Introduction
In this last part, we will continue looking at Brownian motions and Gaussian processes.
We will cover some applications of what we have learned so far.
Example: Counting Zeroes
Example 1 (Counting Zeroes) How many zeroes does B t B_t B t ( 0 , s ) (0, s) ( 0 , s )
If L s L_s L s ( 0 , s ) (0, s) ( 0 , s )
L s s ∼ B e t a ( 1 2 , 1 2 ) , \frac{L_s}{s} \sim \mathrm{Beta}\left(\frac{1}{2}, \frac{1}{2}\right), s L s ∼ Beta ( 2 1 , 2 1 ) , which means that, with probability one, there exists a zero in the interval ( 0 , s ) (0, s) ( 0 , s ) L s L_s L s 0 < L s < s 0 < L_s < s 0 < L s < s
Repeating the argument, we have that 0 < L L s < L s 0 < L_{L_s} < L_s 0 < L L s < L s ( 0 , L s ) ⊂ ( 0 , s ) (0, L_s) \subset (0, s) ( 0 , L s ) ⊂ ( 0 , s )
The conclusion is that, with probability one, there is an infinite number of zeroes in ( 0 , s ) (0, s) ( 0 , s ) s > 0 s > 0 s > 0
Stock Options
Intuition (Stock Options) A (European) stock option is a right (but not an obligation) to buy a stock at a given time t t t K K K
How much can you expect to earn from a stock option at that future time?
We get that (we will derive this),
E [ max ( G t − K , 0 ) ] = G 0 e t ( μ + σ 2 2 ) P ( B 1 > β − σ t t ) − K P ( B 1 > β t ) , \mathbb{E}[\max(G_t - K, 0)] = G_0 e^{t(\mu + \frac{\sigma^2}{2})} P\left(B_1 > \frac{\beta - \sigma t}{\sqrt{t}}\right) - K P\left(B_1 > \frac{\beta}{\sqrt{t}}\right), E [ max ( G t − K , 0 )] = G 0 e t ( μ + 2 σ 2 ) P ( B 1 > t β − σ t ) − K P ( B 1 > t β ) , where β = log ( K / G 0 ) − μ t σ \beta = \frac{\log(K / G_0) - \mu t}{\sigma} β = σ l o g ( K / G 0 ) − μ t
Derivation Firstly, we need to prove the algebraic identity,
e σ x N ( x ; 0 , t ) = e σ 2 t 2 N ( x ; σ t , t ) . e^{\sigma x} \mathcal{N}(x; 0, t) = e^{\frac{\sigma^2 t}{2}} \mathcal{N}(x; \sigma t, t). e σ x N ( x ; 0 , t ) = e 2 σ 2 t N ( x ; σ t , t ) . Proof We have,
e σ x N ( x ; 0 , t ) = 1 2 π t exp ( − x 2 2 t + σ x ) = 1 2 π t exp ( − ( x − σ t ) 2 2 t + σ 2 t 2 ) = e σ 2 t 2 N ( x ; σ t , t ) . ■ \begin{align*}
e^{\sigma x} \mathcal{N}(x; 0, t) & = \frac{1}{\sqrt{2 \pi t}} \exp\left(-\frac{x^2}{2 t} + \sigma x\right) \newline
& = \frac{1}{\sqrt{2 \pi t}} \exp\left(-\frac{(x - \sigma t)^2}{2 t} + \frac{\sigma^2 t}{2}\right) \newline
& = e^{\frac{\sigma^2 t}{2}} \mathcal{N}(x; \sigma t, t). \ _\blacksquare
\end{align*} e σ x N ( x ; 0 , t ) = 2 π t 1 exp ( − 2 t x 2 + σ x ) = 2 π t 1 exp ( − 2 t ( x − σ t ) 2 + 2 σ 2 t ) = e 2 σ 2 t N ( x ; σ t , t ) . ■ Then, defining β = log ( K / G 0 ) − μ t σ \beta = \frac{\log(K / G_0) - \mu t}{\sigma} β = σ l o g ( K / G 0 ) − μ t
E [ max ( G t − K , 0 ) ] ≔ E [ max ( G 0 e μ t + σ B t − K , 0 ) ] = ∫ − ∞ ∞ max ( G 0 e μ t + σ x − K , 0 ) N ( x ; 0 , t ) d x = ∫ β ∞ ( G 0 e μ t + σ x − K ) N ( x ; 0 , t ) d x = G 0 e μ t ∫ β ∞ e σ x N ( x ; 0 , t ) d x − K ∫ β ∞ N ( x ; 0 , t ) d x = G 0 e t ( μ + σ 2 2 ) ∫ β ∞ N ( x ; σ t , t ) d x − K ∫ β ∞ N ( x ; 0 , t ) d x = G 0 e t ( μ + σ 2 2 ) P ( B 1 > β − σ t t ) − K P ( B 1 > β t ) . ■ \begin{align*}
\mathbb{E}[\max(G_t - K, 0)] & \coloneqq \mathbb{E}[\max(G_0 e^{\mu t + \sigma B_t} - K, 0)] \newline
& = \int_{-\infty}^{\infty} \max(G_0 e^{\mu t + \sigma x} - K, 0) \ \mathcal{N}(x; 0, t) \ dx \newline
& = \int_{\beta}^{\infty} (G_0 e^{\mu t + \sigma x} - K) \ \mathcal{N}(x; 0, t) \ dx \newline
& = G_0 e^{\mu t} \int_{\beta}^{\infty} e^{\sigma x} \ \mathcal{N}(x; 0, t) \ dx - K \int_{\beta}^{\infty} \mathcal{N}(x; 0, t) \ dx \newline
& = G_0 e^{t(\mu + \frac{\sigma^2}{2})} \int_{\beta}^{\infty} \mathcal{N}(x; \sigma t, t) \ dx - K \int_{\beta}^{\infty} \mathcal{N}(x; 0, t) \ dx \newline
& = G_0 e^{t(\mu + \frac{\sigma^2}{2})} P\left(B_1 > \frac{\beta - \sigma t}{\sqrt{t}}\right) - K P\left(B_1 > \frac{\beta}{\sqrt{t}}\right). \ _\blacksquare
\end{align*} E [ max ( G t − K , 0 )] : = E [ max ( G 0 e μ t + σ B t − K , 0 )] = ∫ − ∞ ∞ max ( G 0 e μ t + σ x − K , 0 ) N ( x ; 0 , t ) d x = ∫ β ∞ ( G 0 e μ t + σ x − K ) N ( x ; 0 , t ) d x = G 0 e μ t ∫ β ∞ e σ x N ( x ; 0 , t ) d x − K ∫ β ∞ N ( x ; 0 , t ) d x = G 0 e t ( μ + 2 σ 2 ) ∫ β ∞ N ( x ; σ t , t ) d x − K ∫ β ∞ N ( x ; 0 , t ) d x = G 0 e t ( μ + 2 σ 2 ) P ( B 1 > t β − σ t ) − K P ( B 1 > t β ) . ■ 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,
E [ max ( G 3 − 100 , 0 ) ] = 67.3 e 3 ( 0.08 + 0.3 2 2 ) P ( B 1 > β − 0.3 ⋅ 3 3 ) − 100 P ( B 1 > β 3 ) \begin{align*}
\mathbb{E}[\max(G_3 - 100, 0)] & = 67.3 e^{3(0.08 + \frac{0.3^2}{2})} P\left(B_1 > \frac{\beta - 0.3 \cdot 3}{\sqrt{3}}\right) - 100 P\left(B_1 > \frac{\beta}{\sqrt{3}}\right) \newline
\end{align*} E [ max ( G 3 − 100 , 0 )] = 67.3 e 3 ( 0.08 + 2 0. 3 2 ) P ( B 1 > 3 β − 0.3 ⋅ 3 ) − 100 P ( B 1 > 3 β ) where,
β = log ( 100 / 67.3 ) − 0.08 ⋅ 3 0.3 ≈ 1.213. \beta = \frac{\log(100 / 67.3) - 0.08 \cdot 3}{0.3} \approx 1.213. β = 0.3 log ( 100/67.3 ) − 0.08 ⋅ 3 ≈ 1.213. Thus, we can compute this as 67.3 * exp(3 * (0.08 + 0.3^2 / 2)) * (1 - pnorm((1.213 - 0.3 * 3) / sqrt(3), mean = 0, sd = 1)) - 100 * (1 - pnorm(1.213 / sqrt(3), mean = 0, sd = 1)) in R.
Martingales
Definition 1 (Martingale) A stochastic process { Y t } t ≥ 0 \{Y_t\}_{t \geq 0} { Y t } t ≥ 0 t ≥ 0 t \geq 0 t ≥ 0
E [ Y t ∣ Y r , 0 ≤ r ≤ s ] = Y s \mathbb{E}[Y_t \mid Y_r, 0 \leq r \leq s] = Y_s E [ Y t ∣ Y r , 0 ≤ r ≤ s ] = Y s 0 ≤ s ≤ t 0 \leq s \leq t 0 ≤ s ≤ t E [ ∣ Y t ∣ ] < ∞ \mathbb{E}[|Y_t|] < \infty E [ ∣ Y t ∣ ] < ∞ B t B_t B t Proof We have,
E [ B t ∣ B r , 0 ≤ r ≤ s ] = E [ B t − B s + B s ∣ B r , 0 ≤ r ≤ s ] = E [ B t − B s ∣ B r , 0 ≤ r ≤ s ] + E [ B s ∣ B r , 0 ≤ r ≤ s ] = 0 + B s = B s . \begin{align*}
\mathbb{E}[B_t \mid B_r, 0 \leq r \leq s] & = \mathbb{E}[B_t - B_s + B_s \mid B_r, 0 \leq r \leq s] \newline
& = \mathbb{E}[B_t - B_s \mid B_r, 0 \leq r \leq s] + \mathbb{E}[B_s \mid B_r, 0 \leq r \leq s] \newline
& = 0 + B_s \newline
& = B_s.
\end{align*} E [ B t ∣ B r , 0 ≤ r ≤ s ] = E [ B t − B s + B s ∣ B r , 0 ≤ r ≤ s ] = E [ B t − B s ∣ B r , 0 ≤ r ≤ s ] + E [ B s ∣ B r , 0 ≤ r ≤ s ] = 0 + B s = B s . and,
E [ ∣ B t ∣ ] = ∫ − ∞ ∞ ∣ x ∣ N ( x ; 0 , t ) d x = 2 ∫ 0 ∞ x N ( x ; 0 , t ) d x = 2 ∫ 0 ∞ x 1 2 π t exp ( − x 2 2 t ) d x = 2 ⋅ 1 2 π t ⋅ t = 2 t π < ∞ . \begin{align*}
\mathbb{E}[|B_t|] & = \int_{-\infty}^{\infty} |x| \ \mathcal{N}(x; 0, t) \ dx \newline
& = 2 \int_{0}^{\infty} x \ \mathcal{N}(x; 0, t) \ dx \newline
& = 2 \int_{0}^{\infty} x \ \frac{1}{\sqrt{2 \pi t}} \exp\left(-\frac{x^2}{2 t}\right) \ dx \newline
& = 2 \cdot \frac{1}{\sqrt{2 \pi t}} \cdot t \newline
& = \sqrt{\frac{2 t}{\pi}} < \infty. \newline
\end{align*} E [ ∣ B t ∣ ] = ∫ − ∞ ∞ ∣ x ∣ N ( x ; 0 , t ) d x = 2 ∫ 0 ∞ x N ( x ; 0 , t ) d x = 2 ∫ 0 ∞ x 2 π t 1 exp ( − 2 t x 2 ) d x = 2 ⋅ 2 π t 1 ⋅ t = π 2 t < ∞. Further, one can define a martingale with respect to other stochastic processes.
{ Y t } t ≥ 0 \{Y_t\}_{t \geq 0} { Y t } t ≥ 0 { X t } t ≥ 0 \{X_t\}_{t \geq 0} { X t } t ≥ 0 t ≥ 0 t \geq 0 t ≥ 0
E [ Y t ∣ X r , 0 ≤ r ≤ s ] = Y s \mathbb{E}[Y_t \mid X_r, 0 \leq r \leq s] = Y_s E [ Y t ∣ X r , 0 ≤ r ≤ s ] = Y s 0 ≤ s ≤ t 0 \leq s \leq t 0 ≤ s ≤ t E [ ∣ Y t ∣ ] < ∞ \mathbb{E}[|Y_t|] < \infty E [ ∣ Y t ∣ ] < ∞ Example 3 Let Y t ≔ B t 2 − t Y_t \coloneqq B_t^2 - t Y t : = B t 2 − t t ≥ 0 t \geq 0 t ≥ 0 { Y t } t ≥ 0 \{Y_t\}_{t \geq 0} { Y t } t ≥ 0 { B t } t ≥ 0 \{B_t\}_{t \geq 0} { B t } t ≥ 0
Proof We have,
E [ Y t ∣ B r , 0 ≤ r ≤ s ] = E [ B t 2 − t ∣ B r , 0 ≤ r ≤ s ] = E [ ( B t − B s + B s ) 2 − t ∣ B r , 0 ≤ r ≤ s ] = E [ ( B t − B s ) 2 ∣ B r , 0 ≤ r ≤ s ] + 2 B s E [ B t − B s ∣ B r , 0 ≤ r ≤ s ] + B s 2 − t = ( t − s ) + 0 + B s 2 − t = B s 2 − s = Y s . \begin{align*}
\mathbb{E}[Y_t \mid B_r, 0 \leq r \leq s] & = \mathbb{E}[B_t^2 - t \mid B_r, 0 \leq r \leq s] \newline
& = \mathbb{E}[(B_t - B_s + B_s)^2 - t \mid B_r, 0 \leq r \leq s] \newline
& = \mathbb{E}[(B_t - B_s)^2 \mid B_r, 0 \leq r \leq s] + 2 B_s \mathbb{E}[B_t - B_s \mid B_r, 0 \leq r \leq s] + B_s^2 - t \newline
& = (t - s) + 0 + B_s^2 - t \newline
& = B_s^2 - s \newline
& = Y_s.
\end{align*} E [ Y t ∣ B r , 0 ≤ r ≤ s ] = E [ B t 2 − t ∣ B r , 0 ≤ r ≤ s ] = E [( B t − B s + B s ) 2 − t ∣ B r , 0 ≤ r ≤ s ] = E [( B t − B s ) 2 ∣ B r , 0 ≤ r ≤ s ] + 2 B s E [ B t − B s ∣ B r , 0 ≤ r ≤ s ] + B s 2 − t = ( t − s ) + 0 + B s 2 − t = B s 2 − s = Y s . Further,
E [ ∣ Y t ∣ ] = E [ ∣ B t 2 − t ∣ ] ≤ E [ B t 2 ] + t = V a r ( B t ) + ( E [ B t ] ) 2 + t = t + 0 + t = 2 t < ∞ . \begin{align*}
\mathbb{E}[|Y_t|] & = \mathbb{E}[|B_t^2 - t|] \newline
& \leq \mathbb{E}[B_t^2] + t \newline
& = \mathrm{Var}(B_t) + (\mathbb{E}[B_t])^2 + t \newline
& = t + 0 + t \newline
& = 2 t < \infty. \newline
\end{align*} E [ ∣ Y t ∣ ] = E [ ∣ B t 2 − t ∣ ] ≤ E [ B t 2 ] + t = Var ( B t ) + ( E [ B t ] ) 2 + t = t + 0 + t = 2 t < ∞. Thus, { Y t } t ≥ 0 \{Y_t\}_{t \geq 0} { Y t } t ≥ 0 { B t } t ≥ 0 \{B_t\}_{t \geq 0} { B t } t ≥ 0 ■ _\blacksquare ■
Derivation (Geometric Brownian Motion can be a Martingale) Let G t ≔ G 0 e μ t + σ B t G_t \coloneqq G_0 e^{\mu t + \sigma B_t} G t : = G 0 e μ t + σ B t
E [ G t ∣ B r , 0 ≤ r ≤ s ] ≔ E [ G 0 e μ t + σ B t ∣ B r , 0 ≤ r ≤ s ] = E [ G 0 e μ ( t − s ) + σ ( B t − B s ) + μ s + σ B s e μ s + σ B s ∣ B r , 0 ≤ r ≤ s ] = E [ G t − s ] e μ s + σ B s = G 0 e ( t − s ) ( μ + σ 2 2 ) e μ s + σ B s . = G s e ( t − s ) ( μ + σ 2 2 ) . \begin{align*}
\mathbb{E}[G_t \mid B_r, 0 \leq r \leq s] & \coloneqq \mathbb{E}[G_0 e^{\mu t + \sigma B_t} \mid B_r, 0 \leq r \leq s] \newline
& = \mathbb{E}[G_0 e^{\mu(t - s) + \sigma (B_t - B_s) + \mu s + \sigma B_s} e^{\mu s + \sigma B_s} \mid B_r, 0 \leq r \leq s] \newline
& = \mathbb{E}[G_{t - s}] e^{\mu s + \sigma B_s} \newline
& = G_0 e^{(t - s)(\mu + \frac{\sigma^2}{2})} e^{\mu s + \sigma B_s}. \newline
& = G_s e^{(t - s)(\mu + \frac{\sigma^2}{2})}. \newline
\end{align*} E [ G t ∣ B r , 0 ≤ r ≤ s ] : = E [ G 0 e μ t + σ B t ∣ B r , 0 ≤ r ≤ s ] = E [ G 0 e μ ( t − s ) + σ ( B t − B s ) + μ s + σ B s e μ s + σ B s ∣ B r , 0 ≤ r ≤ s ] = E [ G t − s ] e μ s + σ B s = G 0 e ( t − s ) ( μ + 2 σ 2 ) e μ s + σ B s . = G s e ( t − s ) ( μ + 2 σ 2 ) . We see that G t G_t G t B t B_t B t μ + σ 2 2 = 0 \mu + \frac{\sigma^2}{2} = 0 μ + 2 σ 2 = 0 μ = − σ 2 2 \mu = -\frac{\sigma^2}{2} μ = − 2 σ 2
Example 4 (Discounting Future Values of Stocks & The Black-Scholes Formula for Option Pricing) When making investments, there is always a range of choices, some of which are sometimes called “risk-free”.
Such investments may pay a fixed interest.
When interests are compounded frequently, a reasonable model is that an investment of G 0 G_0 G 0 G 0 e r t G_0 e^{rt} G 0 e r t t t t r r r
A common way to take this alternative into account is to instead “discount” all other investments with the factor e − r t e^{-rt} e − r t
For example, the discounted value of a stock can be modeled as,
e − r t G t = e − r t G 0 e μ t + σ B t = G 0 e ( μ − r ) t + σ B t . e^{-rt} G_t = e^{-rt} G_0 e^{\mu t + \sigma B_t} = G_0 e^{(\mu - r) t + \sigma B_t}. e − r t G t = e − r t G 0 e μ t + σ B t = G 0 e ( μ − r ) t + σ B t . A possible assumption about the trend μ \mu μ B t B_t B t
Thus, we get,
μ − r + σ 2 2 = 0 , i.e., μ = r − σ 2 2 . \mu - r + \frac{\sigma^2}{2} = 0, \quad \text{i.e.,} \quad \mu = r - \frac{\sigma^2}{2}. μ − r + 2 σ 2 = 0 , i.e., μ = r − 2 σ 2 . The Black-Scholes formula for option pricing is based on,
Assuming the discounted stock price is a martingale with respect to the Brownian motion.
Using discounting when computing the value of the option.
From this, we get,
e − r t E [ max ( G t − K , 0 ) ] = G 0 P ( B 1 > β − σ t t ) − K e − r t P ( B 1 > β t ) , e^{-rt} \mathbb{E}[\max(G_t - K, 0)] = G_0 P\left(B_1 > \frac{\beta - \sigma t}{\sqrt{t}}\right) - K e^{-rt} P\left(B_1 > \frac{\beta}{\sqrt{t}}\right), e − r t E [ max ( G t − K , 0 )] = G 0 P ( B 1 > t β − σ t ) − K e − r t P ( B 1 > t β ) , where β = log ( K / G 0 ) − ( r − σ 2 2 ) t σ \beta = \frac{\log(K / G_0) - (r - \frac{\sigma^2}{2}) t}{\sigma} β = σ l o g ( K / G 0 ) − ( r − 2 σ 2 ) t