SDE $dX_t= f(t)X_t dt + g(t) dW_t$

Question

Let

• $T>0$
• $(W_t)$ is a standard brownian motion adapted to the filtration
• $f : [0,T] \to \mathbb{R} $ in $L¹([0,T])$
• $g: [0,T] \to \mathbb{R}$ in $ L^2([0,T])$
• $R_t= \int_{0}^t \exp \left( - \int_{0}^{r} f(s) ds \right) g(r) dW_r$
• $Y_t= \exp \left( \int_{0}^{t} f(s) ds \right) R_t$ $(\star)$

$ \begin{cases} dX_t &= f(t)X_t dt + g(t) dW_t\\ X_0 &= 0\\ \end{cases} $

1. Show that $R$ is well-defined
2. Show that $Y$ is a gaussian process
3. Show that $ t \mapsto \int_{0}^t f(s) ds$ has bounded variations
4. Show that if $X$ is defined as $(\star)$, then $X$ solves the SDE
5. let $X$ be a solution of the SDE, show that $X=Y$
6. Can we use the usual theorems of existence and uniqueness ?

We want to solve the previous stochastic differential equation below.

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1. $\int_{0}^{T} g(r) dW_r$ is a gaussian random variable with law $\mathcal{N}(0, {\lVert g \rVert}_2 ^2)$ because $g$ is square integrable.

Let $\phi(r)=\exp \left( - \int_{0}^{r} f(s) ds \right) g(r)$,

we have $\phi(r)^2=\exp( - 2 \int_{0}^r f(s) ds) g(r)^2 \leq \exp( 2 {\lVert f \rVert}_1 ) g(r)^2 $

therefore $\int_0^T \phi(r)^2 dr \leq \exp( 2 {\lVert f \rVert}_1 ) {\lVert g \rVert}_2 ^2$

so $R_t$ is well-defined.

2. $R_t$ si a gaussian process,it is a Wiener integral, then we multiply by a deterministic function.

3. $ \psi :t \mapsto \int_{0}^t f(s) ds$ has for derivative $f$ and the total variation $TV(\psi)= \int_{0}^T \mid \psi' \mid =\int_{0}^T \mid f \mid$ and we know that $f \in L^1$

4. We use Ito formula

$ \begin{align*} d(\exp \left( \int_{0}^{t} f(s) ds \right) R_t ) &=d(\exp \left( \int_{0}^{t} f(s) ds \right) ) R_t +\exp \left( \int_{0}^{t} f(s) ds \right) dR_t +[\exp \left( \int_{0}^{t} f(s) ds \right), R]_t \\ &= f(t)\exp \left( \int_{0}^{t} f(s) ds \right) R_t+\exp \left( \int_{0}^{t} f(s) ds \right) \exp \left( - \int_{0}^{t} f(s) ds \right) g(t)dW_t \\ &= f(t)Y_t +g(t) dW_t \\ \end{align*} $

5. done below

6. ????

Answer 1

Following the enhanced Hull-White equation and the constant variation method

we solve $dZ_t=f(t) Z_t dt$ so $Z_t=c e^{ \int_0^{t} f(s)ds }$, then we solve, with the method of the constant variation :

$ \begin{align*} X_t&= C_t e^{ \int_0^{t} f(s)ds } \\ C_t&= X_t e^{- \int_0^{t} f(s)ds } \\ dC_t&=- f(t) e^{- \int_0^{t} f(s)ds } X_t + f(t) X_t e^{- \int_0^{t} f(s)ds } \left( f(t) X_t dt + g(t) dW_t \right) \\ dC_t &= - f(t) e^{- \int_0^{t} f(s)ds } X_t+ f(t) X_t e^{- \int_0^{t} f(s)ds } dt + e^{- \int_0^{t} f(s)ds } g(t) dW_t \\ dC_t &= e^{- \int_0^{t} f(s)ds } g(t) dW_t \\ C_t &= \int_0^{t} e^{- \int_0^{r} f(s)dr } g(t) dW_t\\ X_t &= e^{ \int_0^{t} f(s)ds } \int_0^{t} e^{- \int_0^{r} f(s)dr } g(t) dW_t \\ \end{align*} $

$$\fbox{$X_t = e^{ \int_0^{t} f(s)ds } \int_0^{t} e^{- \int_0^{r} f(s)dr } g(t) dW_t $ }$$