I'm just getting started with convergence of random variables and while trying to understand it I came across some very confusing notation. It's not the definitions themselves, they make sense, but take a look at this problem:

Let $\lambda = \{w_k|k\in\mathbb{N}\}$ be a set of elementary outcomes of some experiment (the problem is not in English, not sure whether it's 'set of elementary outcomes' or just 'set of outcomes' in English). And let $P(w_k) = \frac{6}{\pi^2k^2}$. Let $(X_n)$ be a sequence of random variables given as $$X_n(w_k) = \begin{cases}n, & k=n \\ 0, &k\neq n \end{cases}$$ Let $(Y_n)$ be a sequence of random variables defined as $$Y_n = 1-\frac{X_n}{n^2}$$ Examine all four types of convergence of the sequences $(X_n)$ and $(Y_n)$.

Okay. For the sake of completeness I wrote down the entire problem, but now let's just disregard the second part (the one where they introduce $Y$). This is my first time dealing with convergence of random variables but I've done a lot of problems with random variables in general and I've never seen a notation such as $$X_n(w_k)$$ Disregarding the whole concept of convergence, I've never seen a notation in the form of random_variable(outcome).

So instead of trying to understand what it meant I went on to look at the solution of the problem and try to maybe catch it in there somehow. The very first line in the solution states:

$$X_n : \bigg(\begin{matrix} 0 & n \\ 1 - \frac{6}{\pi^2n^2} & \frac{6}{\pi^2n^2}\end{matrix} \bigg)$$

I suppose that in some way this could have meant 'the $n$th random variable in the given sequence of random variables can take on the values of either $0$ or $n$' and somehow it connects to the fact that $P(w_k) = \frac{6}{\pi^2k^2}$ because of that $k=n$ thing or something?

The thing is, I'm so confused about this that I don't even know how to explain my extremely informal understanding of it.

So can anyone help me out here? Not looking for the solution to the problem, just a little clarity about notation.



I'm not sure whether this is allowed but I'd like to add just one more question here (but it's kind of similar to the original, still caused by some confusion of what is meant by a random variable at times). From your responses (one in the comments one in the answers) and by reading up on formal definitions of random variables I've managed to understand what is meant by the notation that I asked about. However, this is written below the line that was unclear to me:

$$\bigg(\begin{matrix}0 & n \\ 1-\frac{6}{\pi^2n^2} & \frac{6}{\pi^2n^2} \end{matrix}\bigg) \to_{n->\infty} \bigg(\begin{matrix} 0 & n \\ 1 & 0 \end{matrix}\bigg)$$

That seems clear enough but in the next line they wrote:

$$X_n \to X = 0$$

and then afterwards when checking, for instance the convergence in probability they use:

$$P\{|X_n - X| \geq \epsilon\} = P\{|X_n - 0| \geq \epsilon\} = P\{|X_n| \geq \epsilon\} = ...$$

The $X_n \to X = 0$ is the thing that I don't get, and then subsequently using $X = 0$ as if $X$ was a constant rather than a random variable.

  • $\begingroup$ I think sample space is the best term $\endgroup$ Commented Jan 12, 2020 at 16:26
  • $\begingroup$ By the way about the notation: Some great book (I forgot the name) on probability theory wrote that "random variables are neither random nor variables". In fact, a random variable is defined as a function from the sample space to some measurable space, usually the real numbers. So writing $X(w_k)$ is a good notation $\endgroup$ Commented Jan 12, 2020 at 16:29
  • $\begingroup$ @MaximilianJanisch Yeah, but what does it mean here? It occurred to me that it may mean 'the probability that $w_k$ will happen'. But since it can be equal to $n$ (from that piecewise definition), which can surely be a lot greater than $1$, I'm guessing it's not it? $\endgroup$
    – Koy
    Commented Jan 12, 2020 at 16:31
  • 1
    $\begingroup$ No no no, it means that the function $X_n$ assigns the outcome $w_n$ to the number $n$. (Have a look at these examples: en.wikipedia.org/wiki/Random_variable#Examples). For example, formally we have here for $n\in\mathbb N$ $$\mathsf P(X_n=n)=\mathsf P(\{w_n\})=\frac{6}{\pi^2 n^2}$$ and $$\mathsf P(X_n=0)=1-\frac{6}{\pi^2 n^2},$$ so each $X_n$ is a Bernoulli random variable $\endgroup$ Commented Jan 12, 2020 at 16:34
  • $\begingroup$ You can think of that number as a "score", just like throwing heads on a coin might give you score $+1$ and scoring tails might give you score $-1$ $\endgroup$ Commented Jan 12, 2020 at 16:38

1 Answer 1


Each outcome will correspond to a different sequence $\langle X_1,X_2,X_3,\ldots\rangle$. Based on the definition of the $X_i,$ if $\omega_k$ occurs then the sequence is $\langle 0,0,\ldots,0, k,0,\ldots\rangle$ where the $k$ occurs in the $k$-th spot.

As mentioned in the solution and in the comments, since outcome $\omega_n$ has probability $C/n^2,$ and $X_n=n$ iff that outcome occurs and is zero otherwise, the marginal distribution of $X_n$ is $P(X_n=n)=C/n^2$ and $P(X_n=0)=1-C/n^2.$

However, don't be mislead into thinking that the $\langle X_1,X_2,\ldots\rangle$ is an independent sequence with those marginal distributions. It is not: if we know $X_i=i$ for some $i$, then we know all the rest are zero.


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