Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I´m not sure why this theorem is right, how can i prove it?

Let $X_1,X_2.... $ and $Y_1,Y_2...$ be random variables such that $X_n,Y_n$ are independent for every $n∈\mathbb N$ and such that X, resp Y, are the almost sure limits of $(X_n)$ and $(Y_n)$. Then also X and Y are independant.

Thanks for advice

share

2 Answers 2

up vote 3 down vote accepted

For every continuous bounded functions $u$ and $v$, $u(X_n)\to u(X)$, $v(Y_n)\to v(Y)$ and $u(X_n)v(Y_n)\to u(X)v(Y)$ almost surely. All these random variables are uniformly bounded hence $$ \mathbb E(u(X_n))\to\mathbb E(u(X)),\quad \mathbb E(v(Y_n))\to\mathbb E(v(Y)), $$ and $$ \mathbb E(u(X_n)v(Y_n))\to\mathbb E(u(X)v(Y)). $$ By independence of $X_n$ and $Y_n$, for each $n$, $\mathbb E(u(X_n)v(Y_n))=\mathbb E(u(X_n))\mathbb E(v(Y_n))$. By identification of the limits, $$ \mathbb E(u(X)v(Y))=\mathbb E(u(X))\mathbb E(v(Y)). $$ This holds for every continuous bounded functions $u$ and $v$, hence the distribution of $(X,Y)$ is the product of their distributions, QED.

share
    
thank you, i can work with that –  nordmann Jan 31 '13 at 0:28

Be careful with this answer: it's very possible I messed something up Yes, I did! See comments. Here goes: we want $$\mathbb{P}(X \leqslant a, Y \leqslant b) = \mathbb{P}(X \leqslant a)\mathbb{P}(Y \leqslant a)$$

Suppose first we have sequences of real numbers $a_j \rightarrow a, b_j \rightarrow b$. Then I claim first that $$a \leqslant A, b \leqslant B \Longleftrightarrow \forall n, \exists i | \forall j > i: a_j \leqslant A + \frac{1}{n}, b_j \leqslant B + \frac{1}{n}$$

Applying this criterion to $X_i(\omega) \rightarrow X(\omega), Y_i(\omega) \rightarrow Y(\omega)$ for each $\omega$ (throwing away a set of measure 0), we get $$\{ X \leqslant a, Y \leqslant b \} = \bigcap_n \bigcup_{i \geqslant 0} \bigcap_{j \geqslant i} \{ X_j \leqslant a + \frac{1}{n}, Y_j \leqslant b + \frac{1}{n} \}$$

When evaluating the probability of this, we can drag the $\mathbb{P}$ across each $\cap$ and $\cup$ since the relevant events (everything to the right of said operation) are nested, hence we get $$\mathbb{P}(X \leqslant a, Y \leqslant b) = \lim_n \lim_i \lim_{j \geqslant i} \mathbb{P}(X_j \leqslant a + \frac{1}{n}, Y_j \leqslant b + \frac{1}{n})$$By independence of $X_j$ and $Y_j$, we can rewrite this as $$\lim_n \lim_i \lim_{j \geqslant i} \mathbb{P}(X_j \leqslant a + \frac{1}{n})\mathbb{P}(Y_j \leqslant b + \frac{1}{n}) $$$$= (\lim_n \lim_i \lim_{j \geqslant i} \mathbb{P}(X_j \leqslant a + \frac{1}{n}))(\lim_n \lim_i \lim_{j \geqslant i} \mathbb{P}(Y_j \leqslant b + \frac{1}{n})) = \mathbb{P}(X \leqslant a)\mathbb{P}(Y \leqslant b) $$

share
    
Unfortunately the events $[X_j\leqslant a+1/n,Y_j\leqslant b+1/n]$ are not nested with respect to $j$. –  Did Jan 30 '13 at 11:02
    
@Did absolutely - a good point! –  uncookedfalcon Jan 30 '13 at 17:46

This site is currently not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .