Category Archives: Ito Formula

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

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

Practice with Ito and Integration by parts


\[ X_t =X_0 + \int_0^t B_s dB_s\]

where \(B_t\) is a standard Brownian Motion. Show that \(X_t\) can also be written

\[ X_t=X_0 + \frac12 (B^2_t -t)\]


BDG Inequality

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. Inspired by   problem “Homogeneous Martingales and Hermite Polynomials”  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  verify that \(Y_t\) 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. We know that  \(I^4\) 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}

Ito Variation of Constants

For  functions \(f(x)\) and\( g(x) \) and constant \(\beta>0\),  define \(X_t\) as the solution to the following SDE
\[dX_t = – \beta X_t dt + h(X_t)dt + g(X_t) dW_t\]

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

  1.  Show that \(X_t\) can be written as
    \[X_t = e^{-\beta t} X_0 + \int_0^{t}  e^{-\beta (t-s)} h(X_s) ds +  \int_0^{t}  e^{-\beta (t-s)} g(X_s) dW_s\]
    See exercise:  Ornstein–Uhlenbeck process for guidance.
  2. Assuming that \(|h(x)| < K\)  and \(|g(x)|<K\), show that  there exists a constant \(C(X_0)\) so that
    \[ \mathbf E [|X_t|] < C(X_0) \]
    for all \(t >0\). It might be convenient the remember the Cauchy–Schwarz inequality.
  3. * Assuming that \(|h(x)| < K\)  and \(|g(x)|<K\), show that  for any integer \(p >0\) there exists a constant \(C(p,X_0)\) so that
    \[ \mathbf E [|X_t|^{2p}] < C(p,X_0) \]
    for all \(t >0\). See exercise: Ito Moments for guidance.

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.

Ito Integration by parts

Recall that if \(u(t)\) and \(v(t)\) are deterministic functions which are once differentiable then the classic integration by parts formula states that
\[ \int_0^t u(s) (\frac{dv}{ds})(s)\,ds = u(t)v(t) – u(0)v(0) – \int_0^t v(s) (\frac{du}{ds})(s)\,ds\]

As is suggested by the formal relations

\[ (\frac{dv}{ds})(s)\,ds=dv(s) \qquad\text{and}\qquad (\frac{du}{ds})(s)\, ds=du(s)\]

this can be rearranged  to state

\[ u(t)v(t)- u(0)v(0)=  \int_0^t u(s) dv(s) + \int_0^t v(s) du(s)\]

which holds for more general Riemann–Stieltjes integrals. Now consider two Ito processes \(X_t\) and \(Y_t\) given by

\[dX_t=b_s ds + \sigma_s dW_t \qquad\text{and}\qquad dY_t=f_s ds + g_s dW_t \]

where \(W_t\) is a standard Brownian Motion. Derive the “Integration by Parts formula” for Ito calculus by applying Ito’s formula to \(X_tY_t\). Compare this the the classical formula given above.