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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 $dY_t = dW_t$?

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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 $$f(X_t) -f(X_0) = \int_0^t f'(X_s) \, dX_s + \frac{1}{2} \int_0^t f''(X_s) b^2(X_s) \, ds$$

where

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

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. $Y_t = W_t$.

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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

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

and this is a shorthand notation for the following expression

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

so

$$\frac{1}{X_{t}}-\frac{1}{X_{0}}=-(W_{t}-W_{0})$$

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

$$\frac{1}{X_{t}}-1=-W_{t}$$

so finally

$$X_{t}=\frac{1}{1-W_{t}}$$

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