I know, that BM formula
$$X_{i+1} = X_{i}+\sqrt{\delta t}Z_i,$$ $$Z_i\sim N(0,1),$$
with $X_0\equiv0$ leads to Diffusion equation $$u_t=-\frac{1}{2}u_{xx}$$
for marginal distributions in $X_i=X(t),t=i\delta t,$ in the limit case $\delta t\to 0$.
I want to derive Diffusion equation for Brownian motion with a systematic drift component, i.e.
$$X_{i+1} = X_{i}+\delta tX_{i}+\sqrt{\delta t}Z_i,$$
using Chapmann-Kolmogorov equation and Taylor expression.
How can i derive this? What is the answer?
Any explanation is very appreciated.
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To my calculations:
Let density function for BM $\pi_t (x)$ in time $t$ be defined as $$\pi_t (x)=\frac{1}{\sqrt{2\pi t}} exp \left (\frac{-x^2}{2t}\right).$$
According to properties of BW increments we consider Chapmann-Kolmogorov equation as follow: $$\pi_{t+\delta t} (x')=\int {\frac{exp \left (\frac{-(x'-x)^2}{2\delta t}\right)}{\sqrt{2\pi \delta t}}\pi_t (x) dx}.$$
$x = x'+ \Delta x, \Rightarrow dx\rightarrow d \Delta x,$ $$\Rightarrow \pi_{t+\delta t} (x')= \int {\frac{exp \left (\frac{-(\Delta x)^2}{2\delta t}\right)}{\sqrt{2\pi \delta t}}\pi_t (x'+\Delta x) d\Delta x}.$$
Since $\Delta x = \delta tX_{i}+\sqrt{\delta t}Z_i$ we have Taylor expansion: $$\pi_t (x'+\Delta x) = \pi_t (x') + \frac{d\pi_t(x')}{dx} \Delta x + ... + O(\Delta x^4).$$
Then $$\pi_{t+\delta t} (x')= \int {\frac{exp \left (\frac{-(\Delta x)^2}{2\delta t}\right)}{\sqrt{2\pi \delta t}}\pi_t (x'+\Delta x) d\Delta x}$$
$$=\pi_t (x')+\frac{d\pi_t}{dx} \delta t + \frac{1}{2} \frac{d^2\pi_t}{dx^2}\delta t+\frac{1}{6} \frac{d^3\pi_t}{dx^3}3 \delta t^2 + O(\delta t^3);$$
$$\frac{\pi_{t+\delta t} (x') - \pi_t (x')}{\delta t}=\frac{d\pi_t}{dx} + \frac{1}{2} \frac{d^2\pi_t}{dx^2}+\frac{1}{2}\frac{d^3\pi_t}{dx^3}\delta t + O(\delta t^2);$$
$\delta t \rightarrow 0:$
$$\frac{d}{dt}\pi_t (x')=\frac{d\pi_t}{dx} + \frac{1}{2} \frac{d^2\pi_t}{dx^2}.$$