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I know that $\frac{dW_t}{dt}$, with $W_t$ a brownian motion, does not exist. However, does $\frac{dt}{dW_t}$ exists? Or does it even make sense? I am trying to calculate the quotient of two differentials of Ito processes, specifically, given $V_t$ and $S_t$ Ito processes, I require $\frac{dV_t}{dS_t}$ and the quotient I described appears. I tried to make sense of it by using Ito's rules:

$$ \frac{dt}{dW_t} = \frac{dt \, dW_t}{(dW_t)^2} = \frac{0}{dt} = 0 $$

Is this correct? And if it is, is there an intuition behind it?

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  • $\begingroup$ What are the equations for V and S please? $\endgroup$ – Magic is in the chain May 29 at 17:05
  • $\begingroup$ I think the time derivative of B:M exists in a distributional sense if you wanna learn about that $\endgroup$ – Vlad May 29 at 19:09
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Edits have been made, but the original question asked about $\frac{\mathrm{d}t}{W_t}$.

The random variable does exist

In your [original] question you ask about $\frac{\mathrm{d}t}{W_t}$, although I think you really meant $\frac{\mathrm{d}t}{\mathrm{d}W_t}$. The quantity $\frac{\mathrm{d}t}{W_t}$ does exist and is just a random variable. Be careful though, as for a standard Brownian motion this will be singular at $t = 0$ and I suspect has non zero probability of being singular again within finite time, (I'd need to refresh my knowledge on the properties of Brownian motion). You can do most things you want with this, albeit it's not obvious quite what it's statistics are (I don't know if it even has a mean).

A misinterpretation of SDEs

In your question you say you are after $\frac{\mathrm{d}V_t}{\mathrm{d}W_t}$, but I think you're tackling the problem wrong and misunderstanding SDEs. When we write an ODE we are used to thinking writing something along the lines of $\frac{\mathrm{d}y}{\mathrm{d}t} = f(t)$ and that we can interchangeably write this as $\mathrm{d}y = f(t) \mathrm{d}t$. This tends to give the intuition that we are trying to find a process $y(t)$ which when we differentiate this we get $f(t)$, and that by the fundamental theorem of calculus we have also solved the problem of finding the integral $\int f(t) \mathrm{d}t$. For ODEs were are largely free to switch between these two interpretations, and which one is more correct typically depends on how the problem is set up.

For SDEs there is only one way to write this, and this is as \begin{equation} \mathrm{d}Y_t = f(X_t) \mathrm{d}X_t \end{equation} and never as $\frac{\mathrm{d}Y_t}{\mathrm{d}X_t} = f(X_t)$. (I have not mixed in any finite variation $\mathrm{d}t$ dependence for simplicity). We almost always have some initial condition $Y_0 = y_0$, and hence we must always understand that the SDE as we have written it is strictly shorthand for \begin{equation} Y_t = y_0 + \int_{X_0}^{X_t} f(X_s) \mathrm{d} X_s. \end{equation} Hence when you are dealing with SDEs you almost never have items of the form $\frac{\mathrm{d}\cdot}{\mathrm{d}\cdot}$ (exceptions might be partial derivatives in from Ito's lemma or Radon-Nikodym derivatives).

Abuse of notation

As a side point, your last comment about $(\mathrm{d}W_t)^2 = \mathrm{d}t$ is also an abuse of notation. We also never really produce items of the form $(\mathrm{d}W_t)^2$, but rather $\mathrm{d}\langle W\rangle_t$. This is just another convenient notation, but one where to use it you need to really understand what is going on under the hood, else if you misuse it (as you have) you produce mathematical nonsense. For a good resource on this I recommend Klebaner's book on stochastic calculus.

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    $\begingroup$ Are you sure you've got the first paragraph right? If anything, $\eta(t)$ is $dW_t/dt$ $\endgroup$ – LazyCat May 29 at 16:41
  • $\begingroup$ @LazyCat you're correct, I got muddled in my notes, so will edit this. $\endgroup$ – oliversm May 29 at 17:06

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