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A simple SDE is given as: $$dX_t=\sigma X_tdW_t$$ Ito's lemma confirms that the following is a solution: $$X_t=X_0e^{\sigma W_t-\frac{1}{2}\sigma^2t}$$

I understand how this result is correct according to Ito's lemma, but I don't see why it makes any sense that the $-\frac{1}{2}\sigma^2t$ is there.

For example: Assume that a very unlikely thing happens, namely $W_t=0$ on some interval $t\in[0,a]$. I know that this will Almost Surely not happen, but assume it nontheless (I think the same conclusion should apply if $W_t$ simply coincidentally stays very very close to $0$).

Then we have a situation where $dW_t = 0$ over this interval, and therefore $dX_t=\sigma X_t*0=0$ over this interval. Hence $X_t$ should remain constant over this interval.

Nevertheless, we have $X_a=X_0e^{-\frac{1}{2}\sigma^2 * a}<X_0$, so that $X_t$ does not remain constant on the interval.

If we let go of the assumption that $W_t$ is exactly equal to $0$, but assume that it fluctuates extremely closely around $0$ for a relatively large amount of time, then we still get the seemingly incorrect result that $X_t$ drops to zero over time.

How do we explain this?

Let me formulate my question in less technical terms: If the change of $X$ depends only on the change of the brownian motion $W$, then why does $X$ go to zero over time according to a factor $e^{-\frac{1}{2}\sigma^2}$, regardless of how the Brownian motion moves? i.e. regardless of how close the Brownian motion remains to zero, $X$ still moves exponentially fast towards zero.

user56834
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3 Answers3

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The first equation you've written, namely $$ dX_t = \sigma X_t dW_t \tag{1}$$ is simply a convenient, yet informal, way to denote the corresponding integral equation, it has no proper understanding apart from that. Without getting into strong versus weak solutions (which have to do with what probability spaces the processes are defined on), we say that $X_t$ solves $(1)$, with initial condition $x\in \mathbb{R}$, if it satisfies $$X_t = x + \int_0^t \sigma X_sdW_s \tag{2}$$ for all $t \geq 0,$ almost surely. That is, solutions to the differential equation are only taken to hold on a set of probability one, so the situation you're describing is a moot point. Moreover, it is improper to think about $dX_t(\omega)$ being a path-wise slope, it is simply a notational shorthand.

David
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  • This is helpful technically, but it does not really answer my main concern (I've reformulated it somewhat at the end of the original question). Even with this integral notation, I cannot see why $X$ moves to zero exponentially fast, regardless of how $W$ behaves. – user56834 Mar 30 '17 at 07:31
  • @programmer2134 this seems to be a very different question then before the edit, or at least I'm reading it very differently, I'll try to update my answer later today. – David Mar 30 '17 at 11:02
  • This was really the "intention" behind my question – user56834 Mar 30 '17 at 11:22
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Ito's formula in differential form is $$df(t,x) = f_t dt + f_x dW + \frac{1}{2} f_{xx} dt$$, let's consider a simple situation that $f=f(x)$, denote $g:=f_x$, one get, in integral form $$I_{Riemann} := \int_0^{W(T)} g(x)dx = \int_0^Tg(W(t))dW(t) + \frac{1}{2} g_x(W(t))dt := I_{Ito} + \Delta$$ where $\Delta$ is the adjustment from Riemann integral to Ito, or, $I_{Ito} = I_{Riemann} - \Delta$.

So now the question is, where is the $\Delta$ from?

Let's simplify it more by taking $g(x) = x$, then $$I_{ito} = \int_0^T W(t)dW(t) = I_{Riemann} - \Delta = \int_0^{W(T)} x dx - \frac{1}{2}T = \frac{1}{2} W^2(T) - \frac{1}{2} T$$ , how can we see it intuitively that $\Delta = \frac{1}{2}T$?

Let $t\in [0,T]$ devided to $n^2$ steps, correspondingly, $x\in [0,W(T)]$;

each $n$ steps we call it a march, in the $k$th march, $x$ moves from $[W(T/n*k)$ to $W(T/n*(k+1))]$;

in each march, there's one step that finished this move, while the rest $n-1$ steps are just oscillation;

we further assume there are $(n-1)/2$ times of oscillation, each oscillation contains 2 steps, one step moving out, and one moves back;

each step moves $s$, then $s\sim N(0,\sqrt{T}/n)$.

Now, due to the definition of Ito's integral (that is similar to a Riemann sum, for each time interval $[t_1, t_2]$, take $W(t_1)*(t_2-t_1)$ as the contribution) the oscillation caused a loss:

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The loss of one osccilation is $s^2$, $$E[s^2] = Var(s) = T/n^2$$

So the total loss is $$\Delta = \lim_{n\mapsto \infty} n\cdot \frac{n-1}{2} E[s^2] = \frac{1}{2}T$$, voila.

So, my intuitive understanding is, Ito integral is in time domain ($t$), while the normal Riemann integral is in the random variable value's domain ($x$ or $W(t)$), the Brownian motion has so many oscillation that the loss could not be ignored, the adjustment is the $\frac{1}{2}f_{xx}dt$ in Ito's formula.

athos
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When dealing with Geometric Brownian Motions, one should consider the main fact underlying Ito's algebra, that is, the fact that the Brownian motion ($W(t)$) accumulates quadratic variation at rate one per unit of time. Informally:

$$ dW_t dW_t = dt. $$

This means that, when computing the differential $df(t, x)$, where $x=W(t)$, the Taylor expansion of such function returns a non zero second order differential associated with the pure second derivative w.r.t. $W(t)$, that is:

$$ df(t, W(t)) = f_t dt + f_{x}dx +\frac{1}{2}f_{xx}dt = -\frac{1}{2} \sigma^2X_tdt + \sigma X_tdW(t) + \frac{1}{2} \sigma^2X_tdt, $$

where $x=W(t)$ and the term $\frac{1}{2}f_{xx}dt = \frac{1}{2} \sigma^2X_tdt$ is the term that is due to the quadratic variation of the Brownian motion $W(t)$.

Going back to the Geometric Brownian Motion

$$ X_t=X_0e^{\sigma W(t)-\frac{1}{2} \sigma^2t}, $$

is now clear how the term $-\frac{1}{2} \sigma^2X_tdt$ compensates the quadratic variation of $W(t)$, so that the the GBM can be driftless, that is, with its differential being a martingale.

NicholasLP
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