# Why does [dz(t)]^2 converge to dt over infinitesimally short time periods?

I have some trouble understanding a chapter in George Pennacchi textbook "Asset Pricing". Here the author shows that the square of a Wiener Process $[dz(t)]^2$ converges to $dt$ for infinitesimally short time periods.

By the definition of the Itô-Integral for $\int_0^T[dz(t)]^2$: $$\lim_{n\rightarrow\infty}E_0[(\sum_{i=1}^n [\Delta z_i^2]-\int_0^T[dz(t)]^2)^2]=0$$

Then he states that $$E_0[(\sum_{i=1}^n [\Delta z_i^2] -T)^2]=2T\Delta t$$, which is the equation I cannot derive. The limit for this expression as $\Delta t \rightarrow 0$ is 0 which is obvious. But what I am also not able to grasp is why this result together with the definition of the Itô Integral finally shows that $$\int_0^T[dz(t)]^2=\int_0^T dt$$

Can someone give me some assistance in understanding this important outcome?

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Well, that's not really "the square root", but rather the quadratic variation, and there are many detailed proofs in many sources. This one has, for instance, has a lengthy explanation: fabricebaudoin.wordpress.com/2012/08/24/… –  Georgy Ivanov Jun 14 '14 at 19:33
Thanks Georgy for the link. I edited my post to remove the typo. I was looking for a more intuitive explanation rather than a rigouros proof (with another notation and some vocabulary from measure theory in which I never had a course). –  Roberto Liebscher Jun 21 '14 at 22:05

Beware, oversimplification ahead! (This means that the following is technically not correct, in fact it is false! But: It gives an intuition what is going on!)

If you toss a coin and calculate heads as $-1$ and tails as $1$ you get a mean of $0$ with a variance of $1$. When you add up multiple coin tosses, i.e. create a random process $dz(t)$, the mean stays the same, i.e. $0$ -> this process is a martingale.

Now you square this process: On average you will get $\frac{(-1)^2+1^2}{2}=1$, which means that your mean gets one unit bigger per timestep -> $[dz(t)]^2=dt$

So here you see how a martingale could become a process which grows with time on average when you square it - or in more general terms: how a symmetric process could become asymmetric through a non-linear transformation.

The following is a very nice exposition showing these ideas in a very intuitive manner:
Stochastic Calculus and the Nobel Prize Winning Black-Scholes Equation by Frank Morgan

To address the question why this discrete intuition might make sense even in the continuous case you have to know that the abovementioned stochastic process results in a binomial tree. The basis of this tree is, as the name suggests, the binomial distribution. When you simultaneously reduce the size of and increase the number of time steps this binomial distribution becomes (under some technical conditions) the normal distribution. This is the result of the de Moivre–Laplace theorem. The corresponding continuous stochastic process is a Wiener process (also often called standard Brownian motion), which closes the circle.

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This explain neither convergence, nor why square is the right power. It's really CLT for a random walk. –  LazyCat Jul 2 '14 at 16:10
@LazyCat: Sorry to disagree: The OP asks for some understanding why $[dz(t)]^2$ converges to $dt$, i.e. why a stochastic process, which is a martingale, gives a deterministic function which rises with time when you square it - this is exactly what I address on an intuitive level. –  vonjd Jul 2 '14 at 16:14
@vonjd: no, "i.e. ..." part is wrong. You don't square a martingale, you square the sum before the limit, and the whole point is that 1) it converges 2) only if it's the second power. And both parts are manifestations of CLT. –  LazyCat Jul 2 '14 at 17:48
@LazyCat: Thank you, please see my addendum. –  vonjd Jul 2 '14 at 17:53
Thanks vonjd for giving this nice intuitive explanation of why $[dz(t)]^2$ linearly increases with time. –  Roberto Liebscher Jul 9 '14 at 12:17

Intuitively, because of the central limit theorem: wiener process is a limit of a random walk, and after n steps a random walk moves away from the origin by ~ $\sqrt{n}$

Edit: here is a complete answer. First the formula for the sum. The trick is the following simple observation: if $X_1,.. X_n$ are independent zero mean, then $E(\sum X_i)^2 = \sum{EX_i^2}$. In the case of the formula, independent zero mean incremnents are $d_i=\Delta z_i^2 - T/N,$ since z(t) is Brownian motion.

The sum in question is $$E(\sum{\Delta z_i}^2-T)^2 = E(\sum{d_i})^2 = \sum{E(d_i^2)} = \sum(E\Delta z_i^4 - 2 T / N * E(\Delta z_i^2) + T^2/N^2) = 2 T^2 / N = 2 T * \Delta t$$

Second, for the relation to the last equality: the right hand side is T. The left hand side is in some (precise) sense is the limit of $\sum{\Delta z_i^2}$ One standard way to prove convergences in probability, is to estimate the second moment of the difference. This is the formula we just proved.

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I think you are correct overall, but may you show some sources for it? Also, it is not exactly $\sqrt{n}$ but $\sqrt{n\log\log n}$? –  emcor Jul 2 '14 at 16:39
no, the last asymptotic is from the law of iterated logarithm. Will try to find a good reference. –  LazyCat Jul 2 '14 at 17:52
In any case you should expand your answer because as it stands it is unclear how it relates to the question (at least to me). –  vonjd Jul 2 '14 at 18:29
Judging by the addendum you've made to your answer after reading mine, it's pretty clear to you, how it relates to the question. I'll expand if I have time and there is interest. –  LazyCat Jul 2 '14 at 18:37
For the first, in the original formula $T$ is not inside the summation. The sum has $N$ terms, so instead of subtracting T from the sum, you can subtract T/N from each term. For the second part - $E(\Delta z_i^2) = T/N$, so your second term is $-2 T^2/N^2,$ so the whole thing under the summation sign adds nicely. –  LazyCat Jul 9 '14 at 14:25

Intuitively you are measuring the distance between two random variables which under limiting case turns out to be very small. So you can use one instead of another.

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"Use one instead of other because both are small" . This is not the reason unfortunately –  ash Jul 2 '14 at 14:48
How else who u intuitively explain mean square convergence of Brownian motion increments ? –  Rohit Arora Jul 2 '14 at 14:51
@user2142: Fair question, see my answer :-) –  vonjd Jul 2 '14 at 14:52
That answer has nothing to do with mean square convergence of random variables. That oversimplification I believe is tangent to understanding the first equation. –  Rohit Arora Jul 2 '14 at 15:28
@user2142: Sorry to disagree: The OP asks for some understanding why $[dz(t)]^2$ converges to $dt$, i.e. a stochastic process, which is a martingale, gives a deterministic function which rises with time when you square it - this is exactly what I address on an intuitive level. –  vonjd Jul 2 '14 at 15:57