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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?

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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.

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  • $\begingroup$ 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

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