Please show me how to solve (2) with computation processes.(1) was the initial question which I solved. I show the answers (1) below.
Consider the equation
\begin{eqnarray}
dr_t=(\alpha – \beta r_t ) dt + \sigma \sqrt{r_t} dB_t
\end{eqnarray}
which models the variations of the short rate process $r_t$, where $\alpha, \beta, \sigma $ and $r_0$ are positive parameters. $B_t$ is a S.B.M.
(1) Compute $E[r_t | \mathcal{F}_s ]$ , $0 \le s \le t$.
(2) Compute $E[r_t^2 | \mathcal{F}_s ]$, $0 \le s \le t$.
(Thank you for your help in advance.)
(1) My answer (Revised the Constant Term for $time=t$)
\begin{eqnarray}
r_t &=& r_0 + \alpha t – \beta \int^t_0 r_u du + \sigma \int^t_0 \sqrt{r_u} dB_u\\
E[r_t | \mathcal{F}_s ] &=& E\left[r_0 + \alpha t – \beta \int^t_0 r_u du + \sigma \int^t_0 \sqrt{r_u} dB_u | \mathcal{F}_s \right] \\
&=& r_0 + \alpha t – \beta \int^t_0 E\left[ r_u | \mathcal{F}_s \right]du + \sigma \int^s_0 \sqrt{r_u} dB_u + \sigma \int^t_s \sqrt{r_u} E\left[ dB_u | \mathcal{F}_s \right] \\
&=& \left( r_0 + \sigma \int^s_0 \sqrt{r_u} dB_u \right)+ \alpha t – \beta \int^t_0 E\left[ r_u | \mathcal{F}_s \right]du
\end{eqnarray}
- (Revised the Constant Term for $time=t$) Let $u_t = E[r_t | \mathcal{F}_s ]$ with a condition of $u_s=r_s$, namely $u_s = E[r_s | \mathcal{F}_s ]=r_s$. Then, above equations come to below (O.D.E).
\begin{eqnarray}
E[r_t | \mathcal{F}_s ] &=& \left( r_0 + \sigma \int^s_0 \sqrt{r_u} dB_u \right) + \alpha t – \beta \int^t_0 E\left[ r_u | \mathcal{F}_s \right]du \\
u_t &=& \left( r_0 + \sigma \int^s_0 \sqrt{r_u} dB_u \right)+ \alpha t – \beta \int^t_0 u_u du \\
du_t&=& \alpha dt – \beta u_t dt \\
\frac{du_t}{dt}&=& \alpha – \beta u_t
\end{eqnarray}
-
(1st Step) Consider $u_t/dt = 0 – \beta u_t $.
\begin{eqnarray}
\frac{du_t}{dt}&=& – \beta u_t\\
%\int^t_0 \frac{du_s}{u_s} &=& – \beta \int^t_0 ds\\
%\log \frac{u_t}{u_0} &=& – \beta t\\
%\frac{u_t}{r_0} &=& e^{- \beta t} \\
u_t &=& r_0 e^{- \beta t}
\end{eqnarray} -
(2nd Step) Adjust $u_t$ satisfies the condition of $u_s=r_s$
\begin{eqnarray}
u_t &=& r_0 \left(1 – e^{- \beta (t-s)} \right) + r_s e^{- \beta (t-s)}
\end{eqnarray} - (3rd Step) Consider $u_t/dt = \alpha – \beta u_t $. Adjust $u_t$ as below.
\begin{eqnarray}
u_t &=& \frac{\alpha}{\beta} \left(1 – e^{- \beta (t-s)} \right) + r_s e^{- \beta (t-s)}
\end{eqnarray}
- This satisfies the initial O.D.E. One reaches the solution as below.
\begin{eqnarray}
E[r_t | \mathcal{F}_s ] &=&u_t \\
&=& \frac{\alpha}{\beta} \left(1 – e^{- \beta (t-s)} \right) + r_s e^{- \beta (t-s)}
\end{eqnarray}
$\square$
(2) Thank you for your help in advance.
Best Answer
The ODE approach is fine, but involved. There is a more convenient way to derive the results, especially so for $E_s[r_t^2]$. The trick is to work with $x_t=e^{\beta t}r_t$ and its corresponding sde is simply,
$$dx_t=e^{\beta t}(\alpha dt +\sigma r_t^{\frac{1}{2}} dB_t) $$
Its solution in terms of $x_s$ can then be written as,
$$x_t= x_s + \alpha \int_s^t e^{\beta \tau} d\tau + z_{s,t} $$
where $E_s[z_{s,t}]=0 $. Therefore,
$$E_s[x_t]= x_s + \frac{\alpha}{\beta}(e^{\beta t}-e^{\beta s} ). \tag{1} $$
The process for $x_t^2$ according to Ito's formula is,
$$dx_t^2=e^{\beta t}x_t(2\alpha + \sigma^2) dt + 2\sigma e^{2\beta t} r_t^{\frac{3}{2}} dB_t $$
It's solution in terms of $x_s$ is,
$$x_t^2= x_s^2 + (2\alpha + \sigma^2) \int_s^t e^{\beta \tau}x_{\tau} d\tau + zz_{s,t} $$
where, again, $E_s[zz_{s,t}]=0 $. Then, its expectation becomes,
$$E_s[x_t^2]= x_s^2 + (2\alpha + \sigma^2) \int_s^t e^{\beta \tau}E_s[x_{\tau}] d\tau . $$
Plugging the result (1) to obtain,
$$E_s[x_t^2]= x_s^2 + (2\alpha + \sigma^2) \int_s^t e^{\beta \tau} \left[ x_s + \frac{\alpha}{\beta}(e^{\beta \tau}-e^{\beta s} )\right] d\tau . $$
Carrying out the integration to get,
$$E_s[x_t^2]= x_s^2 + (2\alpha + \sigma^2) \left[ \frac{1}{\beta}(e^{\beta t}-e^{\beta s} )x_s + \frac{\alpha}{2\beta^2}(e^{\beta t}-e^{\beta s} )^2 \right] \tag{2}$$
In the end, rewrite (1) and (2) in terms of the original variable $r_t$ to get the final results,
$$E_s[r_t]= e^{-\beta (t-s)}r_s + \frac{\alpha}{\beta} \left[1-e^{-\beta (t- s)} \right]$$
$$E_s[r_t^2]= e^{-2\beta (t-s)}r_s^2 + (2\alpha + \sigma^2) \left[ \frac{1}{\beta}(e^{-\beta (t-s)}-e^{-2\beta (t-s)} )r_s + \frac{\alpha}{2\beta^2}(1-e^{-\beta(t-s)} )^2 \right].$$