Tag Archives: JCM_math545_HW4_S23

SDE Example: quadratic geometric BM

Show that the solution \(X_t\) of

\[ dX_t=X_t^2 dt + X_t dB_t\]

where \(X_0=1\) and \(B_t\)  is a standard Brownian motion has the representation

\[ X_t = \exp\Big( \int_0^t X_s ds -\frac12 t + B_t\Big)\]

Ornstein–Uhlenbeck process

For \(\alpha \in \mathbf R\) and \(\beta >0\),  Define \(X_t\) as the solution to the following SDE
\[dX_t = – \beta X_t dt + \alpha dW_t\]

where \(W_t\) is a standard Brownian Motion.

  1.  Find \( d(e^{\beta t} X_t)\) using Ito’s Formula.
  2. Use the calculation of   \( d(e^{\beta t} X_t)\) to show that
    \begin{align}  X_t = e^{-\beta t} X_0 + \alpha \int_0^t e^{-\beta(t-s)} dW_s\end{align}
  3. Conclude that \(X_t\) is Gaussian process (see exercise: Gaussian Ito Integrals ). Find its mean and variance at time \(t\).
  4. * Let \(h(t)\) and \(g(t)\) be  deterministic functions of time and let \(Y_t\) solve
    \[dY_t = – \beta Y_t dt + h(t)dt+ \alpha g(t) dW_t\]
    show find a formula analogous to part 2 above for \(Y_t\) and conclude that \(Y_t\) is still Gaussian. Find it mean and Variance.

Homogeneous Martingales and BDG Inequality

Part I

  1. Let \(f(x,y):\mathbb{R}^2 \rightarrow \mathbb{R}\) be a twice differentiable function in both \(x\) and \(y\). Let \(M(t)\) be defined by \[M(t)=\int_0^t \sigma(s,\omega) dB(s,\omega)\]. Assume that \(\sigma(t,\omega)\) is adapted and that \(\mathbb{E} M^2 < \infty\) for all \(t\) a.s. .(Here \(B(t)\) is standard Brownian Motion.) Let \(\langle M \rangle(t)\) be the quadratic variation process of \(M(t)\). What equation does \(f\) have to satisfy so that \(Y(t)=f(M(t),\langle M \rangle(t))\) is again a martingale if we assume that \(\mathbf E\int_0^t \sigma(s,\omega)^2 ds < \infty\).
  2. Set
    f_n(x,y) = \sum_{0 \leq m \leq \lfloor n/2 \rfloor} C_{n,m} x^{n-2m}y^m
    here \(\lfloor n/2 \rfloor\) is the largest integer less than or equal to \(n/2\). Set \(C_{n,0}=1\) for all \(n\). Then find a recurrence relation for \(C_{n,m+1}\) in terms of \(C_{n,m}\), so that \(Y(t)=f_n(B(t),t)\) will be a martingale.Write out explicitly \(f_1(B(t),), \cdots, f_4(B(t),t)\) as defined in the previous item.

Part II

Now consider \(I(t)\) defined by \[I(t)=\int_0^t \sigma(s,\omega)dB(s,\omega)\] where \(\sigma\) is adapted and \(|\sigma(t,\omega)| \leq K\) for all \(t\) with probability one. In light of the above let us set
\begin{align*}Y(t,\omega)=I(t)^4 – 6 I(t)^2\langle I \rangle(t) + 3 \langle I \rangle(t)^2 \ .\end{align*}

  1. Quote  the problem “Ito Moments” to show that \(\mathbb{E}\{ |Y(t)|^2\} < \infty\) for all \(t\). Then use the first part of this problem to conclude that \(Y\) is a martingale.
  2. Show that \[\mathbb{E}\{ I(t)^4 \} \leq 6 \mathbb{E} \big\{ \{I(t)^2\langle I \rangle(t) \big\}\]
  3. Recall the Cauchy-Schwartz inequality. In our language it states that
    \mathbb{E} \{AB\} \leq (\mathbb{E}\{A^2\})^{1/2} (\mathbb{E}\{B^2\})^{1/2}
    Combine this with the previous inequality to show that\begin{align*}\mathbb{E}\{ I(t)^4 \} \leq 36 \mathbb{E} \big\{\langle I \rangle(t)^2 \big\} \end{align*}
  4. As discussed in class \(I^4\) is a submartingale (because \(x \mapsto x^4\) is convex). Use the Kolmogorov-Doob inequality and all that we have just derived to show that
    \mathbb{P}\left\{ \sup_{0\leq s \leq T}|I(s)|^4 \geq \lambda \right\} \leq ( \text{const}) \frac{ \mathbb{E}\left( \int_0^T \sigma(s,\omega)^2 ds\right)^2 }{\lambda}


Associated PDE

Show that if

  1. \[I(t,\omega)=\int_0^t \sigma(s,\omega) dB(s,\omega)\]is a stochastic integral then \[I^2(t)-\int_0^t \sigma^2(s)ds\] is a martingale.
  2. What equation must \(u(t,x)\) satisfy so that
    \[ t \mapsto u(t,B(t))e^{\int_0^t V(B(s))ds} \]
    is a martingale? Here \(V\) is a bounded function. Hint: Set \(Y(t)=\int_0^t V(B(s))ds\) and apply It\^0’s formula to \(Z(t,B(t),Y(t))=u(t,B(t))\exp(Y(t))\).

Exponential Martingale Bound

Let \(\sigma(t,\omega)\) be nonanticipating with \(|\sigma(x,\omega)| < M\) for some  bound  \(M\) . Let \(I(t,\omega)=\int_0^t \sigma(s,\omega) dB(s,\omega)\). Use the exponential martingale \[\exp\big\{\alpha I(t)-\frac{\alpha^2}{2}\int_0^t \sigma^2(s)ds \big\}\] (see the problem here)  and the Kolmogorov-Doob inequality to get the estimate
P\Big\{ \sup_{0\leq t\leq T}|I(t)| \geq \lambda \Big\}\leq 2
\exp\left\{\frac{-\lambda^2}{2M^2 T}\right\}
First express the event of interest in terms of the exponential martingale, then use the Kolmogorov-Doob inequality and after this choose the parameter \(\alpha\) to get the best bound.

Martingale Exit from an Interval – I

Let \(\tau\) be the first time that a continuous martingale \(M_t\) starting from \(x\) exits the interval \((a,b)\), with \(a<x<b\). In all of the following, we assume that \(\mathbf P(\tau < \infty)=1\). Let \(p=\mathbf P_x\{M(\tau)=a\}\).

Find and analytic expression for \(p\) :

  1. For this part assume that \(M_t\) is the solution to a time homogeneous SDE. That is that \[dM_t=\sigma(M_t)dB_t.\] (with \(\sigma\) bounded and smooth.) What PDE should you solve to find \(p\) ? with what boundary data ? Assume for a moment that \(M_t\) is standard Brownian Motion (\(\sigma=1\)). Solve the PDE you mentioned above in this case.
  2. A probabilistic way of thinking: Return to a general martingale \(M_t\). Let us assume that \(dM_t=\sigma(t,\omega)dB_t\) again with \(\sigma\) smooth and uniformly bounded  from above and away from zero. Assume that \(\tau < \infty\) almost surely and notice that \[\mathbf E_x M(\tau)=a \mathbf P_x\{M_\tau=a\} + b \mathbf P_x\{M_\tau=b.\}\] Of course the process has to exit through one side or the other, so \[\mathbf P_x\{M_\tau=a\} = 1 – \mathbf P_x\{M_\tau=b\}\]. Use all of these facts and the Optimal Stopping Theorem to derive the equation for \(p\).
  3. Return to the case when \[dM_t=\sigma(M_t)dB_t\]. (with \(\sigma\) bounded and smooth.) Write down the equations that \(v(x)= \mathbf E_x\{\tau\}\), \(w(x,t)=\mathbf P_x\{ \tau >t\}\), and \(u(x)=\mathbf E_x\{e^{-\lambda\tau}\}\) with \(\lambda > 0\) satisfy. ( For extra credit: Solve them for \(M_t=B_t\) in this one dimensional setting and see what happens as \(b \rightarrow \infty\).)