# Process with negative quadratic variation

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?

• Are you sure i = square root of -1? – David Duarte Jan 23 '20 at 16:25
• Yes, it's the complex number. I am just trying to see how complex Brownian motions work, and their implications, starting with the basics. – Frido Rolloos Jan 23 '20 at 16:35
• $X$ is also complex and has a real part and an imaginary part. – Gordon Jan 23 '20 at 18:21

## 1 Answer

$$i \times \sigma \times W$$ 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

• 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. – Frido Rolloos Jan 24 '20 at 6:41
• How do you know that the variance will be $\sigma^2 \times t$? what is the formula for variance for the OPs equation? – Trajan Feb 6 '20 at 19:25
• 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! – Valometrics.com Feb 6 '20 at 19:35