Currently I am taking a course in probability and encountered the following exercise that deals with different types of convergence of random variables.

The question is as follows. Let $(\Omega, \mathscr{F}, P) = ([0, 1), \mathscr{B}_{[0, 1)}, \lambda_{[0, 1)})$ be a measure space and $X_n = \sqrt{n}\mathbb{1}_{(\frac{1}{n}, \frac{2}{n})}$ be a sequence of random variables. Examine whether the sequence converges in probability, almost surely and convergence in $\mathscr{L}^p$ respectively.

What I know are the relevant definitions.

$(X_n)_{n\geq 1}$ converges to $X$ in probability $:\Leftrightarrow$ $\forall \epsilon > 0$: $\lim\limits_{n\to \infty} P(\vert X_n - X \vert \geq \epsilon) = 0$

$(X_n)_{n\geq 1}$ converges to $X$ almost surely $:\Leftrightarrow$ $P(\lim\limits_{n\to\infty} X_n = X) = 1$

$(X_n)_{n\geq 1}$ converges to $X$ in $\mathscr{L}^p$ $:\Leftrightarrow$ $\lim\limits_{n\to\infty} \mathbb{E}[\vert X_n - X\vert^p] = 0 $

I also know that I can prove convergence a.s. or convergence in $\mathscr{L}^p$ then convergence in probability follows.

What I don't know is how to apply all these definitions practically. I was only given the exercise and we did not have any examples in the lecture.

What I tried is the following. First, I have to figure out the limit $X$. As the set of the indicator function, ($\frac{1}{n}, \frac{2}{n}$), encompasses a smaller and smaller interval with $n \to \infty$ and $\lim\limits_{n \to \infty} (\frac{1}{n}, \frac{2}{n}) = \{0\}$ I assumed $X = 0$. If so, then for convergence in probability we would have

$$\lim\limits_{n \to \infty} P(\vert X_n - X\vert \geq \epsilon) = \lim\limits_{n \to \infty} P(\vert X_n \vert \geq \epsilon) = ?$$

I know what the $X_n$ are but I don't see what probability distribution it is assigned?

Similarly, for convergence a.s. I would, according to the definition, need $\lim X_n = \lim\limits_{n\to\infty} \sqrt{n}\mathbb{1}_{(\frac{1}{n}, \frac{2}{n})}$ but this seems to become infinitely large in $\{0\}$ and zero everywhere else. But, as $\{0\}$ is a set of measure zero, I would conclude convergence almost surely as this excludes sets of measure zero. Correct? If so, then I could conclude convergence in probability as well (though, I would like to see how to prove this with the definition).

Finally, for convergence to 0 in $\mathscr{L}^p$ I consider

$$\lim\limits_{n\to \infty} \mathbb{E}[\vert X_n \vert^p] = \int (\sqrt{n}\mathbb{1}_{(\frac{1}{n}, \frac{2}{n})})^p dP = \int\limits_{(\frac{1}{n}, \frac{2}{n})} n^{\frac{1}{2} + p} dP = n^{\frac{1}{2} + p}\cdot\frac{1}{n} = n^{p - \frac{1}{2}}$$

Hence, I would conclude that this converges for $n \to \infty$ for all $p \geq 1$ and therefore there is no convergence in $\mathscr{L}^p$. Is this correct?


The events your are looking at are subsets of $\Omega = [0,1]$ and the probability on $\Omega$ is the Lebesgue measure. You correctly guessed that the limit in probability is $X=0$, that is because the set $$ \{ | X_n - X | > \epsilon \} = \{ \omega \in [0,1] : \sqrt n 1_{(\frac 1 n , \frac 2 n )}(\omega) > \epsilon \} $$ is $(\textstyle \frac 1 n , \frac 2 n )$ for $\epsilon < \sqrt n$. The probability of this event is the length of the interval (since the probability on $\Omega$ is the Lebesgue measure), so it tends to zero, that is $X_n$ converges in probability to $0$. For the a.s. convergence, you need to consider the Lebesgue measure of the set $\{ \omega \in [0,1] : \sqrt n 1_{(\frac 1 n , \frac 2 n )}(\omega) \underset{n \to \infty}{\rightarrow} 0 \}$ (this set is actually the whole interval $[0,1]$ including $0$). For the convergence in $\mathcal L^p$, the idea is correct but the exponent should be $p/2$ instead of $p - \frac 1 2$.

  • $\begingroup$ Thanks for your reply. I got two questions to your answer: 1. $\Omega$ is actually defined as $\Omega = [0, 1)$, not $\Omega = [0, 1]$. Does that make a difference here? 2. Why is the set $\{\omega \in [0, 1] : \sqrt{n}1_{(\frac{1}{n}, \frac{2}{n})}(\omega) \to 0\}$ the whole interval including zero? Doesn't the indicator function collapse for $n \to \infty$ to the point $\{0\}$, being multiplied with a large $\sqrt{n}$ and hence $\not\to 0$? $\endgroup$ – Taufi May 1 '18 at 12:53
  • $\begingroup$ 1. It doesn't make a difference (think about it) 2. Look carefully at the definition of $X_n$, you have $X_n(0)=0$ for all $n$, so $X_n(0)\to 0$. The limit of the constant sequence of zeroes is zero. $\endgroup$ – Mike Earnest May 1 '18 at 13:07
  • $\begingroup$ Thanks, that clarified it. $\endgroup$ – Taufi May 1 '18 at 14:06

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.