# Convergence in probability and almost surely - examples

Define CDF of random variable $$X_n, n \in \mathbb{N}$$ as: $$F_n(x)= \left\{ \begin{array}{ll} 1-xe^{-nx} & \textrm{x \in [ \frac{1}{n}, \infty)}\\ 0 & \textrm{x \in (-\infty, \frac{1}{n})}\\ \end{array} \right.$$

1. What is the limit of $$\{X_n\}$$ (convergence in probability)?
2. Is $$X_n$$ convergent with probability 1 (almost surely)?

My observations:

1. As $$n \to \infty$$ the CDF looks more and more like a straight line at 1 (starting closer and closer to 0). That intuitively tells me the limit of $${X_n}$$ could be 0 - there will be a tiny interval $$(\frac{1}{n},\frac{1}{n}+\varepsilon$$) having almost all the mass of our distribution. This means that for arbitrary $$\varepsilon > 0, \mathbb{P}(|X_n| \geq \varepsilon)$$ will go to 0 as n tends to infinity. If it is true, how would one prove it formally?
2. This does not converge almost surely though, because when $$x \in (-\infty, \frac{1}{n}]$$(note the closed interval at $$\frac{1}{n}$$) then CDF is 0, which means the limit? point $$0$$ will never actually obtain any mass, thus $$\mathbb{P}(\lim_n X_n = 0) \neq 1$$. Again, if this is correct, how would one prove this rigoriously? This makes no sense after the edit.

Edit: As pointed out in comments, this function has to be right-continuous to be a CDF, thus the point $${\frac{1}{n}}$$ belongs to different interval after the edit.

• If CDF of $X$ is defined as $F_X(x)=\mathbb P(X\leq x)$, then given function is not CDF since it is left-continuous, not right-continuous.
– NCh
Jan 25, 2020 at 17:05
• Uh, this is a question from an old test at my university, which I have found somewhere. I just double checked and it clearly said $1-xe^{-nx}$ when $x > \frac{1}{n}$ and $0$ when $x \leq \frac{1}{n}$. Perhaps it should have been $1-xe^{-nx}, x \geq \frac{1}{n}$ and 0 otherwise as you say. Jan 25, 2020 at 17:10
• Yes, for sure. Not the matter for convergence.
– NCh
Jan 25, 2020 at 17:11

For any $$\varepsilon>0$$, for $$n>\frac1\varepsilon$$, $$F_{X_n}(\varepsilon)=1-\varepsilon e^{-n\varepsilon}\to 1$$ as $$n$$ further increases to infinity.
So $$\mathbb P(|X_n|>\varepsilon)=1-F_{X_n}(\varepsilon)=\varepsilon e^{-n \varepsilon}\to 0$$ as $$n\to\infty$$. There is nothing to proof and your explanation is quite strong.
Hint: for a.s. convergence use Borel-Cantelli lemma. Look at the events $$E_n=\{|X_n|>\varepsilon\}$$ and check whether sum of its probabilities is finite.