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Let $Y$ be a continuous random variable with density function $f_Y(y)=e^{-y}, y > 0$. Consider the sequence $\{X_n\}$, given by $X_n=e^nI_{\{Y>n\}}, n =1,2,\cdots$

How can I show that $\{X_n\} \to 0$ in probability?

My work:

So, I want to show that $P(|X_n-0|>\epsilon) \to 0$ as $n \to \infty$.

$P(|X_n| > \epsilon)=P(X_n > \epsilon)=1-P(X_n \le \epsilon)=1-\int_0^\epsilon\Sigma_1^Y(e^n)dx$

However, I do not think my last equality makes much sense. How would you proceed?

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  • $\begingroup$ What's the $I$ here? $\endgroup$ – jkm Nov 23 '19 at 20:26
  • $\begingroup$ @jkm the indicator function $\endgroup$ – Edison Nov 23 '19 at 20:42
  • $\begingroup$ Draw graphs of some of the $X_n.$ The rest will be obvious. $\endgroup$ – whuber♦ Nov 23 '19 at 22:53
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Consider the following inequality for $\varepsilon \leq 1$

$$P( |X_n| > \varepsilon ) = P(e^n I_{\{Y>n\}}>\varepsilon) \leq P( I_{\{Y>n\}} > \varepsilon) = P(Y > n) = e^{-n} \to 0.$$

Actually we do not need to know the distribution of $Y$. We can use the cumulative function properties $$ P( Y > n) = 1 - F_Y(n) \underset{n\to \infty}{\to} 1 -1 = 0$$

For $\varepsilon > 1$ we get $$ P( |X_n| > \varepsilon) \leq P(|X_n| > 1) \to 0$$

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  • $\begingroup$ why the downvote? $\endgroup$ – Manuel Nov 23 '19 at 20:56
  • $\begingroup$ I'll equal it out for you! I like this idea, but how can you justify the inequality and the equality thereafter? $\endgroup$ – Edison Nov 23 '19 at 21:05
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    $\begingroup$ If $n \geq 1$ then $ e^n I_{Y > n} > I_{Y > n}$. Also An indicator is $1$ or $0$ so the probability of being greater than $\varepsilon$ is equal to the probability ob being 1. $\endgroup$ – Manuel Nov 23 '19 at 21:08
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    $\begingroup$ (+1) for the inequality in the first line. But note that you need to also consider the case $\epsilon\geq 1$ for completeness. $\endgroup$ – gunes Nov 23 '19 at 21:20
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    $\begingroup$ The last line "$X_n$ are positive with probability one": that's not correct, isn't it? $\endgroup$ – StijnDeVuyst Nov 23 '19 at 21:32
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First, $\text{Prob}[Y>n]=e^{-n}$ since $Y$ has an exponential distribution. Therefore, $X_n$ can have only two values, namely $e^n$ with probability $e^{-n}$ and $0$ with probability $1-e^{-n}$. Let's fix some positive value $\epsilon$ and call $n'$ the smallest nonnegative integer bigger than $\ln \epsilon$. Then, for any $n \geqslant n'$, $\text{Prob}[X_n>\epsilon]=e^{-n}$, which clearly goes to 0 for large $n$.

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  • $\begingroup$ I have several questions for you. (1) There are two values for $X_n$ due to the indicator function, correct? (2) Why isn't $P(X_n > \epsilon) = e^{-n} + (1 - e^{-n})$? $\endgroup$ – Edison Nov 23 '19 at 21:09
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    $\begingroup$ (1) correct (2) $\epsilon$ can be small, but never equal to 0. Therefore $\text{Prob}[X_n>\epsilon]=\text{Prob}[X_n=e^n]=e^{-n}$. $\endgroup$ – StijnDeVuyst Nov 23 '19 at 21:17
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$X_n$ can have two values (i.e. either $0$ or $e^n$) with $p=P(X=e^n)=e^{-n}$. For $\epsilon\geq e^{n}$, $P(X_n>\epsilon)$ is always $0$. For $\epsilon<e^n$, $P(X_n>\epsilon)=p=e^{-n}$, and this goes to $0$ as $n$ goes to infinity.

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