Deriving the Second Moment of a Mean-Reverting Square-Root Process

# Moment Ito's Process Proof


# Moment Ito's Process Proof












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!!

## Answer by user16651 (score 1, accepted)

https://quant.stackexchange.com/a/31652

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 $$m(t)=\frac{\sigma^2}{\alpha}X_0e^{-\alpha t}+\left(X^2_0-\frac{\sigma^2}{\alpha}X_0\right)e^{-2\alpha t}$$

## Answer by user16651 (score 3)

https://quant.stackexchange.com/a/31649

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 $$\mathbb{E}[X_t^2]=\frac{\sigma^2}{\alpha}X_0e^{-\alpha t}+(X^2_0-\frac{\sigma^2}{\alpha}X_0)e^{-2\alpha t}\tag{11}$$