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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.

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  • $\begingroup$ 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

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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.

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