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In a 1-dimensional continuous Brownian diffusion process following the SDE: $$ dx(t) = \sigma dW(t) $$ where $W(t)$ represents the Wiener process, what are the units of $\sigma$?

Given that the probability density of, say, $x(s) = 0$ at time $s>0$, given that at time $0$ $x(0) = 0$, is $0 \sim N(0, \sigma^2 s)$, is $\sigma^2$ in $\frac{x^2}{unit \: time}$?

iq447
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1 Answers1

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Involving irrational function in $\mathcal{N}(0,\sigma^2t)$,

We must have in terms of dimension :

$$ [\sigma^2t]=1$$

so

$$ [\sigma]=T^{-1/2}$$

Conclusion

$W(t)$ Wiener process is of a dimension of $[L.T^{1/2}]$ unit of $m.t^{-1/2}$ ISU and $\sigma$ is of a dimension of $[T^{-1/2}]$ unit of $t^{-1/2}$ ISU.

ISU means International System of Unit.

EDX
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  • W itself is in units of length (or whatever $x$'s units are). It is $dW$ whose units involve time. – Ian Jan 26 '21 at 20:00
  • Actually $dW$ is a differential form of $W$ so it is same unit as $W$. Proof : the differential is $\sum \frac{\partial W}{\partial x_i}d{x_i}$ – EDX Jan 27 '21 at 09:58
  • Speaking more carefully, there are different ways to group units depending on how you do the scaling analysis. You can set it up so $\sigma$ is nondimensional, for example, and then $dW$ is just a length (which is a natural thing to want to do, so that the Wiener process itself can be interpreted as the position of a Brownian particle). But to do that you have to scale time up front. – Ian Jan 27 '21 at 12:13
  • Interesting thanks ! – EDX Jan 29 '21 at 10:41