# How to solve the SDE: $dX_t = aX_t dt + b \sqrt{X_t} dW_t$

I am struggling to solve the following SDE, $$dX_t = aX_t dt + b \sqrt{X_t} dW_t$$ where $a$ and $b$ are constants, and $X_0$ is given.

I know to solve $dX_t = aX_t dt + b dW_t$ or $dX_t = a dt + b \sqrt{X_t} dW_t$, but when both $X_t$ and $\sqrt{X_t}$ are introduced, how can I solve it?

• This is a square-root diffusion process, and in finance, it is commonly known as the Cox-Ingersoll-Ross model (CIR) model. It cannot be solved exactly. At best, you can give it's moments like the expected value, variance, and distribution, which I believe is a $\chi^2$ distribution. – Dr. Ikjyot Singh Kohli Mar 4 '18 at 17:03
• Indeed, the complete question is to find $E(X_t)$ and $Var(X_t)$. But when using $d(e^{at}X_t)$, I don't know how to cancel out $X_t$ or $\sqrt{X_t}$ on right hand side. What is the proper way to do that? – SkyDreamer Mar 4 '18 at 17:53
• See answer below. – Dr. Ikjyot Singh Kohli Mar 4 '18 at 18:10

What you can do is the following.

Given $dX = a X dt + b \sqrt{X} dW$ (I have left off the subscript $t$ for ease of typing!), given some initial condition $X(0) = X_0$, the solution to this SDE is found by the following:

Multiply both sides $e^{-at}$ and you get:

$e^{-at} dX = a e^{-at} X dt + b e^{-at} \sqrt{X} dW$,

$e^{-at} dX - ae^{-at} Xdt = be^{-at} \sqrt{X} dW$,

$d[e^{-at} X(t)] = be^{-at} \sqrt{X} dW$

Integrating both sides on some interval, say $[0,t]$, we get:

$e^{-at} X(t) - X_0 = b \int_{0}^{t} e^{-as} \sqrt{X(s)} dW(s)$

You can now use this solution to compute the expectation value and variance.