Tag Archives: JCM_math545_HW8_S14

A modified Wright-Fisher Model


Consider the ODE

\[ \dot x_t = x_t(1-x_t)\]

and the SDE

\[dX_t = X_t(1-X_t) dt + \sqrt{X_t(1-X_t)} dW_t\]

  1. Argue that \(x_t\) can not leave the interval \([0,1]\) if \( x_0 \in (0,1)\).
  2. What is the behavior of \(x_t\) as \(t \rightarrow\infty\) if if \( x _0\in (0,1)\) ?
  3. Can the diffusion \(X_t\) exit the interval \(  (0,1) \) ? Prove your claims.
  4. What do you think happens to \(X_t\) as \(t \rightarrow \infty\) ? Argue as best you can to support your claim.

No Explosions from Diffusion

Consider the following ODE and SDE:

\[\dot x_t = x^2_t \qquad x_0 >0\]

\[d X_t = X^2_t dt + \sigma |X_t|^\alpha dW_t\qquad X_0 >0\]

where \(\alpha >0\) and \(\sigma >0\).

  1. Show that \(x_t\) blows up in finite time.
  2. Find the values of  \(\sigma\) and \(\alpha\) so that \(X_t\) does not explode (off to infinity).

[ From Klebaner, ex 6.12]

Cox–Ingersoll–Ross model

The following model has SDE has been suggested as a model for interest rates:

\[ dr_t = a(b-r_t)dt +  \sigma \sqrt{r_t} dW_t\]

for \(r_t \in \mathbf R\), \(r_0 >0\) and constants \(a\),\(b\), and \(\sigma\).

  1. Find a closed form expression for \(\mathbf E( r_t)\).
  2. Find a closed form expression  for \(\mathrm{Var}(r_t)\).
  3. Characterize the values of parameters of \(a\), \(b\), and \(\sigma\) such that \(r=0\) is an absorbing point.
  4. What is the nature of the boundary at \(0\) for other values of the parameter ?


Ballistic Growth

Consider the SDE
dX(t)=b(X(t))dt +\sigma(X(t))dB(t)
with \(b(x)\to b_0 >0\) as \(x\to\infty\) and with \(\sigma\) bounded and positive. Suppose that \(b\) and \(\sigma\) are such that
\[\lim_{t\to\infty}X(t)=\infty\], with probability one for any starting point. Show that
P_x\Big\{\lim_{t\to\infty}\frac{X(t)}{b_0 t}=1\Big\}=1 \ .
X(t)=x+\int_0^{t}b(X(s))ds +\int_0^{t}\sigma(X(s))dB(s)
and the hypotheses, note that the result follows from showing that
\mathbf P_x\Big\{\lim_{t\to\infty}\frac{1}{t}\int_0^{t}\sigma(X(s))dB(s)=0\Big\}=1 \ .

There are a number of ways of thinking about this. In the end they all come down to essentially the same calculations. One way is to show that for some fixed \(\delta \in(0,1)\) the following statement holds with probability one:

There exist a constants \(C(\omega)\) so that
\int_0^{t}\sigma(X(s))dB(s) \leq Ct^\delta
for all \(t >0\).

To show this partition \([0,\infty]\) into blocks and use the Doob-Kolmogorov inequality to estimate the probability that the max of \( \int_0^{t}\sigma(X(s))ds\) on each block excess \(t^\delta\) on that block. Then use the Borel-Cantelli to show that this happens only a finite number of times.

A different way to organize the same calculation is to estimate
\mathbf P_x\Big\{\sup_{t>a}\frac{1}{t}|\int_0^t \sigma(X(s))dB(s)|>\epsilon\Big\}
by breaking the interval \(t>a\) into the union of intervals of the form \(a2^k <t\leq a2^{k+1}\) for \(k=0,1,\dots\) and using Doob-Kolmogorov Martingale inequality. Then let \(a\to\infty\).

Entry and Exit through boundaries

Consider the following one dimensional SDE.
dX_t&= \cos( X_t )^\alpha dW(t)\\
Consider the equation for \(\alpha >0\). On what interval do you expect to find the solution at all times ? Classify the behavior at the boundaries.

For what values of \(\alpha < 0\) does it seem reasonable to define the process ? any ? justify your answer.

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 \(b \rightarrow \infty\).)

Probability Bridge

For fixed \(\alpha\) and \(\beta\) consider the stochastic differential equation
dY(t)=\frac{\beta-Y(t)}{1-t} dt + dB(t) ~,~~ 0\leq t < 1 ~,~~Y(0)=\alpha.
Verify that \(\lim_{t\to 1}Y(t)=\beta\) with probability one. ( This is called the Brownian bridge from \(\alpha\) to \(\beta\).)
Hint: In the problem “Solving a class of SDEs“,  you found that this equation had the solution
Y_t = a(1-t) + bt + (1-t)\int_0^t \frac{dB_s}{1-s} \quad 0 \leq t <1\; .
To answer the question show that
\lim_{t \rightarrow 1^-} (1-t) \int_0^t\frac{dB_s}{1-s} =0 \quad \text{a.s.}