# Pointwise Convergence of random variables (Normal divided by n)

I am doing a course in mathetmatical statistics ans the lecturer defined pointwise convergence as follows:

Definition: A sequence $(X_{n})_{n \in \mathbb{N}}$ converges point wise to a random variable $X$ if:

$lim_{n \rightarrow \infty} X_{n}(\omega) = X(\omega),\forall \omega \in \Omega$

After defning this convergence he gave the following example, which was not so clear to me.

Example: Consider a sequence of i.i.d random variables $(X_{n})_{n \in \mathbb{N}}$ where $X_{n} \sim N(0,1)$ and let $Y_{n} = \frac{X_{n}}{n}$.

The lecturer argued that $Y_{n}$ does not converge to a limit pointwise since, for example, the sequence $(1,-2,3,-4,...)$ is a legitimate realization of $(X_{n})_{n \in \mathbb{N}}$ and the corresponding realization would be $(1,-1,1,-1,...)$.

Now I am confused because of this because I thought that if $(X_{n})_{n \in \mathbb{N}}$ is iid, it also means that for any $\omega \in \Omega$ we have $X_{n}(\omega) = X_{1}(\omega)$, for all $n \in \mathbb{N}$ , but this apparently not true (otherwise we would have pointwise convergence to $0$). So I guess my question is, can we say something about the random variable $X_{n}$ as a function from the probability space to the reals given that the sequence is iid? Can two random variables can have the same density function but as a function from the probability space to the reals be different functions? What's the connection of the density function to the random variable itself?

I hope my questions are clear. Would appreciate any feedback. Thank you!

## 1 Answer

You can't say a lot about the function $X:\Omega \rightarrow \mathbb{R}$ with only the distribution. For example: $\Omega=\{(0,0),(1,0),(0,1),(1,1)\}$, $P(A)=\frac{|A|}{4}$ and with the power set as the sigma Algebra. Then: $$X_1(\omega):=\begin{cases} 1 \text{ if } \omega=(1,a)\\ 0 \text{ if } \omega=(0,a) \end{cases}\\ X_2(\omega):=\begin{cases} 1 \text{ if } \omega=(a,1)\\ 0 \text{ if } \omega=(a,0) \end{cases}\\ a\in\{0,1\}$$ then they are iid ($X_i \sim Ber(1/2)$) but are of course quite different functions.

And that is the beauty of the construct of random variables. It models an Omega we know nothing of but a few results we can see (outcomes of random variables).

Another example:

$$X:=\begin{cases}-1 &\text{ with probability 1/2}\\ 1 &\text{with probability 1/2}\end{cases}$$ and now look at the distribution of $-X$ it is not independent but is identically distributed.

Addendum: If you have a random variable $X:\Omega \rightarrow \mathbb{R}$ and want to create an arbitrary number of independent random variables with the distribution of X, you can do so by looking at: $\Omega'=\Omega \times...\times\Omega$ and define $X_i:=X\circ\pi_i$ where $\pi_i(x_1,...,x_n)=x_i$ . So $X_i:\Omega'\rightarrow\mathbb{R}$ and as the sigma Algebra you define $\sigma(\{A_1\times ...\times A_n: A_i\in \text{"Old sigma Algebra"}\})$ and as the new meassure you define $P(A_1\times...\times A_n)=P(A_1)\cdot ...\cdot P(A_n)$. Since the Cartesian Product is by definition a Generator of the new sigma algebra (closed under intersections). You get that the meassure is well defined by those sets (Meassure Theory). (Existence is even less trivial). But using those things as fact, showing that those $X_i$ are iid is probably very instructive.