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?
 A: 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 
$$ P( |X_n| > \varepsilon) \leq P(|X_n| > 1) \to 0$$
A: 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 $n$.
A: $X_n$ 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.
