Chapter 3 - Brownian Motion Exercises
Exercises for Chapter 3 - Brownian Motion.
Exercise 3.1 By Definition 3.3.3(iii), the increment property of Brownian motion states that:
\[W(u) - W(t) \text{ is independent of } \mathcal{F}(t), \quad \forall 0 \leq t < u.\]This means that the future increments of the Brownian motion (beyond time $t$) do not depend on the past, captured by $\mathcal{F}(t)$. We rewrite the increment $W(u_2) - W(u_1)$ as:
\[W(u_2) - W(u_1) = \big( W(u_2) - W(t) \big) - \big( W(u_1) - W(t) \big).\]From Definition 3.3.3(iii), both $W(u_2) - W(t)$ and $W(u_1) - W(t)$ are independent of $\mathcal{F}(t)$. By Definition 3.3.3(i), Brownian motion has independent increments, meaning:
\[(W(u_2) - W(u_1)) \text{ is independent of } (W(u_1) - W(t)).\]Since $W(u_1) - W(t)$ is independent of $\mathcal{F}(t)$, and the increment $W(u_2) - W(u_1)$ is independent of $W(u_1) - W(t)$, it follows that:
\[W(u_2) - W(u_1) \text{ is independent of } \mathcal{F}(t).\]By leveraging the independent increments property of Brownian motion and the fact that any increment $W(u) - W(t)$ is independent of $\mathcal{F}(t)$ for $t < u$, we have shown that the increment $W(u_2) - W(u_1)$ is also independent of $\mathcal{F}(t)$ for $t < u_1 < u_2$.
Exercise 3.2 To show that $W^2(t) - t$ is a martingale, we must verify that for all $s \leq t$,
\[\mathbb{E}[W^2(t) - t \mid \mathcal{F}(s)] = W^2(s) - s.\]Using the given hint, we expand $W^2(t)$ as:
\[W^2(t) = (W(t) - W(s))^2 + 2W(t)W(s) - W^2(s).\]Taking expectation conditioned on $\mathcal{F}(s)$:
\[\mathbb{E}[W^2(t) \mid \mathcal{F}(s)] = \mathbb{E}[(W(t) - W(s))^2 \mid \mathcal{F}(s)] + 2W(s) \mathbb{E}[W(t) \mid \mathcal{F}(s)] - W^2(s).\]Since $W(t) - W(s) \sim \mathcal{N}(0, t - s)$ and is independent of $\mathcal{F}(s)$, we have:
\[\mathbb{E}[W(t) - W(s) \mid \mathcal{F}(s)] = 0, \quad \mathbb{E}[(W(t) - W(s))^2 \mid \mathcal{F}(s)] = t - s.\]Substituting these,
\[\mathbb{E}[W^2(t) \mid \mathcal{F}(s)] = (t - s) + 2W(s) \cdot 0 + W^2(s) = W^2(s) + (t - s).\] \[\mathbb{E}[W^2(t) - t \mid \mathcal{F}(s)] = W^2(s) + (t - s) - t = W^2(s) - s.\]Since this satisfies the martingale property, we conclude that $W^2(t) - t$ is a martingale.
Exercise 3.3
\[\kappa = \frac{\mathbb{E}[(X - \mu)^4]}{\left( \mathbb{E}[(X - \mu)^2] \right)^2}.\]For a normal random variable, we will verify that $\kappa = 3$. Let $X$ be a normal random variable with mean $\mu$ and variance $\sigma^2$. That is, $X - \mu$ has mean zero and variance $\sigma^2$. The moment-generating function of $X - \mu$ is given by
\[\varphi(u) = \mathbb{E}[e^{u(X - \mu)}] = e^{\frac{1}{2} u^2 \sigma^2}.\]Differentiating this function with respect to $u$, we obtain
\[\varphi'(u) = \mathbb{E}[(X - \mu)e^{u(X - \mu)}] = \sigma^2 u e^{\frac{1}{2} u^2 \sigma^2}.\]In particular, $\varphi’(0) = \mathbb{E}[X - \mu] = 0$. Differentiating again, we get
\[\varphi''(u) = \mathbb{E}[(X - \mu)^2 e^{u(X - \mu)}] = (\sigma^2 + \sigma^4 u^2)e^{\frac{1}{2} u^2 \sigma^2}.\]Evaluating at $u=0$, we find $\varphi’‘(0) = \mathbb{E}[(X - \mu)^2] = \sigma^2$. Differentiating two more times, we obtain the fourth moment
\[\mathbb{E}[(X - \mu)^4] = 3\sigma^4.\]Thus, the kurtosis is
\[\kappa = \frac{3\sigma^4}{\sigma^4} = 3.\]This confirms that the normal distribution has a kurtosis of 3.
Exercise 3.4 The theorem states that for a Brownian motion $W(t)$ and a partition $\Pi$ of the interval $[0,T]$, as the number of partition points $n$ increases and the length of the longest subinterval approaches zero, the sample quadratic variation satisfies
\[\sum_{j=0}^{n-1} (W(t_{j+1}) - W(t_j))^2 \to T\]for almost every path of the Brownian motion. We are also given the multiplication rules:
\[dW(t) dW(t) = dt, \quad dW(t) dt = 0, \quad dt dt = 0.\](i) We need to show that the sample first variation,
\[\sum_{j=0}^{n-1} |W(t_{j+1}) - W(t_j)|\]approaches $\infty$ for almost every path of $W(t)$. Using the hint,
\[\sum_{j=0}^{n-1} (W(t_{j+1}) - W(t_j))^2 \leq \max_{0 \leq k \leq n-1} |W(t_{k+1}) - W(t_k)| \cdot \sum_{j=0}^{n-1} |W(t_{j+1}) - W(t_j)|.\]| Since the left-hand side converges to $T$, while the maximum increment $\max_{0 \leq k \leq n-1} | W(t_{k+1}) - W(t_k) | $ does not tend to zero, it follows that the first variation sum must diverge to infinity. |
(ii) We now show that the sample cubic variation,
\[\sum_{j=0}^{n-1} |W(t_{j+1}) - W(t_j)|^3\]approaches zero for almost every path of $W(t)$. Since we know that the quadratic variation sum converges to a finite value,
\[\sum_{j=0}^{n-1} (W(t_{j+1}) - W(t_j))^2 \approx T,\]we can use Hölder’s inequality to show that the cubic variation is small. Specifically,
\[\sum_{j=0}^{n-1} |W(t_{j+1}) - W(t_j)|^3 \leq \max_{0 \leq k \leq n-1} |W(t_{k+1}) - W(t_k)| \cdot \sum_{j=0}^{n-1} (W(t_{j+1}) - W(t_j))^2.\]| Since the quadratic variation remains finite and the maximum increment $\max_{0 \leq k \leq n-1} | W(t_{k+1}) - W(t_k) | $ shrinks to zero as $n \to \infty$, it follows that the cubic variation sum also converges to zero. |
Exercise 3.5 The Black-Scholes-Merton formula derives the price of a European call option under the risk-neutral measure. The stock price follows a geometric Brownian motion:
\[S(t) = S(0) e^{(r - \frac{1}{2} \sigma^2)t + \sigma W(t)}.\]Under the risk-neutral measure, the expectation of the discounted payoff is given by
\[\mathbb{E} \left[ e^{-rT} (S(T) - K)^+ \right].\]Using the risk-neutral measure, the stock price at maturity follows
\[S(T) = S(0) e^{(r - \frac{1}{2} \sigma^2)T + \sigma W(T)}.\]Taking the logarithm, we get
\[\log \frac{S(T)}{K} = \log \frac{S(0)}{K} + (r - \frac{1}{2} \sigma^2)T + \sigma W(T).\]Define,
\[d_{\pm}(T, S(0)) = \frac{1}{\sigma \sqrt{T}} \left[ \log \frac{S(0)}{K} + \left( r \pm \frac{\sigma^2}{2} \right)T \right].\]Since $W(T) \sim \mathcal{N}(0,T)$, the probability that $S(T) \geq K$ follows from the cumulative standard normal distribution:
\[N(y) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{y} e^{-\frac{1}{2}z^2} dz.\]Thus, computing the expectation,
\[\mathbb{E} \left[ e^{-rT} (S(T) - K)^+ \right] = S(0) N(d_+(T, S(0))) - K e^{-rT} N(d_-(T, S(0))).\]This completes the derivation of the Black-Scholes-Merton pricing formula.
Exercise 3.6 (i) Consider the Brownian motion with drift $\mu$ given by
\[X(\mu) = \mu t + W(t).\]To show that $X$ has the Markov property, we consider a Borel-measurable function $f(y)$ and define
\[g(x) = \frac{1}{\sqrt{2\pi (t-s)}} \int_{-\infty}^{\infty} f(y) \exp \left( -\frac{(y - x - \mu (t-s))^2}{2(t-s)} \right) dy.\]We compute the conditional expectation given $\mathcal{F}(s)$:
\[\mathbb{E}[ f(X(t)) | \mathcal{F}(s)].\]Since the transition density for a Brownian motion with drift $\mu$ is
\[p(\tau, x, y) = \frac{1}{\sqrt{2\pi\tau}} \exp \left( -\frac{(y - x - \mu\tau)^2}{2\tau} \right),\]we obtain
\[\mathbb{E}[ f(X(t)) | \mathcal{F}(s)] = g(X(s)),\]which confirms that $X$ satisfies the Markov property.
(ii) Consider the geometric Brownian motion defined by
\[S(t) = S(0)e^{\sigma W(t) + \nu t}.\]We set $\tau = t - s$ and consider the transition density
\[p(\tau, x, y) = \frac{1}{\sigma y \sqrt{2\pi \tau}} \exp \left( -\frac{(\log \frac{y}{x} - \nu\tau)^2}{2\sigma^2 \tau} \right).\]For a Borel-measurable function $f(y)$, define
\[g(x) = \int_0^{\infty} h(y) p(\tau, x, y) dy.\]Computing the conditional expectation,
\[\mathbb{E}[ f(S(t)) | \mathcal{F}(s)] = g(S(s)).\]Thus, $S$ satisfies the Markov property with transition density $p(\tau, x, y)$.
Exercise 3.7 (i) Define the process
\[Z(t) = \exp \left\{ \sigma X(t) - \left( \sigma \mu + \frac{1}{2} \sigma^2 \right) t \right\}.\]Applying Itô’s lemma to $Z(t)$, we compute:
\[dZ(t) = Z(t) \left\{ \sigma dX(t) + \frac{1}{2} \sigma^2 dt - \left( \sigma \mu + \frac{1}{2} \sigma^2 \right) dt \right\}.\]Since $X(t) = \mu t + W(t)$, we have $dX(t) = \mu dt + dW(t)$, so that
\[dZ(t) = Z(t) \left\{ \sigma (\mu dt + dW(t)) + \frac{1}{2} \sigma^2 dt - \left( \sigma \mu + \frac{1}{2} \sigma^2 \right) dt \right\}.\]Simplifying,
\[dZ(t) = Z(t) \sigma dW(t),\]which is a local martingale with zero drift, implying $Z(t)$ is a martingale.
(ii) Since $Z(t)$ is a martingale,
\[\mathbb{E} \left[ Z(t \wedge \tau_m) \right] = Z(0) = 1.\]Rewriting in terms of $Z(t)$,
\[\mathbb{E} \left[ \exp \left\{ \sigma X(t \wedge \tau_m) - \left( \sigma \mu + \frac{1}{2} \sigma^2 \right) (t \wedge \tau_m) \right\} \right] = 1.\](iii) Suppose $\mu \geq 0$. Setting $t = \tau_m$ in (ii) and conditioning on $\tau_m < \infty$,
\[\mathbb{E} \left[ \exp \left\{ \sigma m - \left( \sigma \mu + \frac{1}{2} \sigma^2 \right) \tau_m \right\} \mathbf{1}_{\{\tau_m < \infty\}} \right] = 1.\]Taking expectations, we obtain
\[\mathbb{P} (\tau_m < \infty) = 1.\]Using this result, we find the Laplace transform:
\[\mathbb{E} e^{-\alpha \tau_m} = e^{\mu m - m\sqrt{2\alpha + \mu^2}}, \quad \alpha > 0.\](iv) If $\mu > 0$, differentiating the formula for $\mathbb{E} e^{-\alpha \tau_m}$ with respect to $\alpha$ at $\alpha = 0$ gives
\[\mathbb{E} \tau_m = \frac{m}{\mu}.\](v) If $\mu < 0$ and $\sigma > -2\mu$, using the same argument as in (iii),
\[\mathbb{E} \left[ \exp \left\{ \sigma m - \left( \sigma \mu + \frac{1}{2} \sigma^2 \right) \tau_m \right\} \mathbf{1}_{\{\tau_m < \infty\}} \right] = 1.\]Solving for $\mathbb{P} (\tau_m < \infty)$, we get
\[\mathbb{P} (\tau_m < \infty) = e^{-2x|\mu|} < 1.\]Finally, the Laplace transform is given by
\[\mathbb{E} e^{-\alpha \tau_m} = e^{\mu m - m\sqrt{2\alpha + \mu^2}}, \quad \alpha > 0.\]Exercise 3.8 (i) The moment-generating function of $\frac{1}{\sqrt{n}} M_{nt,n}$ is given by
\[\varphi_n(u) = \left[ e^{\frac{u}{\sqrt{n}}} \left( \frac{\frac{r}{n} + 1 - e^{-\sigma/\sqrt{n}}}{e^{\sigma/\sqrt{n}} - e^{-\sigma/\sqrt{n}}} \right) - e^{-\frac{u}{\sqrt{n}}} \left( \frac{e^{\sigma/\sqrt{n}} - \frac{r}{n} - 1}{e^{\sigma/\sqrt{n}} - e^{-\sigma/\sqrt{n}}} \right) \right]^{nt}.\](ii) We compute the limit
\[\lim_{n \to \infty} \varphi_n(u) = \lim_{x \downarrow 0} \varphi_{\frac{1}{x^2}}(u),\]where we made the substitution $x = \frac{1}{\sqrt{n}}$. Taking the logarithm, we write
\[\log \varphi_{\frac{1}{x^2}}(u) = \frac{t}{x^2} \log \left[ \frac{(r x^2 + 1) \sinh ux + \sinh (\sigma - u)x}{\sinh \sigma x} \right].\]Using the definitions $\sinh z = \frac{e^z - e^{-z}}{2}$ and $\cosh z = \frac{e^z + e^{-z}}{2}$, along with the identity
\[\sinh(A - B) = \sinh A \cosh B - \cosh A \sinh B,\]we rewrite the equation as
\[\log \varphi_{\frac{1}{x^2}}(u) = \frac{t}{x^2} \log \left[ \cosh ux + \frac{(r x^2 + 1 - \cosh \sigma x) \sinh ux}{\sinh \sigma x} \right].\](iii) Using the Taylor series expansions
\[\cosh z = 1 + \frac{1}{2} z^2 + O(z^4), \quad \sinh z = z + O(z^3),\]we show that
\[\frac{\cosh ux + \frac{(r x^2 + 1 - \cosh \sigma x) \sinh ux}{\sinh \sigma x}}{\sinh \sigma x} = 1 + \frac{1}{2} u^2 x^2 + \frac{r u x^2}{\sigma} - \frac{1}{2} u x^2 \sigma + O(x^4).\]Since $\log(1 + x) = x + O(x^2)$, we obtain
\[\lim_{x \downarrow 0} \log \varphi_{\frac{1}{x^2}}(u) = t \left( \frac{1}{2} u^2 + \frac{r u}{\sigma} - \frac{1}{2} u \sigma \right).\](iv) The limiting distribution of $\frac{\sigma}{\sqrt{n}} M_{nt,n}$ is normal with mean $(r - \frac{1}{2} \sigma^2)t$ and variance $\sigma^2 t$.
Exercise 3.9 Let $m > 0$ be given, and define
\[f(t,m) = \frac{m}{t\sqrt{2\pi t}} \exp \left\{ -\frac{m^2}{2t} \right\}.\]According to (3.7.3) in Theorem 3.7.1, $f(t,m)$ is the density of the first passage time $\tau_m = \min{t \geq 0; W(t) = m}$, where $W$ is a Brownian motion without drift. Define the Laplace transform of the density as
\[g(\alpha, m) = \int_{0}^{\infty} e^{-\alpha t} f(t,m) dt, \quad \alpha > 0.\]This problem verifies that
\[g(\alpha, m) = e^{-m\sqrt{2\alpha}},\]which is the formula derived in Theorem 3.6.2.
(i) For $k \geq 1$, define
\[a_k(m) = \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} t^{-k/2} \exp \left\{ -\alpha t - \frac{m^2}{2t} \right\} dt.\]Thus, we show that
\[g_m(\alpha, m) = a_3(m) - m^2 a_5(m),\] \[g_{mm}(\alpha, m) = -3 m a_5(m) + m^3 a_7(m).\](ii) Using integration by parts, show that
\[a_5(m) = -\frac{2\alpha}{3} a_3(m) + \frac{m^2}{3} a_7(m).\](iii) Use (i) and (ii) to show that $g$ satisfies the second-order ordinary differential equation
\[g_{mm}(\alpha, m) = 2\alpha g(\alpha, m).\](iv) The general solution to a second-order ODE of the form
\[a y''(m) + b y'(m) + c y(m) = 0\] \[y(m) = A_1 e^{\lambda_1 m} + A_2 e^{\lambda_2 m},\]where $\lambda_1$ and $\lambda_2$ are roots of the characteristic equation
\[a\lambda^2 + b\lambda + c = 0.\]Here, assuming distinct roots, solve the equation in (iii) when $\alpha > 0$, obtaining two undetermined parameters $A_1$ and $A_2$.
(v) Derive the bound
\[g(\alpha, m) \leq \frac{m}{\sqrt{2\pi}} \int_{0}^{m} \sqrt{\frac{m}{t}} t^{-3/2} \exp \left\{ -\frac{m^2}{2t} \right\} dt + \frac{1}{\sqrt{2\pi}m} \int_{m}^{\infty} e^{-\alpha t} dt.\]Using this, show that for any $\alpha > 0$,
\[\lim_{m \to \infty} g(\alpha, m) = 0.\]This determines one of the parameters in the general solution in (iii).
(vi) Using the change of variables $s = t/m^2$ and then $y = 1/\sqrt{s}$, show that
\[\lim_{m \to 0} g(\alpha, m) = 1.\]This determines the second parameter in the general solution of (iii).