So we have an Ito Stochastic Differential Equation with $b$ as a constant: $$dX_t = (bX_t +1)dt +2 \sqrt{X_t}dW_t $$

I then am told to let $Y_t = \sqrt{X_t} $ and thus derive the Ito stochastic differential equation $dY_t = A(Y_t) dt + B(Y_t)dW_t$ and to then determine $A(Y_t), B(Y_t)$.

I can see that this is supposed to be an application of Ito's lemma. Which states that if we have the Ito Stochastic Differential Equation $X_t$, we could then define a new Ito stochastic process on $Y)y = g(X_t,t)$ which obeys $dY_t=dg(X_t,t)=d\sqrt{X_t}$.

Anything further from this is where my notes stop and I can't see any examples of how to actually implement this lemma?

I thought of substituting $Y_t$ into the equation to get: \begin{align}dY_t & = A(Y_t)dt + B(Y_t)dW_t \\ & = b \sqrt{X_t}dt+ \sqrt{X_t}dX_t \\ &= b(\frac{1}{b})\sqrt{X_t}dt + \sigma \sqrt{X_t}dW_t \\ & = b(\frac{1}{b}\sqrt{X_t}dt + (\sqrt{X_t})^2dW_t \end{align}

Although I am pretty sure this is incorrect. Can anyone help, thank you.


1 Answer 1


To see how this follows directly from Itô, write $f(x,t)=\sqrt{x}$. Then, the partial derivatives are: $f_t= 0$, $f_x=\frac{1}{2\sqrt{x}}$ and $f_{xx}=-\frac{1}{4x^{3/2}}$. So since $f : \mathbb{R}^2 \to \mathbb{R}$ is $C^2$, and $X_t$ is semimartingale, so is $Y_t = f(X_t,t)$, with: \begin{equation} Y_t = Y_0 + \int_{0}^{t} \dfrac{\partial f}{\partial t}(X_u,u) du + \int_{0}^{t} \dfrac{\partial f}{\partial x}(X_u,u) dX_u + \dfrac{1}{2} \int_{0}^{t} \dfrac{\partial^2 f}{\partial x^2}(X_u,u) d[X]_u. \end{equation} Using inherent characterisation of stochastic integration and bilinearity of the quadratic variation process, you can calculate that $d[X]_t = 4X_tdt$ (only the Brownian motion term contributes to this quadratic variation). Therefore, in differential form, we have: \begin{equation} dY_t = \dfrac{1}{2\sqrt{X_t}} dX_t- \frac{1}{8X_{t}^{\frac{3}{2}}}d[X]_t = \\ \frac{1}{2}\left(b \sqrt{X_t} + \frac{1}{\sqrt{X_t}}\right )dt + dW_t-\frac{1}{2 \sqrt{X_t}}dt, \end{equation} i.e. \begin{equation} dY_t =\frac{1}{2} bY_t dt + dW_t. \end{equation}


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.