Introduction
In this part, we will continue looking at Brownian motions and Gaussian processes.
Namely, we will firstly look at zeroes of Brownian motions and introduce the Brownian bridge.
Then, we will discuss Brownian motion with a drift term and geometric Brownian motion.
Zeroes of Brownian Motion
Intuition (Zeroes of Brownian Motion) Let L L L ( 0 , 1 ) (0, 1) ( 0 , 1 ) L = max { t : 0 < t < 1 , B t = 0 } L = \max \ \{t : 0 < t < 1, B_t = 0\} L = max { t : 0 < t < 1 , B t = 0 }
L ∼ B e t a ( 1 2 , 1 2 ) . L \sim \mathrm{Beta}\left(\frac{1}{2}, \frac{1}{2}\right). L ∼ Beta ( 2 1 , 2 1 ) . Proof (Outline of Proof) P ( L > s ) = ∫ − ∞ ∞ P ( L > s ∣ B s = t ) N ( t ; 0 , s ) d t = 2 ∫ − ∞ 0 P ( L > s ∣ B s = t ) N ( t ; 0 , s ) d t = 2 ∫ − ∞ 0 P ( M 1 − s > − t ) N ( t ; 0 , s ) d t = 2 ∫ 0 ∞ 2 P ( B 1 − s > t ) N ( t ; 0 , s ) d t = 4 ∫ 0 ∞ ∫ t ∞ N ( r ; 0 , 1 − s ) N ( t ; 0 , s ) d r d t = … = 1 π ∫ s 1 1 x ( 1 − x ) d x = ∫ s 1 B e t a ( x ; 1 2 , 1 2 ) d x . \begin{align*}
P(L > s) & = \int_{-\infty}^{\infty} P(L > s \mid B_s = t) \ \mathcal{N}(t; 0, s) \ dt \newline
& = 2 \int_{-\infty}^{0} P(L > s \mid B_s = t) \ \mathcal{N}(t; 0, s) \ dt \newline
& = 2 \int_{-\infty}^{0} P(M_{1 - s} > -t) \ \mathcal{N}(t; 0, s) \ dt \newline
& = 2 \int_{0}^{\infty} 2 P(B_{1 - s} > t) \ \mathcal{N}(t; 0, s) \ dt \newline
& = 4 \int_{0}^{\infty} \int_{t}^{\infty} \mathcal{N}(r; 0, 1 - s) \ \mathcal{N}(t; 0, s) \ dr \ dt \newline
& = \ldots
& = \frac{1}{\pi} \int_s^1 \frac{1}{\sqrt{x(1 - x)}} \ dx \newline
& = \int_s^1 \mathrm{Beta}\left(x; \frac{1}{2}, \frac{1}{2}\right) \ dx. \newline
\end{align*} P ( L > s ) = ∫ − ∞ ∞ P ( L > s ∣ B s = t ) N ( t ; 0 , s ) d t = 2 ∫ − ∞ 0 P ( L > s ∣ B s = t ) N ( t ; 0 , s ) d t = 2 ∫ − ∞ 0 P ( M 1 − s > − t ) N ( t ; 0 , s ) d t = 2 ∫ 0 ∞ 2 P ( B 1 − s > t ) N ( t ; 0 , s ) d t = 4 ∫ 0 ∞ ∫ t ∞ N ( r ; 0 , 1 − s ) N ( t ; 0 , s ) d r d t = … = ∫ s 1 Beta ( x ; 2 1 , 2 1 ) d x . = π 1 ∫ s 1 x ( 1 − x ) 1 d x We can then, let L t L_t L t ( 0 , t ) (0, t) ( 0 , t )
L t t ∼ B e t a ( 1 2 , 1 2 ) . \frac{L_t}{t} \sim \mathrm{Beta}\left(\frac{1}{2}, \frac{1}{2}\right). t L t ∼ Beta ( 2 1 , 2 1 ) . Note The probability that a Brownian motion has at least one zero in ( r , t ) (r, t) ( r , t ) 0 ≤ r < t 0 \leq r < t 0 ≤ r < t 1 − P ( L t < r ) 1 - P(L_t < r) 1 − P ( L t < r )
Further, the cumulative distribution for the B e t a ( 1 2 , 1 2 ) \mathrm{Beta}\left(\frac{1}{2}, \frac{1}{2}\right) Beta ( 2 1 , 2 1 )
P ( L t < r ) = ∫ 0 r t B e t a ( s ; 1 2 , 1 2 ) d s = 2 π arcsin ( r t ) . P(L_t < r) = \int_0^{\frac{r}{t}} \mathrm{Beta}\left(s; \frac{1}{2}, \frac{1}{2}\right) \ ds = \frac{2}{\pi} \arcsin\left(\sqrt{\frac{r}{t}}\right). P ( L t < r ) = ∫ 0 t r Beta ( s ; 2 1 , 2 1 ) d s = π 2 arcsin ( t r ) . Brownian Bridge
Definition 1 (Brownian Bridge) Define a Gaussian process X t X_t X t B t B_t B t B 1 = 0 B_1 = 0 B 1 = 0 X t X_t X t
If 0 < s < t < 1 0 < s < t < 1 0 < s < t < 1 ( B s , B t , B 1 ) (B_s, B_t, B_1) ( B s , B t , B 1 )
E [ ( B s , B t , B 1 ) ] = ( 0 , 0 , 0 ) V a r ( ( B s , B t , B 1 ) ) = Σ = ( s s s s t t s t 1 ) . \begin{align*}
\mathbb{E}[(B_s, B_t, B_1)] & = (0, 0, 0) \newline
\mathrm{Var}((B_s, B_t, B_1)) & = \Sigma =
\begin{pmatrix}
s & s & s \newline
s & t & t \newline
s & t & 1 \newline
\end{pmatrix}. \newline
\end{align*} E [( B s , B t , B 1 )] Var (( B s , B t , B 1 )) = ( 0 , 0 , 0 ) = Σ = s s s s t t s t 1 . Conditioning on B 1 = 0 B_1 = 0 B 1 = 0 E [ X t ] = 0 \mathbb{E}[X_t] = 0 E [ X t ] = 0
C o v ( X s , X t ) = s ( 1 − t ) = s − s t . \mathrm{Cov}(X_s, X_t) = s(1 - t) = s - st. Cov ( X s , X t ) = s ( 1 − t ) = s − s t . It follows that this is identical to the Brownian bridge defined above.
Brownian Motion with Drift and Geometric Brownian Motion
Definition 2 (Brownian Motion with Drift) For any real μ > 0 \mu > 0 μ > 0 σ > 0 \sigma > 0 σ > 0 X t X_t X t
X t = μ t + σ B t . X_t = \mu t + \sigma B_t. X t = μ t + σ B t . This is a Brownian motion with a draft, and is often a more useful model than standard Brownian motion.
Note Note that X t X_t X t μ t \mu t μ t σ 2 t \sigma^2 t σ 2 t
This is a Gaussian process with continuous paths and stationary and independent increments.
Definition 3 (Geometric Brownian Motion) The stochastic process,
G t = G 0 e μ t + σ B t , G_t = G_0 e^{\mu t + \sigma B_t}, G t = G 0 e μ t + σ B t , where G 0 > 0 G_0 > 0 G 0 > 0 μ \mu μ σ 2 \sigma^2 σ 2
log ( G t ) \log(G_t) log ( G t ) log ( G 0 ) + μ t \log(G_0) + \mu t log ( G 0 ) + μ t σ 2 t \sigma^2 t σ 2 t
Note One can show that,
E [ G t ] = G 0 e t ( μ + 1 2 σ 2 ) V a r ( G t ) = G 0 2 e 2 t ( μ + σ 2 ) ( e σ 2 t − 1 ) . \begin{align*}
\mathbb{E}[G_t] & = G_0 e^{t(\mu + \frac{1}{2} \sigma^2)} \newline
\mathrm{Var}(G_t) & = G_0^2 e^{2 t(\mu + \sigma^2)} (e^{\sigma^2 t} - 1). \newline
\end{align*} E [ G t ] Var ( G t ) = G 0 e t ( μ + 2 1 σ 2 ) = G 0 2 e 2 t ( μ + σ 2 ) ( e σ 2 t − 1 ) . Proof Using that log ( G t ) ∼ N ( log ( G 0 ) + μ t , σ 2 t ) \log(G_t) \sim \mathcal{N}(\log(G_0) + \mu t, \sigma^2 t) log ( G t ) ∼ N ( log ( G 0 ) + μ t , σ 2 t )
E [ G t ] = E [ e log ( G t ) ] = ∫ − ∞ ∞ e x N ( x ; log ( G 0 ) + μ t , σ 2 t ) d x = ∫ − ∞ ∞ 1 2 π σ 2 t exp ( − ( x − ( log ( G 0 ) + μ t ) ) 2 2 σ 2 t + x ) d x = ∫ − ∞ ∞ 1 2 π σ 2 t exp ( − ( x − ( log ( G 0 ) + μ t − σ 2 t ) ) 2 2 σ 2 t + log ( G 0 ) + μ t + 1 2 σ 2 t ) d x = e log ( G 0 ) + μ t + 1 2 σ 2 t ∫ − ∞ ∞ 1 2 π σ 2 t exp ( − ( x − ( log ( G 0 ) + μ t − σ 2 t ) ) 2 2 σ 2 t ) d x = e log ( G 0 ) + μ t + 1 2 σ 2 t = G 0 e t ( μ + 1 2 σ 2 ) . \begin{align*}
\mathbb{E}[G_t] & = \mathbb{E}[e^{\log(G_t)}] \newline
& = \int_{-\infty}^{\infty} e^x \ \mathcal{N}(x; \log(G_0) + \mu t, \sigma^2 t) \ dx \newline
& = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2 t}} \exp\left(-\frac{(x - (\log(G_0) + \mu t))^2}{2 \sigma^2 t} + x\right) \ dx \newline
& = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2 t}} \exp\left(-\frac{(x - (\log(G_0) + \mu t - \sigma^2 t))^2}{2 \sigma^2 t} + \log(G_0) + \mu t + \frac{1}{2} \sigma^2 t\right) \ dx \newline
& = e^{\log(G_0) + \mu t + \frac{1}{2} \sigma^2 t} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2 t}} \exp\left(-\frac{(x - (\log(G_0) + \mu t - \sigma^2 t))^2}{2 \sigma^2 t}\right) \ dx \newline
& = e^{\log(G_0) + \mu t + \frac{1}{2} \sigma^2 t} \newline
& = G_0 e^{t(\mu + \frac{1}{2} \sigma^2)}. \newline
\end{align*} E [ G t ] = E [ e l o g ( G t ) ] = ∫ − ∞ ∞ e x N ( x ; log ( G 0 ) + μ t , σ 2 t ) d x = ∫ − ∞ ∞ 2 π σ 2 t 1 exp ( − 2 σ 2 t ( x − ( log ( G 0 ) + μ t ) ) 2 + x ) d x = ∫ − ∞ ∞ 2 π σ 2 t 1 exp ( − 2 σ 2 t ( x − ( log ( G 0 ) + μ t − σ 2 t ) ) 2 + log ( G 0 ) + μ t + 2 1 σ 2 t ) d x = e l o g ( G 0 ) + μ t + 2 1 σ 2 t ∫ − ∞ ∞ 2 π σ 2 t 1 exp ( − 2 σ 2 t ( x − ( log ( G 0 ) + μ t − σ 2 t ) ) 2 ) d x = e l o g ( G 0 ) + μ t + 2 1 σ 2 t = G 0 e t ( μ + 2 1 σ 2 ) . Similarly, we can compute E [ G t 2 ] \mathbb{E}[G_t^2] E [ G t 2 ] V a r ( G t ) = E [ G t 2 ] − ( E [ G t ] ) 2 \mathrm{Var}(G_t) = \mathbb{E}[G_t^2] - (\mathbb{E}[G_t])^2 Var ( G t ) = E [ G t 2 ] − ( E [ G t ] ) 2
E [ G t 2 ] = E [ e 2 log ( G t ) ] = ∫ − ∞ ∞ e 2 x N ( x ; log ( G 0 ) + μ t , σ 2 t ) d x = ∫ − ∞ ∞ 1 2 π σ 2 t exp ( − ( x − ( log ( G 0 ) + μ t ) ) 2 2 σ 2 t + 2 x ) d x = ∫ − ∞ ∞ 1 2 π σ 2 t exp ( − ( x − ( log ( G 0 ) + μ t − 2 σ 2 t ) ) 2 2 σ 2 t + log ( G 0 ) + μ t + 2 σ 2 t ) d x = e log ( G 0 ) + μ t + 2 σ 2 t ∫ − ∞ ∞ 1 2 π σ 2 t exp ( − ( x − ( log ( G 0 ) + μ t − 2 σ 2 t ) ) 2 2 σ 2 t ) d x = e log ( G 0 ) + μ t + 2 σ 2 t = G 0 2 e t ( 2 μ + 2 σ 2 ) . \begin{align*}
\mathbb{E}[G_t^2] & = \mathbb{E}[e^{2 \log(G_t)}] \newline
& = \int_{-\infty}^{\infty} e^{2x} \ \mathcal{N}(x; \log(G_0) + \mu t, \sigma^2 t) \ dx \newline
& = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2 t}} \exp\left(-\frac{(x - (\log(G_0) + \mu t))^2}{2 \sigma^2 t} + 2x\right) \ dx \newline
& = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2 t}} \exp\left(-\frac{(x - (\log(G_0) + \mu t - 2 \sigma^2 t))^2}{2 \sigma^2 t} + \log(G_0) + \mu t + 2 \sigma^2 t\right) \ dx \newline
& = e^{\log(G_0) + \mu t + 2 \sigma^2 t} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2 t}} \exp\left(-\frac{(x - (\log(G_0) + \mu t - 2 \sigma^2 t))^2}{2 \sigma^2 t}\right) \ dx \newline
& = e^{\log(G_0) + \mu t + 2 \sigma^2 t} \newline
& = G_0^2 e^{t(2 \mu + 2 \sigma^2)}. \newline
\end{align*} E [ G t 2 ] = E [ e 2 l o g ( G t ) ] = ∫ − ∞ ∞ e 2 x N ( x ; log ( G 0 ) + μ t , σ 2 t ) d x = ∫ − ∞ ∞ 2 π σ 2 t 1 exp ( − 2 σ 2 t ( x − ( log ( G 0 ) + μ t ) ) 2 + 2 x ) d x = ∫ − ∞ ∞ 2 π σ 2 t 1 exp ( − 2 σ 2 t ( x − ( log ( G 0 ) + μ t − 2 σ 2 t ) ) 2 + log ( G 0 ) + μ t + 2 σ 2 t ) d x = e l o g ( G 0 ) + μ t + 2 σ 2 t ∫ − ∞ ∞ 2 π σ 2 t 1 exp ( − 2 σ 2 t ( x − ( log ( G 0 ) + μ t − 2 σ 2 t ) ) 2 ) d x = e l o g ( G 0 ) + μ t + 2 σ 2 t = G 0 2 e t ( 2 μ + 2 σ 2 ) . Thus,
V a r ( G t ) = G 0 2 e t ( 2 μ + 2 σ 2 ) − ( G 0 e t ( μ + 1 2 σ 2 ) ) 2 = G 0 2 e 2 t ( μ + σ 2 ) ( e σ 2 t − 1 ) . \begin{align*}
\mathrm{Var}(G_t) & = G_0^2 e^{t(2 \mu + 2 \sigma^2)} - \left(G_0 e^{t(\mu + \frac{1}{2} \sigma^2)}\right)^2 \newline
& = G_0^2 e^{2 t(\mu + \sigma^2)} (e^{\sigma^2 t} - 1). \newline
\end{align*} Var ( G t ) = G 0 2 e t ( 2 μ + 2 σ 2 ) − ( G 0 e t ( μ + 2 1 σ 2 ) ) 2 = G 0 2 e 2 t ( μ + σ 2 ) ( e σ 2 t − 1 ) . ■ _\blacksquare ■
Modeling with Brownian Motion and Geometric Brownian Motion
Example 1 (Stock Prices) From what we have seen so far, one can model the price of a stock with:
Use a continuous-time stochastic model.
Consider the factor with which it changes, not the differences in prices.
Consider the log-normal distribution for such factors.
Use a parameter for the trend of the price (drift), and one for the volatility (variance) of the price.
Make a Markov assumption (although, we have to reflect on this choice).
This leads to using a geometric Brownian motion as a model,
G t = G 0 e μ t + σ B t . G_t = G_0 e^{\mu t + \sigma B_t}. G t = G 0 e μ t + σ B t . In this context σ \sigma σ
Example 2 A stock price is modeled with G 0 = 67.3 G_0 = 67.3 G 0 = 67.3 μ = 0.08 \mu = 0.08 μ = 0.08 σ = 0.3 \sigma = 0.3 σ = 0.3
Solution We want to compute,
P ( G 3 > 100 ) ≔ P ( G 0 e μ ⋅ 3 + σ B 3 > 100 ) = P ( e μ ⋅ 3 + σ B 3 > 100 G 0 ) = P ( μ ⋅ 3 + σ B 3 > log ( 100 G 0 ) ) = P ( B 3 > log ( 100 G 0 ) − μ ⋅ 3 σ ) \begin{align*}
P(G_3 > 100) & \coloneqq P\left(G_0 e^{\mu \cdot 3 + \sigma B_3} > 100\right) \newline
& = P\left(e^{\mu \cdot 3 + \sigma B_3} > \frac{100}{G_0}\right) \newline
& = P\left(\mu \cdot 3 + \sigma B_3 > \log\left(\frac{100}{G_0}\right)\right) \newline
& = P\left(B_3 > \frac{\log\left(\frac{100}{G_0}\right) - \mu \cdot 3}{\sigma}\right) \newline
\end{align*} P ( G 3 > 100 ) : = P ( G 0 e μ ⋅ 3 + σ B 3 > 100 ) = P ( e μ ⋅ 3 + σ B 3 > G 0 100 ) = P ( μ ⋅ 3 + σ B 3 > log ( G 0 100 ) ) = P B 3 > σ log ( G 0 100 ) − μ ⋅ 3 Since B 3 ∼ N ( 0 , 3 ) B_3 \sim \mathcal{N}(0, 3) B 3 ∼ N ( 0 , 3 ) 1 - pnorm((log(100 / 67.3) - 0.08 * 3) / 0.3, mean = 0, sd = sqrt(3)) in R.