# How can I show that $X_n=e^n I_{\{Y>n\}} \to 0$ in probability?

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?

• What's the $I$ here?
– jkm
Nov 23, 2019 at 20:26
• @jkm the indicator function Nov 23, 2019 at 20:42
• Draw graphs of some of the $X_n.$ The rest will be obvious.
– whuber
Nov 23, 2019 at 22:53

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

• why the downvote? Nov 23, 2019 at 20:56
• I'll equal it out for you! I like this idea, but how can you justify the inequality and the equality thereafter? Nov 23, 2019 at 21:05
• 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. Nov 23, 2019 at 21:08
• (+1) for the inequality in the first line. But note that you need to also consider the case $\epsilon\geq 1$ for completeness. Nov 23, 2019 at 21:20
• The last line "$X_n$ are positive with probability one": that's not correct, isn't it? Nov 23, 2019 at 21:32

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 $$n$$.

• I have several questions for you. (1) There are two values for $X_n$ due to the indicator function, correct? (2) Why isn't $P(X_n > \epsilon) = e^{-n} + (1 - e^{-n})$? Nov 23, 2019 at 21:09
• (1) correct (2) $\epsilon$ can be small, but never equal to 0. Therefore $\text{Prob}[X_n>\epsilon]=\text{Prob}[X_n=e^n]=e^{-n}$. Nov 23, 2019 at 21:17

$$X_n$$ 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 $$\epsilon, $$P(X_n>\epsilon)=p=e^{-n}$$, and this goes to $$0$$ as $$n$$ goes to infinity.