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?
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$\begingroup$ Are you sure i = square root of -1? $\endgroup$– David DuarteJan 23, 2020 at 16:25
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$\begingroup$ Yes, it's the complex number. I am just trying to see how complex Brownian motions work, and their implications, starting with the basics. $\endgroup$– user34971Jan 23, 2020 at 16:35
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$\begingroup$ $X$ is also complex and has a real part and an imaginary part. $\endgroup$– GordonJan 23, 2020 at 18:21
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1 Answer
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$i \times \sigma \times W$ 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
answered Jan 23, 2020 at 20:06
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$\begingroup$ 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. $\endgroup$– user34971Jan 24, 2020 at 6:41
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$\begingroup$ How do you know that the variance will be $\sigma^2 \times t$? what is the formula for variance for the OPs equation? $\endgroup$– TrajanFeb 6, 2020 at 19:25
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$\begingroup$ 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