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

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

Learning probability by doing !

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

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

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Posted in Ito Formula, SDE examples

Tagged JCM_math545_HW4_S23

Define

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

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

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

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Posted in Ito Formula, SDE examples

Tagged JCM_math545_HW3_S23

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*}

- 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.
- Show that \[\mathbb{E}\{ I(t)^4 \} \leq 6 \mathbb{E} \big\{ \{I(t)^2\langle I \rangle(t) \big\}\]
- Recall the Cauchy-Schwartz inequality. In our language it states that

\begin{align*}

\mathbb{E} \{AB\} \leq (\mathbb{E}\{A^2\})^{1/2} (\mathbb{E}\{B^2\})^{1/2}

\end{align*}

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*} - 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

\begin{align*}

\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}

\end{align*}

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Posted in Ito Formula, Ito Integrals, Martingales

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.

- 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. - 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. - * 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.

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Posted in Ito Formula, SDE examples, Stochastic Calculus

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.

- Find \( d(e^{\beta t} X_t)\) using Ito’s Formula.
- 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} - Conclude that \(X_t\) is Gaussian process (see exercise: Gaussian Ito Integrals ). Find its mean and variance at time \(t\).
- * 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.

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Posted in Ito Formula, SDE examples

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.

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Posted in Ito Formula