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# Category Archives: SDE examples

## A modified Wright-Fisher Model

Consider the ODE

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

and the SDE

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

- Argue that \(x_t\) can not leave the interval \([0,1]\) if \( x_0 \in (0,1)\).
- What is the behavior of \(x_t\) as \(t \rightarrow\infty\) if if \( x _0\in (0,1)\) ?
- Can the diffusion \(X_t\) exit the interval \( (0,1) \) ? Prove your claims.
- 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 \(\sigma >0\).

- Show that \(x_t\) blows up in finite time.
- 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 \(\sigma\).

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

## 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

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

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

## 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.

## 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.

- 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.

## Exponential Martingale

Let \(\sigma(t,\omega)\) be adapted to the filtration generated by a standard Brownian Motion \(B_t\) such that \(|\sigma(x,\omega)| < K\) for some bound \(K\) . Let \(I(t,\omega)=\int_0^t \sigma(s,\omega) dB(s,\omega)\).

- Show that

\[M_t=\exp\big\{\alpha I(t)-\frac{\alpha^2}{2}\int_0^t \sigma^2(s)ds \big\}\]

is a martingale. It is called the exponential martingale. - Show that \(M_t\) satisfies the equation

\[ dM_t =\alpha M_t dI_t = \alpha M_t \sigma_t dB_t\]

## 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 \( t\in[0,1] \).

## 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 \ .

\]

From

\[

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

\begin{align*}

\mathbf P_x\Big\{\lim_{t\to\infty}\frac{1}{t}\int_0^{t}\sigma(X(s))dB(s)=0\Big\}=1 \ .

\end{align*}

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

\begin{align*}

\int_0^{t}\sigma(X(s))dB(s) \leq Ct^\delta

\end{align*}

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 \(a\to\infty\).

## 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 \(\text{Cov}(X_t,Y_t)=\mathbf{E} (X_t Y_t) – \mathbf{E} (X_t) \mathbf{E}( Y_t)\).

## Hyperbolic SDE

Consider

\begin{align*}

dX_t=& Y_t dB_t + \frac12 X_t dt\\

dY_t=& X_t dB_t + \frac12 Y_t dt

\end{align*}

Show that \(X_t^2-Y_t^2\) is constant for all \(t\).