SDE $Y_t=Y_0+W_t +\int_{0}^t \gamma(Y_s) ds$

Question

• Let $a,b :\mathbb{R} \to \mathbb{R}$
• Suppose $a$ to be of class $C^1$ and $b$ continuous.
• Suppose that $a(x) >0$ for every $x \in [0,T]$
• Let $h :\mathbb{R} \to \mathbb{R}$ defined by $h(x)= \int_{0}^x \frac{1}{a(y)}dy$
• Suppose $ \int_{- \infty}^0 \frac{1}{a(y)}dy = \int_{0}^ {\infty} \frac{1}{a(y)}dy$
• $\gamma= \left( \frac{b}{a} - \frac{a'}{2} \right) \circ h ^ {-1}$

Let $X$ be a solution of $X_t =X_0 + \int_{0}^t a(X_s) dW_s + \int_{0}^t b(X_s) ds, t \in [0,T]$

1. Show that $h : \mathbb{R} \to \mathbb{R}$ is a bijection of class $C^2$
2. Show that $Y=h(X_t)$ solves $Y_t=Y_0+W_t +\int_{0}^t \gamma(Y_s) ds$
3. Determine $\tilde{b} : [0,T] \times \mathbb{R}$ such that $X$ solves $X_t= X_0 + \int_{0}^t a(X_s) \circ dWs + \int_{0}^t \tilde{b}(X_s) ds , t \in [0,T]$,

$\circ$ means here the Stratonovitch equation.

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1. $h'= \frac{1}{a} >0$ sp $h$ strictly increasing so bijective, and $\frac1a$ derivable, so $h$ is class $C^²$

$ \begin{align*} dY_t&= h'(X_t) + \frac12 h''(X_t) d[X]_t \\ &= h'(X_t) dX_t +\frac12 \left( - \frac{a'(X_t)}{a^2(X_t)} a^2(X_t) dt \right) \\ &= \frac{a(X_t)}{a(X_t)} dW_t + [ \dfrac{b(X_t) }{a(X_t)} - \dfrac{a'(X_t)}{2} ] ds\\ &= dW_t + [ \frac{b}{a} - \frac{a'}{2}](X_t) dt \\ &= dW_t + [ \frac{b}{a} - \frac{a'}{2}]\circ h^{-1}(Y_t) dt \\ \end{align*} $ \

Answer 1

$ \begin{align*} [a(X),W]_t&= \int_{0}^t a'(X_s) d[X,W]\\ &= \int_{0}^t a'(X_s) a(X_s)d[W,W] \\ &=\int_{0}^t a'(X_s) a(X_s)ds \\ \int_{0}^{t} a(X_s) \circ dW&=\int_{0}^{t} a(X_s) dWs + \dfrac12 [a(X),W]_s\\ &=\int_{0}^{t} a(X_s) dWs + \dfrac12 a'(X_s) a(X_s)ds \\ X_t&= X_0+ \int_{0}^{t} a(X_s) dWs + \dfrac12 \int_0^t a'(X_s) a(X_s)ds - \dfrac12 \int_0^t a'(X_s) a(X_s)ds+\int_0^t b(X_s)ds \\ &=X_0+ \int_{0}^{t} a(X_s) \circ dWs + \int_0^t - \dfrac12 a'(X_s) a(X_s) + b(X_s) ds \\ \tilde{b}(X_s)&= - \dfrac12 a'(X_s) a(X_s) + b(X_s) \end{align*} $

Is it correct ? Thanks.