Let $T$ be a random variable with $Exp(\lambda)$ distribution for $\lambda >0$.

I want to show that $T < \infty$ a.s.

In order to do that, we need to show that $P(T < \infty) =1$.

So I thought of different ways of proving that but as I am not that good with limits and continuous distributions, I am a little bit confused.

1) So my first idea was to show that $$P(T \leq t)= 1- e^{- \lambda t} \underset{t \rightarrow \infty}{\rightarrow} 1,$$ but I am not sure whether this implies that $P(T \leq \infty)=1$ or $P(T < \infty)=1$. If only the first one, then this property is trivial.

2) My second idea was to work with $P(T > t)=e^{-\lambda t}$. If for every fixed $t$ we have $P(T > t) = e^{-\lambda t}$, then when $t \rightarrow \infty$ we have $$P(T = \infty) = \lim_{t \rightarrow \infty} P(T > t)= \lim_{t \rightarrow \infty} e^{-\lambda t} = 0,$$ but again I am not sure (and it is somehow basically the same as in 1)).

So can you help me by telling me what $P(T=\infty)$ means expressed in terms of $P(T\leq t)$?

My guess?

So is $\lim_{t \rightarrow \infty} P(T > t) = P(T = \infty)$ and $\lim_{t \rightarrow \infty} P(T \leq t) = P(T < \infty)$? Is that true? Are these also the same events?

Thank you very much and please excuse the stupid questions.

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Note that we have $$ \bigcup_{n=0}^\infty \{T \le n\} = \{T < \infty\}$$ as $T(\omega) < \infty$ for $\omega \in \Omega$ iff $T(\omega) \le n$ for some $n \in \def\N{\mathbf N}\N$. Now, by the so called continuity of the measure, we have $$ P(\{T \le n\}) \to P(\{T < \infty\})$$ To prove that, write \begin{align*} P(T < \infty) &= P\left(\{T \le 0\} \cup \bigcup_{n \in \N} (\{T \le n+1\} - \{T \le n\})\right)\\ &= P(T\le 0) + \sum_{n\in\N} P(\{T \le n+1\} - \{T \le n\})\\ &= P(T\le 0) + \lim_{n \to \infty}\sum_{k=0}^{n-1} P(T \le k+1) - P(T\le k)\\ &= \lim_{n\to \infty} P(T \le n) \end{align*} Taking complements, we have $P(T > n) \to P(T = \infty)$.

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  • $\begingroup$ Wow. Thank you very much. This was extremely helpful and exactly what I am was looking for. Moreover, it is written in such a way that I am able to understand these things (as a non-mathematician). Thank you very much!! That helped a lot! $\endgroup$ – user136457 Mar 27 '15 at 9:42
  • $\begingroup$ I read your answer and was really convinced. But now I am asking myself: Does this really work for continuous random variables? $\endgroup$ – user136457 Mar 27 '15 at 10:08
  • 2
    $\begingroup$ Yes, it does. The argument does not make any assumptions on $T$ besides being a random variable. $\endgroup$ – martini Mar 27 '15 at 11:30
  • $\begingroup$ Could I ask you what it means to have $\{T \leq n+1\} - \{T \leq n\}$ in the first line above? It seems that each is an event so I am not sure what you mean by the difference of events. $\endgroup$ – user321627 Oct 27 '16 at 12:09
  • $\begingroup$ @user321627 It is the usual set difference, $$ A - B = \{x \in A : x \not\in B\}$$ $\endgroup$ – martini Oct 27 '16 at 13:10

If $P\{X=+\infty\}>0$ then $P\{X<+\infty\}<1$.

Then $\lim_{t\rightarrow\infty}F_X(t)=1$ implies that for some $t\in \mathbb R$ we have $P\{X\leq t\} >P\{X<+\infty\}$ contradicting that $\{X\leq t\}\subseteq\{X<+\infty\}$.

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