I have a following stochastic integral - related problem that I have difficulty to solve:

Given \begin{equation} dX_t = -\alpha X_tdt+\sigma\sqrt{X_t}dW_t \end{equation}

and the second moment of $X_t$ is denoted by $m^{(2)}_t=\mathbb{E}(X_t^2)$.

Can you prove that $m^{(2)}_t$ has the following expression: \begin{equation} m^{(2)}_t=\frac{\sigma^2}{\alpha}X_0\exp(-\alpha t)+(X^2_0-\frac{\sigma^2}{\alpha}X_0)\exp(-2\alpha t) \end{equation}

I can give you the expression of $d(X_t^2)$ just to save some time: \begin{equation} d(X_t^2)=-2\alpha X^2_t dt+2\sigma X_t\sqrt{X_t} dWt+\sigma^2 X_t dt \end{equation}

Many thanks!!


Another way \begin{equation} X_t^2=X_0^2-2\alpha \int_{0}^{t}X^2_s ds+2\sigma\int_0^t X_s\sqrt{X_s} dW_s+\sigma^2\int_{0}^{t} X_s ds \end{equation} thus $$\mathbb{E}[X_t^2]=X_0^2-2\alpha\int_{0}^{t}\mathbb{E}[X^2_s] ds+\sigma^2\int_{0}^{t} \mathbb{E}[X_s] ds$$ As you saw in the first answer, $\mathbb{E}[X_s]=X_0^2e^{-\alpha s}$, thus $$\mathbb{E}[X_t^2]=X_0^2-2\alpha\int_{0}^{t}\mathbb{E}[X^2_s] ds-\frac{\sigma^2}{\alpha}X_0^2\left(e^{-\alpha t}-1\right) $$

Set $m(t)=\mathbb{E}[X_t^2]$, we have $$m(t)=X_0^2-2\alpha \int_0^t m(s)ds-\frac{\sigma^2}{\alpha}X_0\left(e^{-\alpha t}-1\right)$$ Take differentiate with respect to time, $$m'(t)=-2\alpha\,m(t)+\sigma^2 X_0\,e^{-\alpha t}$$ In other words $$m'(t)+2\alpha\,m(t)=\sigma^2 X_0\,e^{-\alpha t}$$ This ODE is a First-order equation , $$\mu(t)=e^{\int {2\alpha } dt}=e^{2\alpha t}$$ and $$m(t)=\frac{1}{e^ {2\alpha t}}\left(\sigma^2 X_0 \int e^ {2\alpha t}e^ {-\alpha t}dt+c\right)$$ where $c$ is a constant, we have $$m(t)=e^ {-2\alpha t}\left(\frac{\sigma^2 X_0}{\alpha}e^{\alpha t} +c\right)$$ on the other hand $m(0)=X_0^2$ and $m(0)=\frac{\sigma^2 X_0}{\alpha}$, therefor $c=X_0^2-\frac{\sigma^2 X_0}{\alpha}$. Finall we have $$m(t)=\frac{\sigma^2}{\alpha}X_0e^{-\alpha t}+\left(X^2_0-\frac{\sigma^2}{\alpha}X_0\right)e^{-2\alpha t}$$

  • $\begingroup$ @Donkey_JOHN I solved this question again . Two methods are correct. $\endgroup$ – user16651 Dec 25 '16 at 10:38
  • $\begingroup$ No problem, clear and sound! Thank you for your help! @BehrouzMaleki $\endgroup$ – Donkey_JOHN Dec 25 '16 at 12:11
  • $\begingroup$ Please. good luck $\endgroup$ – user16651 Dec 25 '16 at 12:17

Set $f(t,x)=xe^{\alpha t}\in\mathbb{C}\left([0,\infty)\times\mathbb{R}\right)$. By application of Ito's lemma, we have $$d\left(X_te^{\alpha t}\right)=\alpha e^{\alpha t}X_t dt+e^{\alpha t}dX_t +\underbrace{d[e^{\alpha t},X_t]}_{0}\tag 1$$ thus $$d\left(X_te^{\alpha t}\right)=\sigma e^{\alpha t}\sqrt{X_t}dW_t\,. \tag 2$$ By Integration on $[0,t]$, we have $$X_te^{\alpha t}=X_0+\sigma \int_{0}^{t}e^{\alpha s}\sqrt{X_s}dW_s \tag 3$$ therefore $$X_t=X_0e^{-\alpha t}+\sigma \int_{0}^{t}e^{-\alpha (t-s)}\sqrt{X_s}dW_s \, .\tag 4$$ Now calculate $\mathbb{E}[X_t]$ and $\text{Var}(X_t)$ and apply $$\mathbb{E}[X_t^2]=\text{Var}(X_t)+\mathbb{E}[X_t]^2\tag 6$$ Note $$\mathbb{E}[X_t]=X_0e^{-\alpha t}\tag 7$$ and $$\text{Var}(X_t)=\mathbb{E}\left[\left(\sigma \int_{0}^{t}e^{-\alpha (t-s)}\sqrt{X_s}dW_s\right)^2\right]=\sigma^2 \int_{0}^{t}e^{-2\alpha (t-s)}\mathbb{E}\left[X_s\right]ds\tag 8$$ therefore $$\text{Var}(X_t)=\frac{X_0\sigma^2}{\alpha}(e^{-\alpha t}-e^{-2\alpha t})+X_0^2e^{-2\alpha t}\tag 9$$ More details $$\mathbb{E}[X_t^2]=\text{Var}(X_t)+\mathbb{E}[X_t]^2=\frac{X_0\sigma^2}{\alpha}(e^{-\alpha t}-e^{-2\alpha t})+X_0^2e^{-2\alpha t}+X_0^2e^{-2\alpha t}\tag{10}$$ then $$\mathbb{E}[X_t^2]=\frac{\sigma^2}{\alpha}X_0e^{-\alpha t}+(X^2_0-\frac{\sigma^2}{\alpha}X_0)e^{-2\alpha t}\tag{11}$$

  • $\begingroup$ In equation $8$ I use the Ito's isometry formula. See it en.wikipedia.org/wiki/It%C3%B4_isometry $\endgroup$ – user16651 Dec 24 '16 at 16:31
  • $\begingroup$ Hi can you solve the problem without solving $X_t$? $\endgroup$ – Donkey_JOHN Dec 24 '16 at 16:31
  • $\begingroup$ Hi . No, because we need $\mathbb{E}[X_t]$ $\endgroup$ – user16651 Dec 24 '16 at 16:32
  • $\begingroup$ Solve the integral in equation 8 and insert the result to equation 6 $\endgroup$ – user16651 Dec 24 '16 at 16:36
  • $\begingroup$ The dynamic of the $m^{(2)}_t$ can be expressed as follow: $dm^{(2)}_t=-2\alpha m^{(2)}_t+\sigma^2 m^{(1)}_t$, could you solve directly with this? since you have the expression of $m^{(1)}$ $\endgroup$ – Donkey_JOHN Dec 24 '16 at 16:37

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.