Today seems to be question day for me, sorry.

The complex process

$$ dX = i\sigma dW $$

where $i = \sqrt{-1}$ and $dW$ is a standard (real-valued) Brownian motion will have a negative variance correct?

  • $\begingroup$ Are you sure i = square root of -1? $\endgroup$ Jan 23, 2020 at 16:25
  • $\begingroup$ Yes, it's the complex number. I am just trying to see how complex Brownian motions work, and their implications, starting with the basics. $\endgroup$
    – user34971
    Jan 23, 2020 at 16:35
  • $\begingroup$ $X$ is also complex and has a real part and an imaginary part. $\endgroup$
    – Gordon
    Jan 23, 2020 at 18:21

1 Answer 1


$i \times \sigma \times W$ is a solution of your equation. Its variance at time $t$ is equal to $\sigma^2 \times t$ which is positive.

Please check this page for more details about how to compute variance for complex random variables:

Wikipedia: complex random variables

The variance is always a nonnegative real number. It is equal to the sum of the variances of the real and imaginary part of the complex random variable

  • $\begingroup$ Thank you. Yes I think it is almost a matter of definition. The wikipedia page appears to define variance as $\langle Z, \bar{Z} \rangle$, however $\langle Z,Z \rangle$ could give a negative quantity. It does make sense I suppose to define variance as something positive. $\endgroup$
    – user34971
    Jan 24, 2020 at 6:41
  • $\begingroup$ How do you know that the variance will be $\sigma^2 \times t$? what is the formula for variance for the OPs equation? $\endgroup$
    – Trajan
    Feb 6, 2020 at 19:25
  • $\begingroup$ The defintion of variance for complex random variable $X$ is: $var(Re(X))+var(Im(X))$ Have a look on wikipedia link for more details! $\endgroup$ Feb 6, 2020 at 19:35

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