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I'm proving monotone convergence for random variables, i.e.

If random variables $X_n$ are positive and increasing almost surely to $X$, then $$\lim_{n\rightarrow \infty}\mathbb{E}\left\{X_n\right\}=\mathbb{E}\left\{X\right\}$$

The proof so far goes like this:

For each fixed $n$, choose an increasing sequence $Y_{n,k}$ of simple random variables increasing to $X_n$, and set $Z_k=\max_{n\leq k}Y_{n,k}$. Then $\left(Z_k\right)_{k\geq 1}$ is a non-decreasing sequence of positive simple random variables, and thus it has a limit $Z=\lim_{k\to\infty} Z_k$. Also, $$Y_{n,k}\leq Z_k \leq X_k \leq X\hspace{30pt}\text{almost surely}\hspace{20pt}\text{for }n\leq k$$ which implies that $$X_n \leq Z \leq X\hspace{30pt}\text{almost surely}.$$ Next, if we let $n\to\infty$, we have $Z=X$ almost surely...

I don't understand how the inequality $X_n \leq Z \leq X\text{ almost surely}$ can be deduced from the previous inequality.

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1 Answer 1

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Since $Y_{n,k}\le Z_k\le X$, it follows that $$ \lim_{k\to\infty} Y_{n,k}\le \lim_{k\to\infty} Z_k\le \lim_{k\to\infty} X $$ This is a property of limits. The definitions of $Y_{n,k}$ and $Z$ then make the above become $$X_n\le Z\le X.$$

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  • $\begingroup$ Gah, thanks, obvious. I forgot that $\lim_{k\to \infty}Y_{n,k}=X_n$. $\endgroup$ Oct 6, 2014 at 1:15

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