Let $\mathbb{R}^{\infty}$ be the space of infinite sequences $(x_1, \dots, )$. Equip the space with $\sigma$-algebra $\mathcal B(\mathbb{R}^{\infty})$ generated by finite dimensional rectangles. We say two random sequences $X = (X_1, \dots )$ on $(\Omega, \mathcal F, P)$ and $Y = (Y_1, \dots)$ on $(\Omega', \mathcal F', P')$ have the same distribution if for all $B \in \mathcal B(\mathbb R^{\infty})$, $$ P(X \in B) = P'(Y \in B) .$$

Suppose $X = (X_1, \dots)$ and $Y=(Y_1, \dots)$ have the same distribution and $\{X_n\}$ converges almost surely (almost everywhere) to some random variable $X_{\infty}$ on $(\Omega, \mathcal F, P)$. Then $\{Y_n\}$ converges almost surely to some random variable $Y_{\infty}$ on $(\Omega', \mathcal F', P')$ and $X_{\infty}$ and $Y_{\infty}$ have the same distribution.

I am stuck with the problem for a while. I am new to probability theory and to be honest, I don't quite know how to apply the condition with same distribution. Any help will be appreciated.

Edit: Thanks to the hint by @Kavi Rama Murthy. I post my solution here. Any critics are welcome.

Let $\{b_j\}_1^n \subseteq \mathbb R^{\infty}$ and $B = \{ \{b_j\} \colon \lim_{j \to \infty} b_j \text{ exits }\}$. We observe \begin{align*} \displaystyle B = \bigcap_{k=1}^{\infty} \bigcup_{N \in \mathbb N} \bigcap_{m,n \ge N} \{ \{b_j\} \colon |b_m - b_n| < \frac{1}{k}\}. \end{align*} It is clear $B \subseteq \mathcal B^{\infty}$. Then \begin{align*} X^{-1}(B) = \{ \omega \colon \{X_n(\omega)\} \text{ is Cauchy }\} \\ Y^{-1}(B) = \{ \omega \colon \{Y_n(\omega)\} \text{ is Cauchy }\}. \end{align*} Since $\{X_n(\omega)\}$ converges to $X_{\infty}(\omega)$ almost everywhere, $P(X^{-1}(B)) = 1$. It follows $P'(Y^{-1}(B)) = 1$. This is, the set $\Omega_0' = \{\omega \colon \lim_{n \to \infty } Y_n(\omega) \text{ exits }\}$ has measure $1$. Let $Y_{\infty} = \lim_{n \to \infty} Y_n(\omega)$ on $\Omega_0'$ and be any number in $\mathbb R$ on $\Omega \setminus \Omega_0$. $Y_n \to Y_{\infty}$ almost everywhere. By excluding the sets with measure $0$ in both spaces, we may assume $X_n \to X_{\infty}$ and $Y_n \to Y_{\infty}$ pointwise everywhere. For any Borel set $B \subseteq \mathcal B( \mathbb R)$, \begin{align*} \displaystyle X_{\infty}^{-1}(B) = \limsup_{n \to \infty} X_n^{-1}(B) = \bigcap_{k=1}^{\infty} \bigcup_{n =k}^{\infty} X_n^{-1}(B) \\ Y_{\infty}^{-1}(B) = \limsup_{n \to \infty} Y_n^{-1}(B) = \bigcap_{k=1}^{\infty} \bigcup_{n =k}^{\infty} Y_n^{-1}(B) \\ \end{align*} Clearly we have $P( X^{-1}(B)) = P'( Y^{-1}(B))$.

I am not convinced the proof to that $X_{\infty}$ has the same distribution as $Y_{\infty}$ is correct. Would anyone give suggestions on how to approach this part?


You only have to observe that $B=\{\{a_n\}:\{a_n\} converges\}$ is a Borel set in $\mathbb R^{\infty}$. (Use definition of limit prove this fact). When you apply the hypothesis to this set you get exactly what you want.

  • $\begingroup$ Can you elaborate more please? The set where The limit exists Is certainly measurable. But we still need to show The measure Is 1. $\endgroup$ – user1101010 Oct 13 '17 at 7:07
  • $\begingroup$ What is given is $P\{X^(-1)(B)\}=1$ What you need is $P\{Y^(-1)(B)\}=1$. This is immediate from the hypothesis. $\endgroup$ – Kavi Rama Murthy Oct 13 '17 at 7:11
  • $\begingroup$ Sorry about poor technical typing. The inverse image under X of B is precisely {w:lim:X_n(w) exists} and this event has probability 1. By the hypothesis we can change X to Y. I hope this is clear enough. $\endgroup$ – Kavi Rama Murthy Oct 13 '17 at 8:00

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.