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Why is $dB^2=dt$? Every online source I've come across lists this as an exercise or just states it, but why isn't this ever explicitly proved? I know that $dB=\sqrt{dt}Z$, but I don't know what squaring a Gaussian random variable means.

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What do you mean by every book? Could you list a couple? –  cardinal Nov 14 '11 at 2:58
    
My mistake. I meant every online source I've come across from googling. This was stated in a Mathematical Finance class without justification, and I've been spending hours trying to figure out how this comes about. –  j.diddland Nov 14 '11 at 6:13
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Sorry. The point of my question, which may not have been clear, was to get a feel for the level at which you were expecting an answer. What textbook does the course use? Do you know about quadratic variation? At the level of say, S. Shreve, Stochastic Calculus for Finance or, maybe, Karatzas & Shreve, Brownian Motion and Stochastic Calculus? Or, maybe, the course is more at the level of J. C. Hull, Options, Futures, and Other Derivatives? Providing this kind of info will help me or someone else provide an answer at the appropriate level. Cheers. :) –  cardinal Nov 14 '11 at 9:43
    
I think the answer to my bounty clarification request is that a simple calculation shows that the standard deviation of $dB^2$ is actually of the order of $dt^{3/2}$, while its expectation is of the order of $dt$. So the randomness can be ignored. –  max Sep 26 '13 at 6:24

3 Answers 3

$$dB_t^2 = dt, \qquad (dt)^2 = 0, \qquad dB_t \, dt = 0 \tag{1}$$ are basically rules to simplify the calculation of the quadratic (co)variation of Itô processes - and nothing more:

Let $(B_t)_{t \geq 0}$ a one-dimensional Brownian motion and $(X_t)_{t \geq 0}$ an Itô process, i.e.

$$dX_s = \sigma(s) \, dB_s + b(s) \, ds$$

Then, by Itô's formula,

$$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s + \int_0^t f''(X_s) \, \sigma^2(s) \, ds. \tag{2}$$

The point is: If we simply apply the rules in $(1)$, we obtain

$$dX_s^2 = (\sigma(s) \, dB_s + b(s) \, ds)^2 = \sigma^2(s) \underbrace{dB_s^2}_{ds} + 2b(s) \sigma(s) \underbrace{ds B_s}_{0} + b^2(s) \, \underbrace{ds^2}_{0} \\ = \sigma^2(s) \, ds.$$

Therefore, we can rewrite $(2)$ in the following way:

$$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s + \int_0^t f''(X_s) \, dX_s^2$$

i.e. Itô's formula justifies the calculation rules in $(1)$.


The (mathematical) reason why this works fine can be seen while proving Itô's formula. Actually, one can show that

$$\sum_{j=1}^n g(B_{t_{j-1}}) \cdot (B_{t_j}-B_{t_{j-1}})^2 \tag{3}$$

converges to

$$\int_0^t g(B_s) \, ds$$

as the mesh size $|\Pi|$ of the partition $\Pi=\{0=t_0<\ldots<t_n=t\}$ tends to $0$. This convergence is based on the fact that $B_t^2-t$ is a martingale. On the other hand, comparing $(3)$ with the definition of Riemann-Stieltjes integrals, it's natural to define

$$\int_0^t g(B_s) \, dB_s^2 := \lim_{|\Pi| \to 0}\sum_{j=1}^n g(B_{t_{j-1}}) \cdot (B_{t_j}-B_{t_{j-1}})^2.$$

Consequently,

$$\int_0^t g(B_s) \, dB_s^2 = \int_0^t g(B_s) \, ds.$$

A similar reasoning applies to Itô processes. Note that these integrals, i.e. integrals of the form

$$\int_0^t g(X_s) \, dX_s^2,$$

are exactly the integrals appearing in Itô's formula $(2)$ ($g \hat{=} f''$).

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In simple words, does it make sense to say that $$dB_t^2$$ is non-stochastic? If so, is there any intuitive reason for that, which can be explained without strict definitions? –  max Sep 30 '13 at 0:37
    
@max First of all, we have to define $dB_t^2$ to give integrals of the form $$\int_0^s f(s) \, dB_s^2$$ a meaning. Actually, one can use semimartingale theory to do so, but that's a different matter. As I tried to explain, (the proof of) Itô's formula motivates to define $$\int_0^t f(X_s) \, dB_s^2 := \int_0^t f(X_s) \, ds$$ where $X$ is an Itô process - for this particular class of integrals! –  saz Sep 30 '13 at 6:32
    
The only heuristical explanation I know is the following: By scaling property, $B_t \sim \sqrt{t} B_1$. Thus (heuristically!) $$dB_t^2 = B_1^2 \, dt$$ Since $\mathbb{E}B_1^2 = 1$, we have $dB_t^2 = dt$. This might give you some intuition why this works, but as @Did mentioned several times the formula is based on a much deeper result. –  saz Sep 30 '13 at 6:35

For independent random variables, the variance of the sum equals the sum of the variances. So $\mathbb{E}((\Delta B)^2)=\Delta t$, i.e. if you increment $t$ a little bit, then the variance of the value of $B$ before that increment plus the variance of the increment equals the variance of the value of $B$ after the increment.

Or you could say $$ \frac{\mathbb{E}((\Delta B)^2)}{\Delta t} = 1. $$ That much follows easily from the first things you hear about the Wiener process. I could then say "take limits", but that might be sarcastic, so instead I'll say that for a fully rigorous answer, I'd have to do somewhat more work.

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Sorry but the nonrigorous shorthand $dB_t^2=dt$ refers to a much deeper result than the (true) fact that $B_{t+s}-B_t$ is centered with variance $s$, which your answer reduces to. As such, one may find said answer rather misleading. Or, since the question is badly formulated and the OP never answered @cardinal's fully justified demand for background, one can also consider all this rather moot... –  Did Dec 11 '12 at 6:13
    
@Did : Your clarification about the "deeper result" is precisely what I was hoping to see in one of the answers... Could someone possibly provide such an answer at the most intuitive / least rigorous level that you feel is possible? Thx.. –  max Sep 22 '13 at 9:01
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@max I have trouble understanding the combo "bounty+comments" to revive this old question. Why not ask a carefully worded new question explaining unambiguously the points you want to see dealt with and addressing the concerns user cardinal voiced Nov 14 '11 at 9:43? By the way, the bounty mention that "The current answer(s) ... require revision given recent changes" is odd since the question underwent zero "recent changes". –  Did Sep 22 '13 at 9:11
    
@Did : sorry I thought if I asked a new question, mods would close it as a duplicate. My questions on stackoverflow were closed a few times as duplicates, even when I tried to explain why (in my opinion) they weren't... As to the wording of the bounty, there were only a few options, and none of them fit the situation. I figured I could loosely interpret "recent changes" to mean "comments added recently" (by you). –  max Sep 22 '13 at 9:14
    
@max I was not aware of the restricted choice of options one is given when offering a bounty, sorry about that. // Yeah, dup or not dup, that is the question... :-) In the case at hand, as I said, to reformulate carefully the question, adding the information cardinal asked for and the reference you put in the bounty, would make for a new question significantly different from (and better than) the present one. At least, I would be ready to argue it is... (Unrelated: "Recently" = 10 months ago?) –  Did Sep 22 '13 at 9:20

Obviously $dB_t^2 \neq dt$, since $dB_t \sim \mathcal{N} (0, dt)$ is a random variable, while $dt$ is deterministic.

As Michael Hardy said, they really meant to say $\mathbb{E} \left[ dB_t^2 \right] = dt$. To convince yourself, compute $$ \mathbb{E} \left[ dB_t^n \right] = \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi dt}} \exp\left(-\frac{x^2}{2 dt}\right) x^n dx \, .$$

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Sorry but the nonrigorous shorthand $dB_t^2=dt$ refers to a much deeper result than the (true) fact that $B_{t+s}-B_t$ is centered with variance $s$, which your answer reduces to. As such, one may find said answer rather misleading. Or, since the question is badly formulated and the OP never answered @cardinal's fully justified demand for background, one can also consider all this rather moot... –  Did Dec 11 '12 at 6:13
    
Certainly not. First, because neither $E(dB_t^4)$ nor $dt^2$ are well defined objects. Second, and even more importantly, because what the shorthand $dB_t^2=dt$ refers to is the whole class of Doob's semimartingale decompositions which Itô's formula provides (for example, to stay at the level of a toy example, the fact that $t\mapsto B_t^2-t$ is a martingale, which is not reducible to the fact that $E(B_t^2)=t$). –  Did Dec 11 '12 at 6:44
    
I am aware of a fairly rigorous analysis that shows a term containing $dB_t^2$ in Ito's lemma, for example, converges to one that contains $dt$ almost surely, using the Borel-Cantelli lemma. See pages 4 to 6 in the lecture math.nyu.edu/faculty/goodman/teaching/StochCalc2012/notes/… . –  William S. Wong Dec 11 '12 at 6:47
    
...Where the author takes care to repeat regularly that the derivation is "informal". –  Did Aug 13 at 9:11

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