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.

  • $\begingroup$ 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. $\endgroup$
    – NCh
    Jan 25, 2020 at 17:05
  • $\begingroup$ 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. $\endgroup$
    – blahblah
    Jan 25, 2020 at 17:10
  • $\begingroup$ Yes, for sure. Not the matter for convergence. $\endgroup$
    – NCh
    Jan 25, 2020 at 17:11

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.