb{N}}x_n,\inf_{n\in\mathbb{N}}y_n\right]=[x,y].$$ Since $P$ is a probability, it follows that $P$ is continuous from above, so $$\lim P(I_n)=P(\lim (I_n))=P([x,y]).$$

Second solution:

If you're not familiar with these concepts, you can solve it by using the following property:

$If A_1\supseteq A_2\supseteq A_3\supseteq \cdots$ is a countable sequence of events such that $\cap A_n=\emptyset$, then $$\lim_{n\rightarrow \infty} P(A_n)=0$$.

So, $$\lim_{n\rightarrow \infty} P([x_n,y_n])=\lim_{n\rightarrow \infty} \{P([x_n,x))+P([x,y])+P((y,y_n])\}$$

First, notice that for every $n\in \mathbb{N}$ $[x_n,x)\supseteq [x_{n+1},x)$ since $x_n$ is increasing. Similarly, $(y,y_n]\supseteq (y,y_{n+1}]$. Also, since $x=\lim x_n=\sup x_n$ and $y=\lim y_n=\inf y_n$, we get $$\bigcap_{n\in \mathbb{N}}[x_n,x)=\bigcap_{n\in\mathbb{N}} (y,y_n]=\emptyset.$$ The result follows using the property on the sets $[x_n,x)$ and $(y,y_n]$. We have now $$\lim_{n\rightarrow \infty} P([x_n,x))=\lim_{n\rightarrow \infty} P(y,y_n]=0.$$

With this in mind, the previous equality becomes $$\lim_{n\rightarrow \infty} P[x_n,y_n]=0+P([x,y])+0=P([x,y]),$$ the desired result.

If you drop the hypothesis of convergence but keep monotonicity,you will have unbounded sequences, so instead of $[x,y]$ you'll get the emptyset since $[x_n,y_n]$ will degenerate for large $n$.

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  • $\begingroup$ this is true if i change that xn now is decreasing and yn increasing? $\endgroup$ – Rachel Feb 21 '17 at 21:26
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    $\begingroup$ It is not generally true, you may get $[x,y]$, $(x,y)$, $[x,y)$ or $(x,y]$ $\endgroup$ – Knull Feb 24 '17 at 8:09

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