Prove $(X_n)n≥1$, $X_n \sim Exp(n)$ converges in distribution to $0$

Question

Prove $(X_n)n≥1$, $X_n \sim Exp(n)$ converges in distribution to $0$ using the following property:

$X_n \to X$ in distribution $\iff F_{X_n}(x)\to F_X(x) \forall$ continuity points of $F_X$

My attempt

If $x<0, F_{X_n}(x)=0 \to 0,$ as $n\to +\infty$

If $x\ge0, F_{X_n}(x)=1-e^{-nx} \to 1$ as $n\to +\infty$

From these results how can I identify I am getting the cumulative distribution function of a constant 0? It looks that I got a step function at 0. What is the cumulative distribution function of a constat random variable to start with? It doesn't even have much of "random".

Note: I know I can use a stronger convergence to imply the result but the exercise asks to use this property explicitly.

Answer 1

If $X$ is the constant random variable $c$ then $F_X(x)=P(X\leq x)=1$ if $ x \geq c$ and $0$ if $x<c$ (simply because the empty set has probability $0$ and the sample space $\Omega$ has probability $1$). Taking $c=0$ we see that $F_{X_n}(x) \to F_X(x)$ for $x>0$ as well as $x<0$. Convergence in distribution requires convergence only at points of continuity of the limit function, so convergence at $0$ is not required.