I have a Gaussian pdf defined as $$f_X(x) =\frac{1}{\sigma\sqrt{2\pi}}\exp\left\{-\frac{(x-\mu)^2}{2\sigma^2}\right\}$$
whose $\mu = \frac{d^2}{6D}$, where $d$ is distance parameter and $D$ is the diffusion coefficient.
How can $\sigma^2$ be derived from $\mu$? And will it be a function of $d$ and $D$ as well?
This is what I have come up with.
To my understanding, $$\sigma^2=Var(X)$$ $$=E(X^2)-E(X)^2$$ where $E(X)^2 = \mu^2 = (\frac{d^2}{6D})^2$.
How can I proceed from here? How can $=E(X^2)$ be obtained?
or Is this a wrong approach? If so, what should I do?
The mean and variance parameters in a normal distribution are both free parameters, so setting the mean parameter does not impose any restrictions on the variance parameter. Hence, without more information on any outside restrictions, there is no way to "derive" the variance from the mean. It appears in your question that you have a formula for the mean parameter that comes from exogenous variables. Unless you also have a formula for the variance from these exogenous variables (which is not specified), you are out of luck.