# Types of convergence for sequence of random variables

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 $p - \frac 1 2$.
• 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 $\not\to 0$? – Taufi May 1 '18 at 12:53
• 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. – Mike Earnest May 1 '18 at 13:07