Bivariate Normal Distribution
If \(\small x,y\) follow a normal distribution with correlation coefficient \(\small \rho\), mean \(\small \mu_1,\mu_2\), and variance \(\small \sigma_1,\sigma_2\), the probability density function is given by
\[ \small p(x,y) = \frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\exp\left(-\frac{1}{2(1-\rho^2)}\left[\frac{(x-\mu_1)^2}{\sigma_1^2}+\frac{(y-\mu_2)^2}{\sigma_2^2}-2\rho\frac{x-\mu_1}{\sigma_1}\frac{y-\mu_2}{\sigma_2} \right]\right). \]
This can be obtained by steadily calculating the probability density function of the multivariate normal distribution:
\[ \small p(X) = \frac{1}{\sqrt{(2\pi)^n|\Sigma|}}\exp\left(-\frac{1}{2}(X-\mu)\Sigma^{-1}(X-\mu)^T \right) \]
by setting the variance-covariance matrix \(\small \Sigma\) as:
\[ \small \Sigma = \left[\begin{array}{cc} \sigma_1^2 & \rho\sigma_1\sigma_2 \\ \rho\sigma_1\sigma_2 & \sigma_2^2 \end{array} \right]. \]
As a special case, the above probability density function cannot be used to calculate the case where \(\small \rho = \pm 1\). In this case, once \(\small x\) is determined, \(\small y\) is definite, so
\[ \small \begin{align} &\rho=1 \Rightarrow x = \mu_1+\epsilon_1, \;y = \mu_2+\frac{\sigma_2}{\sigma_1}\epsilon_1 \Rightarrow y = \mu_2 + \frac{\sigma_2}{\sigma_1}(x-\mu_1) \\ &\rho=-1 \Rightarrow x = \mu_1+\epsilon_1, \;y = \mu_2-\frac{\sigma_2}{\sigma_1}\epsilon_1 \Rightarrow y = \mu_2 – \frac{\sigma_2}{\sigma_1}(x-\mu_1) \ \end{align} \]
holds in each case. In these cases, the probability density function could be calculated as:
\[ \small p_{\rho=\pm 1}(x,y) = \frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(x-\mu_1)^2}{2\sigma_1^2} \right)1_{\left\{y=\mu_2 \pm\frac{\sigma_2}{\sigma_1}(x-\mu_1)\right\}}. \]
This relationship can also be inverted and expressed as:
\[ \small p_{\rho=\pm 1}(x,y) = \frac{1}{\sqrt{2\pi\sigma_2^2}}\exp\left(-\frac{(y-\mu_2)^2}{2\sigma_2^2} \right)1_{\left\{x=\mu_1 \pm\frac{\sigma_1}{\sigma_2}(y-\mu_2)\right\}}. \]
Complex Normal Distribution
Since the probability distribution of the normal distribution of two variables is given in the previous section, we can consider the probability distribution for complex numbers in a similar way if we put \(\small z=x+iy\). Since the calculations become complicated in general settings, we assume \(\small \rho=0\) and \(\small \sigma = \sigma_1=\sigma_2\). The problem is to find the probability distribution that \(\small z\) follows when the probability distribution for complex numbers is:
\[ \small \begin{align*} &x = \mu_1 + \epsilon_1, \quad \epsilon_1\sim N(0, \sigma^2) \\ &y = \mu_2 + \epsilon_2, \quad \epsilon_2\sim N(0, \sigma^2). \end{align*} \]
Note that the probability distribution of complex numbers is essentially a bivariate probability distribution, and is therefore expressed as the joint probability distribution of \(\small z,z^\ast\).
Because the relationship between the real part, the imaginary part, and the complex number is
\[ \small \begin{align*} &x = \frac{z+z^\ast}{2} \\ &y = \frac{z-z^\ast}{2i}, \ \end{align*} \]
simply substituting the formula for the probability density function of the bivariate normal distribution gives
\[ \small p(z,z^\ast) = \frac{1}{2\pi\sigma^2} \exp\left(-\frac{1}{2}\left[\frac{(\frac{z+z^\ast}{2}-\mu_1)^2}{\sigma^2}+\frac{(\frac{z-z^\ast}{2i}-\mu_2)^2}{\sigma^2} \right]\right). \]
To simplify the equation, let define
\[ \small \begin{align*} &w = z-\mu_1-\mu_2i=z-\mu_z \\ &w^\ast = z^\ast-\mu_1+\mu_2i=z^\ast-\mu_{z^\ast}, \end{align*} \]
it can be replaced with
\[ \small \begin{align*} \frac{z+z^\ast}{2}-\mu_1 = \frac{w+w^\ast}{2} \\
\frac{z-z^\ast}{2i}-\mu_2 = \frac{w-w^\ast}{2i}. \end{align*} \]
Hence,
\[ \small \frac{(\frac{z+z^\ast}{2}-\mu_1)^2}{\sigma^2}+\frac{(\frac{z-z^\ast}{2i}-\mu_2)^2}{\sigma^2}=\frac{ww^\ast}{\sigma^2} = \frac{(z-\mu_z)(z^\ast-\mu_{z^\ast})}{\sigma^2} \]
holds true. The final probability distribution can be summarized as:
\[ \small p(z,z^\ast) = \frac{1}{2\pi\sigma^2} \exp\left(-\frac{(z-\mu_z)(z^\ast-\mu_{z^\ast})}{2\sigma^2}\right). \]
When considering the standard normal distribution for complex numbers, it seems that \(\small \sigma=1/\sqrt{2}\), and the probability density function of the complex standard normal distribution is expressed as:
\[ \small p(z,z^\ast) = \frac{1}{\pi} \exp\left(-|z|^2 \right). \]
Complex Brownian Motion
Let us express a complex random variable in the form of a stochastic process and consider a stochastic process:
\[ \small \begin{align*} &dx(t) = \mu_1 dt + \sigma dB_1, \quad B_1(t)\sim N(0, t) \\
&dy(t) = \mu_2 dt + \sigma dB_2, \quad B_2(t)\sim N(0, t). \end{align*} \]
Let define
\[ \small \mu_z = \mu_1 + \mu_2i \]
and it is easy to guess that
\[ \small p(z,z^\ast,t) = \frac{1}{2\pi\sigma^2t} \exp\left(-\frac{(z-\mu_z t)(z^\ast-\mu_{z^\ast}t)}{2\sigma^2t}\right) \]
is the probability density function of complex Brownian motion. Let us try to find the partial differential equation that this probability distribution satisfies.
If we calculate the partial differential coefficient, we get
\[ \small \begin{align*} &\frac{\partial p(z,z^\ast, t)}{\partial t} = \Biggl(-\frac{1}{t}+\frac{\mu_{z}(z^\ast-\mu_{z^\ast} t)+\mu_{z^\ast}(z-\mu_z t)}{2\sigma^2t} \\ &\qquad\qquad\qquad\quad+\frac{\sigma^2}{2}\frac{(z-\mu_z t)(z^\ast-\mu_{z^\ast} t)}{\sigma^4t^2} \Biggr)p(z,z^\ast,t) \\ &\frac{\partial p(z,z^\ast, t)}{\partial z} = -\frac{z^\ast-\mu_{z^\ast} t}{2\sigma^2t}p(z,z^\ast,t) \\ &\frac{\partial p(z,z^\ast, t)}{\partial z^\ast} = -\frac{z-\mu_{z} t}{2\sigma^2t}p(z,z^\ast,t) \\ &\frac{\partial^2 p(z,z^\ast, t)}{\partial z\partial z^\ast} = \left(-\frac{1}{2\sigma^2t}+\frac{(z-\mu_{z} t)(z^\ast-\mu_{z^\ast} t)}{4\sigma^4t^2}\right)p(z,z^\ast,t). \end{align*} \]
Therefore,
\[ \small \frac{\partial p(z,z^\ast, t)}{\partial t} = -\mu_z\frac{\partial p(z,z^\ast, t)}{\partial z}-\mu_{z^\ast}\frac{\partial p(z,z^\ast, t)}{\partial z^\ast}+2\sigma^2\frac{\partial^2 p(z,z^\ast, t)}{\partial z\partial z^\ast} \]
is the Fokker-Planck equation for complex Brownian motion. Because the dimension is two-dimensional and the volatility of a complex number is usually multiplied by \(\small 1/\sqrt{2}\), but it is calculated as 1, the coefficient multiplied by the second derivative is 2. I regretted not calculating \(\small \rho\) and \(\small \sigma_1,\sigma_2\) at this point, but I think the Fokker-Planck equation including these probably won’t be a neat equation…
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