Consider Hull White model $dr(t)=[\theta(t)-\alpha(t)r(t)]dt+\sigma(t)dW(t)$ when we solve the SDE above we have $r(t)=e^{-\alpha t}r(0)+\frac{\theta}{\alpha}(1-e^{-\alpha t})+\sigma e^{-\alpha t}\int_{0}^{t}e^{\alpha u}dW(u) $ and when we take expectation and variance we have $r(t) \sim N(e^{-\alpha t}r(0)+\frac{\theta}{\alpha}(1-e^{-\alpha t}),\frac{\sigma^2}{2\alpha}(1-e^{-\alpha t}))$.

I know the calculate how find SDE and find expectation or variance but I don't understand why $r(t)$ has normal distribution.



2 Answers 2


For simplicity, we assume that $\alpha$ is a positive constant. You need to show that, for any $t>0$, \begin{align*} M_t = \int_0^t e^{\alpha u} dW_u \end{align*} is normally distributed, where $\{W_t, \, t \ge 0\}$ is a standard Brownian motion with respect to the filtration $\{\mathscr{F}_t,\, t \ge 0\}$. Here, we employ the time-changed Brownian motion technique. For $t\ge 0$, let $\mathscr{G}_t = \mathscr{F}_{\frac{1}{2}\ln(1+2t)}$. Consider the process $X=\{X_t, t \geq 0\}$, where \begin{align*} X_t = \int_0^{\frac{1}{2}\ln(1+2t)} e^{\alpha u} dW_u. \end{align*} Then $X$ is a continuous martingale with respect to the filtration $\{\mathscr{G}_t,\, t \ge 0\}$. Moreover, \begin{align*} \langle X, X\rangle_t &= \langle M, M\rangle_{\frac{1}{2}\ln(1+2t)}\\ &=\int_0^{\frac{1}{2}\ln(1+2t)} e^{2u} du =t. \end{align*} By Levy's martingale characterization of Brownian motion, $\{X_t, t \ge 0\}$ is a Brownian motion. That is, for $t >0$, $X_t$ is normally distributed. Consequently, for any $t >0$, \begin{align*} M_t &= \int_0^t e^{\alpha u} dW_u\\ &=X_{\frac{1}{2}(e^{2t}-1 )} \end{align*} is normally distributed, and $r_t$ is also normally distributed.

  • $\begingroup$ $X_t$ is the product of a function of $t$ and an Ito integral wich is a martingale. How you say that $X_t$ is a martingale too? $\endgroup$
    – ab94
    Commented Sep 5, 2019 at 17:48
  • $\begingroup$ Thanks @ab94. There are indeed some mistakes in the previous answer. See the revision above. $\endgroup$
    – Gordon
    Commented Nov 26, 2019 at 19:20

This is a special case of the question of why $$ \int_0^T f(t) dW_t $$ is normally distributed for a continuous function $f(t).$ This Ito integral can be approximated by a sum $$ \sum_{i=0}^{N-1} f(i T/N) (W_{(i+1)T/N} - W_{i T/N}) .$$ The Brownian increments $(W_{(i+1)T/N} - W_{i T/N})$ are independent normally distributed random variables. The key point is that the sum of independent normally distributed variables is again normally distributed.

  • 4
    $\begingroup$ The second key point is that if a sequence of normal variable converges in law then the limit is also normal. $\endgroup$
    – AFK
    Commented May 18, 2015 at 21:17
  • 2
    $\begingroup$ The sum, which is normal, converges to the ito integral in $L^2(P)$, and the key point is that the $L^2$ limit of a sequence of normal random variables is still normal. $\endgroup$
    – Gordon
    Commented May 18, 2015 at 22:36

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