Using Itos Lemma To Derive an Ito Stochastic Differential Equation

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.

• It seems that Karson Velez and donnius are likely to be the same person what a strange behavior ...math.stackexchange.com/questions/3670352/… Commented May 13, 2020 at 9:21
• Karson is actually friend who I study with, both of us were struggling with the same question :) Commented May 20, 2020 at 21:01
• Commented Apr 19, 2022 at 12:27

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: $$$$Y_t = Y_0 + \int_{0}^{t} \dfrac{\partial f}{\partial t}(X_u,u) du + \int_{0}^{t} \dfrac{\partial f}{\partial x}(X_u,u) dX_u + \dfrac{1}{2} \int_{0}^{t} \dfrac{\partial^2 f}{\partial x^2}(X_u,u) d[X]_u.$$$$ Using inherent characterisation of stochastic integration and bilinearity of the quadratic variation process, you can calculate that $$d[X]_t = 4X_tdt$$ (only the Brownian motion term contributes to this quadratic variation). Therefore, in differential form, we have: $$$$dY_t = \dfrac{1}{2\sqrt{X_t}} dX_t- \frac{1}{8X_{t}^{\frac{3}{2}}}d[X]_t = \\ \frac{1}{2}\left(b \sqrt{X_t} + \frac{1}{\sqrt{X_t}}\right )dt + dW_t-\frac{1}{2 \sqrt{X_t}}dt,$$$$ i.e. $$$$dY_t =\frac{1}{2} bY_t dt + dW_t.$$$$