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This is a question about a proof I saw in a script about stochastic processes. First I state a theorem which is needed in the proof. After that there are two questions, which are highlighted. Between the questions I provide my thoughts / work so far.

$\mathbf{Theorem}$ For any sequence $(X_n)$ of positive r.v. there exists a sequence $(\tilde{X}_n)$ with $\tilde{X}_n\in\operatorname{conv}(X_k;k\ge n)$ (the convex hull) such that $\tilde{X}_n\to X$ $P$-a.s. The r.v. variable $X$ can take values in $[0,\infty].$

This theorem is also known als Komlos theorem. The convex hull is the set $\operatorname{conv}(X_k;{k\ge n})=\{\sum_{k=n}^\infty \lambda_k X_k| \sum_{k=n}^\infty \lambda_k=1,\forall k\ge n:\lambda_k\ge 0,\lambda_k\not=0 \mbox{ for finitely many } \}$ and does not depend on $\omega$, i.e. it is deterministic.

Suppose I have a sequence of r.v. $g_n$ all positive which converge in probability to $g$. Furthermore, I know that for every $n$ there is a stochastic process $Z^n$, such that $g_n\le Z^n_T$. The stochastic processes $Z^n$ are positive, RCLL and indexed by $t\in[0,T]$.

$\mathbf{Question}$ They claim, using a diagonal argument one can find convex combinations $(\tilde{g}_n)$ and $(\tilde{Z}^n_q)$ for all rationals $q\in[0,T]$ converging $P$-a.s. to $g$ and $Z^\infty_q$ respectively.

Is it meant, that we use the same sequence $\Lambda\subset\mathbb{N}$ for $(\tilde{g}_n)$ and $(\tilde{Z}^n)$? Clearly I can find such a sequence for $(g_n)$ using the theorem above. For $(Z^n)$ I would take a numeration of $\mathbb{Q}$, i.e. $q_1,q_2,\dots$ Then looking at $(Z^n_{q_1})$, we can find a $(\tilde{Z}^n_{q_1})$, where $n\in\Lambda_1$, such that $\tilde{Z}^n_{q_1}\to Z^\infty_{q_1}$. For $q_2$ I take a subsequence $\Lambda_2\subset\Lambda_1$ such that $\tilde{Z}^n_{q_2}\to Z^\infty_{q_2}$ and so on. Therefore a general $\tilde{Z}^n_{q_j} \in \operatorname{conv}(Z^k_{q_j};k\ge n)$, for $n\in \Lambda_j$, right?

If it is meant that we use the same sequence $\Lambda$ for both $(\tilde{g}_n)$ and $(\tilde{Z}^n)$, then I would just start with $(g_n)$ and the go to $(Z^n_{q_1})$ etc.

$\mathbf{Question}$ They state, we have $g_n\le Z^n_T$ and we use same convex combinations for $(\tilde{g}_n)$ and all $\tilde{Z}^n_q$, so $g\le Z^\infty_T$.

Why we are using the same convex combinations? If so, then then we can conclude:

$$\tilde{g}_n=\sum_{k=n}^\infty\lambda_kg_k\le\sum_{k=n}^\infty\lambda_kZ_T^k=\tilde{Z}^n_T$$ hence $g\le Z^\infty_T$. But what guarantees that we can use the same convex combinations, i.e. the same $\lambda_k$'s? I'm very thankful for your help!

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+1 for the structure and showing some effort. –  user8 Apr 24 '13 at 9:43
    
I'm about to type an answer, but I noticed there's an error in your statement of Komlos theorem. You have $\tilde X_n\in\mathop{conv}(X_k;k\geq n)$ where it should be $\tilde X_n\in\mathop{conv}(X_k;k\leq n)$. –  Tim May 2 '13 at 9:03
    
@Tim no this is not an error. See for example the book "The Mathematics of Arbitrage" by Delbaen/Schachermayer Lemma 9.8.1. –  user20869 May 2 '13 at 9:14
    
I got the statement from here math.stackexchange.com/questions/89263/… I don't have the book you mention but Googling "Komlos Theorem" gives me at least three further references on the first page. It's possible there are two different versions of the theorem or even a typo in the book. –  Tim May 2 '13 at 9:24
    
@hulik: OK, I've figured it out, your statement is a very minor alteration of the usual statement, which would require a 2 or 3 line proof. A $\leq$ would also be a minor alteration of the usual statement, which would not require a 2 or 3 line proof. It doesn't make any difference to the answer to the question. –  Tim May 2 '13 at 9:40
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up vote 1 down vote accepted

It looks like the argument is a slight twist on the standard diagonalization argument, because the text is talking about convex combinations and not subsequences.

Your approach is pretty much spot on except that you take a subsequence $\Lambda_2\subset\Lambda_1$ instead of a sequence of convex combinations, which leads to the wrong conclusions. There's a couple of extra arguments you need to add for the convex combinations.

So you have to be a little bit careful constructing the sequence, once you do so it's obvious that you can use the "same convex combinations" of $g_n$ and $Z^n_t$.

I'll start the argument from the beginning but try to follow your notation.

First as $g_n$ converges in probability we may choose a subsequence $g_{n_k}$ that converges almost surely with $g_{n_k} \leq Z^{n_k}_t$. Therefore we may assume without loss of generality that $g_n$ converges almost surely.

Now consider a stochastic process $\hat Z_t\in \operatorname{conv}(Z_t^k:k\geq n)$ indexed by $t\in[0,T]$ with $$\hat Z_t = \sum_{k=n}^\infty \hat\lambda_k Z^k_t$$

Then by definition for every $s\in[0,T]$ we have $\hat Z_s = \sum_{k=n}^\infty \hat\lambda_k Z^k_s$. In particular for fixed $s\in[0,t]$ and every random variable $X\in\operatorname{conv}(Z_s^k:k\geq n)$ we may choose a stochastic process $\hat Z_t\in\operatorname{conv}(Z_t^k:k\geq n)$ with $\hat Z_s = X$ almost surely.

Now, order the rationals $\{q_i:i\in\mathbb N\}$ , I want to choose a sequence of stochastic processes $\tilde Z_t^n \in \operatorname{conv}(Z_t^k:k\geq n)$ and a stochastic process $Z^\infty_t$ such that for every $i \in\mathbb N$ we have $\tilde Z_{q_i}^n \to Z_{q_i}^\infty$ almost surely.

First by Komlos theorem we may choose a sequence of random variables $X^1_n\in\operatorname{conv}(Z_{q_i}^k:k\geq n)$ and a random variable $Z^\infty_{q_1}$ such that $X^1_n\to Z^\infty_{q_1}$ almost surely.

Hence we may choose a sequence of stochastic processes $\tilde Z_t^{n,1} \in \operatorname{conv}(Z_t^k:k\geq n)$ such that $\tilde Z_{q_1}^{n,1} = X^1_n\to Z^\infty_{q_1}$ almost surely.

Now let us suppose inductively that I have constructed a sequence $\tilde Z_t^{n,i} \in \operatorname{conv}(Z_t^k:k\geq n)$ such that $\tilde Z_{q_j}^{n,i}\to Z^\infty_{q_j}$ as $n\to\infty$ almost surely for every $j=i$.

As $\tilde Z_{q_{i+1}}^{n,i}$ is a sequence of random variables I may choose $X^{i+1}_n \in\operatorname{conv}(\tilde Z_{q_i+1}^{k,i}:k\geq n)$ such that $X^{i+1}_n\to Z^\infty_{q_{i+1}}$ almost surely as $n\to\infty$ for some random variable $Z^\infty_{q_{i+1}}$ and a sequence of stochastic processes
$$\tilde Z_{t}^{n,i+1} \in \operatorname{conv}(\tilde Z_t^{k,i}:k\geq n)$$ Such that $\tilde Z_{q_{i+1}}^{n,i+1} = X^{i+1}_n\to Z^\infty_{q_{i+1}}$ as $n\to\infty$ almost surely.

Now $\tilde Z_{t}^{n,i+1}$ is a convex combination of a finite number of elements of $\operatorname{conv}(Z_t^k:k\geq n)$. Hence it is itself a convex combination of a finite number of elements of $\{Z_t^k:k\geq n\}$. Therefore $\tilde Z_{t}^{n,i+1}\in \operatorname{conv}(Z_t^k:k\geq n)$ and it remains to show that $\tilde Z_{q_j}^{n,i+1}$ converges almost surely for $j\leq i+1$.

For every $\varepsilon>0$ $\mathbb P$ almost every $\omega$ and each $j\leq i$ I may choose $N(\omega)$ large enough that $\left| Z^\infty_{q_j}(\omega) -\tilde Z_{q_{j}}^{N+k,i}(\omega)\right| < \varepsilon$ hence $\left|Z_{q_j} - Z^\infty_{q_j}\right|$ for every $Z_t\in \operatorname{conv}(\tilde Z_t^{k,i}:k\geq N)$.

In particular $\tilde Z_{q_j}^{n,i+1}\to Z^\infty$ for every $j\leq i+1$.

Now we may consider the diagonal sequence $$\tilde Z_t^{n,n} = \sum_{k=n}^{\infty} \lambda_{n,k} Z_t^n\in\operatorname{conv}(Z_t^k:k\geq n).$$

And set

$$\tilde g_n = \sum_{k=n}^{\infty} \lambda_{n,k} g_{k}\in\operatorname{conv}(g_k:k\geq n).$$

So $\tilde g_n \leq \tilde Z_t^{n,n} $ and $\tilde g_n$ converges almost surely to $g$ so we have $g\leq Z^\infty_{t}$ for every rational $t$.

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I think it should be $\hat{Z}_{q_1}^{n,1}\in\operatorname{conv}(Z_{q_1}^k;k\ge n)$ and so on for $q_i$. –  user20869 May 2 '13 at 14:07
    
@hulik: Where, and instead of what? –  Tim May 2 '13 at 14:47
    
where you wrote: "Order the rationals $q_i$...". But also later, when you choose your sequence according to $q_i$ –  user20869 May 2 '13 at 14:48
    
@hulik: No, that would break the argument. I'll try and think of a good, short, explanation and add it in. –  Tim May 2 '13 at 15:16
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yes...I know. But why should you be able to choose such a thing uniformly in $t$? –  user20869 May 2 '13 at 15:20
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