I need to solve the SDE: $$ dX_t = (X_t)^3 dt + (X_t)^2 dW_t ; X(0)=1 $$

Now what I found is this is an SDE of the form: $$dXt =a(X_t)dt+b(X_t)dW_t$$ where $a(x) = \frac{1}{2} b(x)b′(x)$

Using the substitution $y = h(x) = \int_{x} {\frac{ds}{b(s)}}$

we get the reduced linear scalar SDE $dY_t = dW_t$

Hence $X_t= \frac{1}{1-W_t}$.

Now my problem is I am getting $dY_t=X_t dt + dW_t$. Can anyone explain how is $dY_t = dW_t$?


Set $a(x) := x^3$ and $b(x) := x^2$, then

$$Y_t := \int_{X_0}^{X_t} \frac{ds}{b(s)} = \left[ - \frac{1}{x} \right]_{X_0}^{X_t} = 1- \frac{1}{X_t}.$$

Now recall that Itô's formula states $$f(X_t) -f(X_0) = \int_0^t f'(X_s) \, dX_s + \frac{1}{2} \int_0^t f''(X_s) b^2(X_s) \, ds$$


$$\int_0^t f'(X_s) \, dX_s = \int_0^t f'(X_s) \, a(X_s) \, ds + \int_0^t f'(X_s) b(X_s) \, dW_s \tag{1}$$

for any "nice" function $f$. For $f(x) := - x^{-1}$, we obtain

$$\begin{align*} Y_t &= 1- \frac{1}{X_t} = f(X_t)-f(X_0) \\ &= \int_0^t X_s^{-2} \, dX_s + \frac{1}{2} \int_0^t (-2 X_s^{-3}) b^2(X_s) \, ds \\ &= \int_0^t X_s^{-2} \, dX_s - \int_0^t X_s \, ds. \end{align*}$$

Finally, using $(1)$, we get

$$Y_t = \int_0^t \, dW_s + \underbrace{ \int_0^t X_s \, ds - \int_0^t X_s \, ds}_{0},$$

i.e. $Y_t = W_t$.


Let $f(x)=\frac{1}{x}$. Then $f'(x)=-\frac{1}{x^{2}}$ and $f''(x)=\frac{2}{x^{3}}$. Now, we use Ito Lemma for function $f(x)$ and stochastic process $X_{t}$ with the following dynamics $$ dX_t =X_{t}^{3} dt + X_{t}^{2} dW_t, \qquad X_{0}=1 $$

We have

$$df(X_{t})=f'(X_{t})dX_{t}+\frac{1}{2}f''(X_{t})dX_{t}dX_{t}$$ $$=-\frac{1}{X_{t}^{2}}\cdot dX_{t}+\frac{1}{2}\cdot\frac{2}{X_{t}^{3}}\cdot dX_{t}dX_{t}$$ $$=-\frac{1}{X_{t}^{2}}\biggl(X_{t}^{3} dt + X_{t}^{2} dW_t\biggr)+\frac{1}{2}\cdot\frac{2}{X_{t}^{3}}\cdot X_{t}^{4}dt=-dW_{t}$$

because $dX_{t}dX_{t}=X_{t}^{4}dt$. It means that


and this is a shorthand notation for the following expression

$$\int_{0}^{t} d\biggl(\frac{1}{X_{s}}\biggr)=-\int_{0}^{t} dW_{s}$$



Now, using that $X_{0}=1$ and $W_{0}=0$, we have


so finally



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.