The definition I have been given of the $\limsup\limits_{n \to \infty} Y_n$ where the $Y_n$ are random variables is that it is another random variable defined as $(\limsup\limits_{n \to \infty} Y_n)(\omega) = \limsup\limits_{n \to \infty} Y_n(\omega)$ for any $\omega \in \Omega$ where the $\lim\sup$ on the R.H.S is the standard $\lim\sup$ of a sequence of reals (as that is the definition of $\lim\sup$ for a sequence of measurable functions).

However I am then given the question to prove that if we have $X_i$ as IID standard $N(0,1)$ random variables then we get $\limsup\limits_{n \to \infty} \frac{X_n}{\sqrt{2\log(n)}} = 1$ almost surely. In this case if I use the definition given above we get that $X_n(\omega) = X_m(\omega)$ for any $n,m$ and $\omega$ as all the $X_i$ have the same distribution. So our sequence is then just decaying to $0$ so the $\lim\sup$ should be $0$ and not $1$ almost surely.

I'm sure that I'm the one who has a conceptual error somewhere but I can't see what. I know the question is not wrong as other people have asked it here. Please help me clear up this confusion.


Two random variables having the same distribution does not mean they take the same value on each $\omega$. Indeed, independence already kill that possibility unless the random variable is constant (a.s.).

  • $\begingroup$ I'm confused, can't we define each $X_n$ as being a measurable function from $\mathbb{R} \to \mathbb{R}$ with the Borel $\sigma$-algebra such that the preimage has the measure given by the C.D.F of the standard normal variable? Then we would have $X_n(\omega)$ is the same for every $n$? Also if I can't think of the $\limsup$ in that way what is the best way to think of the $\limsup$ of a sequence of random variables? $\endgroup$ – Abdul Hadi Khan Sep 16 '18 at 15:42
  • $\begingroup$ Your $X_i$ would not be independent, contradicting the i.i.d. assumption. The way to think about them is the same as how you think about real number case, except now each $\omega$ can (and generally will) get a different sequence of real numbers, i.e. $\limsup Y_n$ is just another random variable. $\endgroup$ – user10354138 Sep 16 '18 at 15:50
  • $\begingroup$ Usually iid sequence of random variables is to take a product probability space $\Omega:=(\Omega_1)^\mathbb{N}$, where $\Omega_1$ is a probability space modelling a single member of that iid random variable, and each member of that sequence corresponds to a different projection map. What you have described is just the diagonal of this product, which typically is null unless there are atoms. $\endgroup$ – user10354138 Sep 16 '18 at 15:54

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.