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Discovering the Bessel Process
Let \(W_t=(W^{(1)}_t,\dots,W^{(n)}_t) \) be an \(n\)-dimensional Brownian motion with \( W^{(i)}_t\) standard independent 1-dim brownian motions and \(n \geq 2\).
Let
\[X_t = \|W_t\| = \Big(\sum_{i=1}^n (W^{(i)}_t)^2\Big)^{\frac12}\]
be the norm of the brownian motions. Even though the absolute value is not differentiable at zero we can still apply Itos formula since Brownian motion never visits the origin if the dimension is greater than zeros.
- Use Ito’s formula to show that \(X_t\) satisfies the Ito process
\[ dX_t = \frac{n-1}{2 X_t} dt + \sum_{i=1}^n \frac{W^{(i)}_t }{X_t} dW^{(i)}_t \] - Using the Levy-Doob Theorem show that
\[Z_t =\sum_{i=1}^n \int_0^t \frac{W^{(i)}_t }{X_t} dW^{(i)}_t \]
is a standard Brownian Motion. - In light of the above discussion argue that \(X_t\) and \(Y_t\) have the same distribution if \(Y_t\) is defined by
\[ dY_t = \frac{n-1}{2 Y_t} dt + dB_t\]
where \(B_t\) is a standard Brownian Motion.
Take a moment to reflect on what has been shown. \(W_t\) is a \(\mathbf R^n\) dimensional Markov Process. However, there is no guarantee that the one dimensional process \(X_t\) will again be a Markov process, much less a diffusion. The above calculation shows that the distribution of \(X_{t+h}\) is determined completely by \(X_t\) . In particular, it solves a one dimensional SDE. We were sure that \(X_t\) would be an Ito process but we had no guarantee that it could be written as a single closed SDE. (Namely that the coefficients would be only functions of \(X_t\) and not of the details of the \(W^{(i)}_t\)’s.
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.
- 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.
Solving a class of SDEs
Let us try a systematic procedure for solving SDEs which works for a class of SDEs. Let
\begin{align*}
X(t)=a(t)\left[ x_0 + \int_0^t b(s) dB(s) \right] +c(t) \ .
\end{align*}
Assuming \(a\), \(b\), and \(c\) are differentiable, use Ito’s formula to find the equation for \(dX(t)\) of the form
\begin{align*}
dX(t)=[ F(t) X(t) + H(t)] dt + G(t)dB(t)
\end{align*}
were \(F(t)\), \(G(t)\), and \(H(t)\) are some functions of time depending on \(a,b\) and maybe their derivatives. Solve the following equations by matching the coefficients. Let \(\alpha\), \(\gamma\) and \(\beta\) be fixed numbers.
Notice that
\begin{align*}
X(t)=a(t)\left[ x_0 + \int_0^t b(s) dB(s) \right] +c(t)=F(t,Y(t)) \ .
\end{align*}
where \(dY(t)=b(t) dB(t)\). Then you can apply Ito’s formula to this definition to find \(dX(t)\).
- First consider
\[dX_t = (-\alpha X_t + \gamma) dt + \beta dB_t\]
with \(X_0 =x_0\). Solve this for \( t \geq 0\) - Now consider
\[dY(t)=\frac{\beta-Y(t)}{1-t} dt + dB(t) ~,~~ 0\leq t < 1 ~,~~Y(0)=\alpha.\]
Solve this for \( t\in[0,1] \). - \begin{align*}
dX_t = -2 \frac{X_t}{1-t} dt + \sqrt{2 t(1-t)} dB_t ~,~~X(0)=\alpha
\end{align*}
Solve this for \( t\in[0,1] \).
Around the Circle
Consider the equation
\begin{align}
dX_t &= -Y_t dB_t – \frac12 X_t dt\\
dY_t &= X_t dB_t – \frac12 Y_t dt
\end{align}
Let \((X_0,Y_0)=(x,y)\) with \(x^2+y^2=1\). Show that \(X_t^2 + Y_t^2 =1\) for all \(t\) and hence the SDE lives on the unit circle. Does this make intuitive sense ?
Correlated SDEs
Let \(B_t\) and \(W_t\) be standard Brownian motions which are
independent. Consider
\begin{align*}
dX_t&= (-X_t +1)dt + \rho dB_t + \sqrt{1-\rho^2} dW_t\\
dY_t&= -Y_t dt + dB_t \ .
\end{align*}
Find the covariance of \(\text{Cov}(X_t,Y_t)=\mathbf{E} (X_t Y_t) – \mathbf{E} (X_t) \mathbf{E}( Y_t)\).