My problem is the following Ito differential equation: $$\,\mathrm dX_t = (bX_t +1)\,\mathrm dt + 2 \sqrt{X_t} \,\mathrm dW_t$$ where $b$ is a constant.
I need to find the expectation value which I know is in the form $\mu(t) = c_1 e^{bt} +c_2$ where the constants $c_1$ and $c_2$ are written in terms of $\mathbb{E}[X_0] $ and $b$.
My lecture notes don't really explain how I would find the solution and I can't find any similar examples online but my attempt so far is as follows:
$$\mathrm dX_t = (bX_t+1)\,\mathrm dt+2\sqrt{X_t}\,\mathrm dW_t\\
\mathrm dX_t = b\left(\frac{1}{b} + X_t\right)\,\mathrm dt+2\sqrt{X_t}\,\mathrm dW_t$$
Say $Y_t=X_t e^{bt}$ then$$\mathrm dY_t = b X_t e^{bt}\,\mathrm dt+e^{kt}\,\mathrm dX_t= b·\frac{1}{b} e^{kt}\,\mathrm dt + \sigma e^{bt}\,\mathrm dW_t.$$
My idea was to then integrate, which I can then use the fact that the expectation value of a stochastic integral is equal to 0 to find my result. However I don't believe I have done this correct, does anyone have any suggestions?
Thanks.
Best Answer
This might help. As it turns out if you let $Y_t = \sqrt{X_t}$ use can use Ito's formula to show that $$dY_t = \frac{b}{2}Y_tdt + dW_t$$ (see this question). This SDE is not so bad and has the solution $$Y_t = e^{bt/2}\left(Y_0 + \int_0^t e^{-bs/2}dW_s\right).$$
This solution is obtained by the use of an integrating factor (just like with a linear ODE). Observe that moving the $dt$ part to the other side resembles the product rule. This suggests the integrating factor of $e^{-bt/2}$ So, that we have $$e^{-bt/2}dW_t = e^{-bt/2}dY_t -\frac{b}{2}e^{-bt/2}Y_tdt = d\left(e^{-bt/2}Y_t\right)$$ We can then integrate both sides to obtain $$ e^{-bt/2}Y_t = Y_0 + \int_0^t e^{-bs/2}dW_s. $$
If I understand the question we seek $\mathbb{E}X_t$. So, since $X_t = Y_t^2$ \begin{align} \mathbb{E}X_t &= e^{bt}\mathbb{E}\left[\left(Y_0 + \int_0^t e^{-bs/2}dW_s\right)^2\right] \\ &= e^{bt}\mathbb{E}\left[Y_0^2 + 2Y_0\int_0^t e^{-bs/2}dW_s + \left(\int_0^t e^{-bs/2}dW_s\right)^2\right] \\ &=e^{bt}\mathbb{E}\left[X_0\right] + 0 + e^{bt}\mathbb{E}\left[\left(\int_0^t e^{-bs/2}dW_s\right)^2\right]\\ &=e^{bt}\mathbb{E}\left[X_0\right] + e^{bt}\int_0^t e^{-bs}ds\\ &=e^{bt}\mathbb{E}\left[X_0\right] + e^{bt}\int_0^t e^{-bs}ds\\ &=e^{bt}\mathbb{E}\left[X_0\right] - \frac{e^{bt}}{b}\left(e^{-bt} - 1\right)\\ &= \left(\mathbb{E}\left[X_0\right] + \frac{1}{b}\right)e^{bt} -\frac{1}{b} \end{align} In there the fourth equal sign is drawn by the Ito isometry (we have a deterministic integrand).
You will also see the common assumption that the initial value of the process is independent with (of?) the noise.