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.
Wiener Process $dB^2=dt$
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2What do you mean by *every book*? Could you list a couple? – 2011-11-14
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0My 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. – 2011-11-14
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1Sorry. 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. :) – 2011-11-14
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0I 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. – 2013-09-26
4 Answers
$$dB_t^2 = dt, \qquad (dt)^2 = 0, \qquad dB_t \, dt = 0 \tag{1}$$ 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
$$\int_0^t g(B_s) \, dB_s^2 := \lim_{|\Pi| \to 0}\sum_{j=1}^n g(B_{t_{j-1}}) \cdot (B_{t_j}-B_{t_{j-1}})^2.$$
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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0In 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? – 2013-09-30
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0@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! – 2013-09-30
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0The 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. – 2013-09-30
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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0Sorry 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... – 2012-12-11
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0@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.. – 2013-09-22
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1@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". – 2013-09-22
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0@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). – 2013-09-22
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0@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?) – 2013-09-22
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0@Did : probably too late to re-do this as a new question at this point - only would cause more confusion. Unless moderators agree and help. – 2013-09-22
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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0Sorry 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... – 2012-12-11
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0Certainly 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 $E(B_t^2)=t$). – 2012-12-11
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0I 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 http://www.math.nyu.edu/faculty/goodman/teaching/StochCalc2012/notes/Week6.pdf . – 2012-12-11
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2...Where the author takes care to repeat regularly that the derivation is "informal". – 2014-08-13