Fanyu Zhao – Page 5 – Learning and Ideas

Black and Scholes, 1973

For,

$$ dS_t = \mu S_t \ dt +\sigma S_t \ dW_t $$

In calculating $d f(S_t)$ , we would get, (by Taylor Expansion)

$$ df(S_t) = \bigg( \frac{\partial f}{\partial t} + \frac{\partial f}{\partial S_t}\mu S_t +\frac{1}{2}\frac{\partial^2 f}{\partial S_t^2}\sigma^2 S_t^2 \bigg)dt + \frac{\partial f}{\partial S_t}\sigma S_t \ dW_t $$

A Special Form of $f(\cdot)$ is $f(S) = log(S)$ ,

$$ d\ log(S_t) = \bigg( \mu – \frac{1}{2}\sigma^2 \bigg)dt + \sigma \ dW_t $$

We get $Y_t = log S_t$ is the price of a derivative security with respect to $S_t$ and $t$ and then,

$$ dY_t= \bigg( \frac{\partial Y_t}{\partial t} + \frac{\partial Y_t}{\partial S_t}\mu S_t +\frac{1}{2}\frac{\partial^2 Y_t}{\partial S_t^2}\sigma^2 S_t^2 \bigg)dt + \frac{\partial Y_t}{\partial S_t}\sigma S_t \ dW_t $$

Consider a portfolio $\Pi$ is constructed with (1) short one derivative, and (2) long some fraction of stocks, $\Delta$ , such that the portfolio is risk natural. ( $\Delta = \frac{\partial Y}{\partial S}$ )

$$ \Pi_t = -Y +\Delta \ S_t $$

Differentiate it,

$$ d\Pi_t = -dY_t +\frac{dY}{dS}dS_t $$

Subtitute $dY_t$ and $dS_t$ into the above equation, we would then get the stochastic process of portfolio, by Ito’s Lemma.

$$ d\Pi_t =-\bigg( \frac{\partial Y}{\partial t} + \frac{1}{2}\frac{\partial^2 Y}{\partial S^2} \sigma^2 S_t^2 \bigg) dt $$

The diffusion term $dW_t$ disappears, and that means the portfolio is riskless during the interval $dt$ . Under a no arbitrage assumption, this portfolio can only earn the riskless return, $r$ .

$$ d\Pi_t =r\Pi_t \ dt $$

Subtitute $d\Pi_t$ and $\Pi_t$ into, we would get the Partial Differential Equation (PDE) / Black-Scholes equation,

$$ – \bigg( \frac{\partial Y}{\partial t} + \frac{1}{2}\frac{\partial^2 Y}{\partial S^2} \sigma^2 S_t^2 \bigg) dt = r\bigg(- Y_t + \frac{\partial Y}{\partial S}S_t \bigg)dt $$

$$ rY_t = \frac{\partial Y}{\partial t} + \frac{1}{2}\frac{\partial^2 Y}{\partial S^2} \sigma^2 S_t^2 + \frac{\partial Y}{\partial S}S_t $$

Then, we guess (where $U(.)$ is a function of $S_t$ at time t=T),

$$ Y_t = e^{-r(T-t)} U(S_T) $$

For a European call with strike price, K, $U(S_T)$ would be the payoff at maturity,

$$ U(S_T) = max( S – K , 0 ) $$

Finally, through a series of process to find a specific solution of the PDE, we can solve the value of call, ( $\Phi(\cdot)$ is the cumulative standard normal distribtion)

$$ c = S_t\Phi(d_1) – K e^{-r (T-t)}\Phi(d_2) $$

with,

$$ d_1 =\frac{ log(S_t/K) + (r-\frac{1}{2}\sigma^2)(T-t)}{\sigma \sqrt{T-t}} $$

$$ d_1 = \sigma \sqrt{T-t} $$

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Volatility Forecast – Ornstein Uhlenbeck Process

VolatilityForecast_OrnsteinUhlenbeck Download

See notes and the realisation.

We assume the volatility of returns follows a stochastic process. Define it as the Ornstein-Uhlenbeck process,

$$dX_t = \kappa (\theta – X_t)dt +\sigma \ dW_t$$

What’s the distribution function of the Ornstein-Uhlenbeck process?

We could apply the MLE estimation to find the parameter of the above random process.

Recall our Vasicek Form Ornstein-Uhlenbeck process is like the following:

$$ dX_t = \kappa (\theta – X_t)dt +\sigma \ dW_t $$

Multiply both sides by $e^{kt}$ , then we get,

$$ e^{kt}dX_t = \kappa e^{kt} \theta \ dt – \kappa e^{kt} X_t \ dt +\sigma e^{kt}\ dW_t $$

We know that $d( e^{kt} X_t )=e^{kt}dX_t + k e^{kt}X_t dt$ , and substitute it inside. Then, we get,

$$ d(e^{kt}X_t)=e^{kt}\theta \ dt + e^{kt}\sigma \ dW_t $$

Take an integral from [0,T],

$$ \int_0^T d(e^{kt}X_t)= \int_0^T e^{kt}\theta \ dt + \int_0^T e^{kt}\sigma \ dW_t $$

$$ X_T = X_0 e^{-kT} +\theta (1-e^{-kT}) + \int_0^T e^{-k(T-t)}\sigma \ dW_t $$

$\int_0^T e^{-k(T-t)}\sigma \ dW_t \sim N(0, \sigma^2\int_0^T e^{-2k(T-t)}dt)$

We then find $\mathbb{E}(X_T)$ and $Var(X_T)$ .

$ \mathbb{E}(X_T) =\mathbb{E}\bigg( X_0 e^{-kT} +\theta (1-e^{-kT}) + \int_0^T e^{-k(T-t)}\sigma \ dW_t \bigg)$

$ \mathbb{E}(X_T)= X_0 e^{-kT} +\theta (1-e^{-kT}) $

$Var(X_T) = \mathbb{E}\bigg( \big( X_T – \mathbb{E}(X_T) \big)^2 \bigg)$

$Var(X_T)= \mathbb{E}\bigg( \big( \int_0^T e^{-k(T-t)}\sigma \ dW_t \big)^2 \bigg) $

By Ito’s Isometry: $I(t)=\int_0^t \Delta(s)dW_s$ , then

$Var[I(t)]=\mathbb{E}[(I^2(t))]=\int_0^t \Delta^2(s)ds$

Then,

$ Var(X_T)= \int_0^T e^{-2k(T-t)}\sigma^2 \ dt $

$ Var(X_T)= \frac{\sigma^2}{2k}\big( 1-e^{-2kT} \big) $

Therefore, we finally get,

$ X_T \sim N\bigg(X_0 e^{-kT} +\theta (1-e^{-kT}), \frac{\sigma^2}{2k}\big( 1-e^{-2kT} \big) \bigg)$

MLE of Ornsterin-Uhlenbeck Process

$X_t \sim N(\cdot,\cdot)$

,where $\mathbb{E}(X_{t+\delta t})=X_0 e^{-k\ \delta t} +\theta (1-e^{-k\ \delta t})$ , and $Var(X_{t+\delta t})= \frac{\sigma^2}{2k}(1-e^{-2k\ \delta t})$ .

$$ f_{\theta}(x_{t+\delta t}|\theta)=\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$

We replace $\mu = \mathbb{E}(X_{t+\delta t})$ and $\sigma^2 = Var(X_{t+\delta t})$ insider.

Code realisation could be found in notes.

Maximum Likelihood Estimator

Joint Probability Density Function or Likelihood Function:

$f(x_1, x_2, …, x_n|\theta) = f(x_1|\theta) f(x_2|\theta) … f(x_N| \theta) $

$f(x_1, x_2, …, x_n|\theta) = \prod_{i=2}^n f(X_i | \theta) =L(\theta)$

A likelihood function is the density function regarded as a function of $\theta$ .

$ L(\theta |x) = \prod f(x|\theta) $, $\theta \in \Theta$

As we know the sample, we need to ‘guess’ the parameter $\theta$ to let the joint prob distribution have a maximum probability.

The Maximum Likelihood Estimator (MLE) is :

$ \hat{\theta}(x) = arg\max_{\theta} L(\theta|x) $

To simplify the calculation, we normally apply a logarithm transformation,

$$ l(\theta)= log\ L $$

$$log L(\theta |x) =log \prod f(x|\theta)= \sum_{i=1}^N log\ f_{\theta} (y_i) $$

As the logarithm transformation is monotonous,

$$ \hat{\theta}(x) = arg\max_{\theta} L(\theta|x) \Leftrightarrow $$

$$\hat{\theta}(x) = arg\max_{\theta} l(\theta|x) $$

To review it, we aim to estimate the parameters. The logic of the MLE estimator is that we guess a probability distribution, fitting into our data (sample). Then, we find or guess the value of parameters such that the joint probability function (likelihood function) achieves the maximum value.

In another word, we guess the probability dist and parameters to make sample data achieve highest possibility.

To find the $argmax_{\theta} l(\theta)$ is equivalent to $argmin_{\theta} -l(\theta)$

A Normal Distribution Example

A normal distribution example,for random sample $x_i$ , i=1, 2, 3, …, N

$$f_{\theta}(x) = \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$

We substitute $f_{\theta}(x)$ into the log-likelihood distribution, then we find,

$$ l(\mu, \sigma^2)=log\ L(\mu,\sigma^2) $$

$$= -\frac{n}{2}\big( log2\pi + log\sigma^2 \big) -\frac{1}{2\sigma^2}\sum_{i=1}^n (x_i -\mu)^2 $$

Taking the First Order Conditions, F.O.C.

By setting $\frac{\partial l}{\partial \mu}= 0$ and $\frac{\partial l}{\partial \sigma^2}= 0$ , we solve $\mu_{MLE}$ and $\sigma^2_{MLE}$ .

$$ \mu_{MLE} =\bar{x}$$

$$\sigma^2_{MLE} =\frac{1}{n} \sum_{i=1}^n (x_i – \bar{x})$$

P.S. We can prove that there are the local maximum, coz the second partial derivatives are negative.

def MLE_norm(x):
    mu_hat = np.mean(x)
    sigma2_hat = np.var(x)
    return mu_hat, sigma2_hat

#----------

mu = 5
sigma = 2.5
N = 10000

np.random.seed(0)
x = np.random.normal( mu, sigma, size=(N,) )

#----------

mu_hat, sigma2_hat = MLE_norm(x)

for_mu_hat = ' $\hat{\mu} = '+format(round(mu_hat,2))+'$ '
for_sigma2_hat = ' $\hat{\sigma} = '+format(round(np.sqrt(sigma2_hat),2))+'$ '
print('The MLE for data is:')
display(Latex(for_mu_hat))
display(Latex(for_sigma2_hat))

By our calculation, we know that mathematically the MLE estimator of mu and sigma are the mean and variance.

Perform MLE Numerically

If the log-likelihood function wasn’t continuous or differentiable. Can solve numerically through an optimisation problem where the objective function is …

In other words, sometimes we cannot calculate the MLE estimator by hand, because the log-likelihood function might not be differentiable. What can we do?

We apply the computer to interactively ‘guess’ values (Optimizer) that can end up with the maximum log-likelihood (or a minimum of negative log-likelihood).

The maximum likelihood estimator (MLE):

$$\hat{\theta}(x) = arg max_{\theta} L(\theta|x)$$

$$L(\theta) = \prod_{i=1}^N f(X_i | \theta) $$

$$ log\ f(X|\theta)=\sum_{i=1}^N log\ f(x_i|\theta)$$

$$ arg\max_{\theta} l(\theta) \Leftrightarrow arg\min_{\theta} -l(\theta) $$

Why we use the minimizer instead? There is only minimizer available in the scipy package, and those mini&max are basically equivalent.

def log_likelihood(theta, x):
    mu = theta[0]
    sigma = theta[1]
    l_theta = np.sum( np.log( sc.stats.norm.pdf(x, loc=mu, scale=sigma) ) )
    # there is a negative sign in front. We minimise it later.
    return -l_theta 

# Constraint Function that Restrict sigma to be positive.
def sigma_pos(theta):
    sigma = np.abs(theta[1])
    return sigma

#--------

cons_set = {'type':'ineq', 'fun': sigma_pos}

theta0 = [2,3] # inital guess. there could be any random numbers, coz it is just the intital value inputed into the optimiser
# find the minimum value of the log likelihood function
opt = sc.optimize.minimize(fun=log_likelihood, x0=theta0, args=(x,), constraints=cons_set)

for_mu_hat = ' $\hat{\mu} = '+format(round(opt.x[0],2))+'$ '
for_sigma2_hat = ' $\hat{\sigma} = '+format(round(opt.x[1],2))+'$ '

print('The MLE for data is:')
display(Latex(for_mu_hat))
display(Latex(for_sigma2_hat))

Maximum-Likelihood-Estimator Download

Risk-Neutral Pricing

Our basic logic is,

$$ \text{Option Value} =\text{Discounted Expectation of the Payoff} $$

However, there is not a perfect correlation between the Option and the Underlying Asset. That fact encourage us to do the risk-netural method for valuing an option.

Geometric Brownian Motion

Non-dividend paying stock, ( $S_0>0$ )

$$ dS_t = \mu S_t\ dt+\sigma S_t \ dW_t $$

Money market, or bank account, ( $B_0 = 1$ )

$$ dB_t = r B_t\ d_t $$

and,

$$ \{ W_t \}_{t\in [0,T]} : \mathbb{P} \sim Brownian \ Motion $$

$ \{S_t\} $ is not a martingals under $\mathbb{P}$ because $\mu S_t$ is not zero.

So, we apply a transformation to $S_t$ and to make it be a martingal (eliminate the drift term).

We here look at $\{ \frac{S_t}{B_t} \}$ ,

$$ d\big( \frac{S_t}{B_t} \big)=-r\frac{S_t}{B_t}dt +\frac{1}{B_t} dS_t +\frac{1}{2}dS_t (-r\frac{1}{B_t}dt) $$

$$ d\big( \frac{S_t}{B_t} \big)=-r\frac{S_t}{B_t}dt +\bigg( \frac{1}{B_t}-\frac{1}{2}r\frac{1}{B_t}dt \bigg)\bigg( \mu S_t\ dt +\sigma S_t \ dW_t \bigg) $$

The cross terms dt & dW would decay to zero quickly in the stochastic integration. We therefore would get,

$$ d\big( \frac{S_t}{B_t} \big) = \sigma ( \frac{S_t}{B_t} )\underbrace{\bigg( \frac{\mu – r}{\sigma} dt + dW_t \bigg)}_{d\tilde{W_t}} $$

The drift disappears, instead we get $d\big( \frac{S_t}{B_t} \big)d\tilde{W_t}$

$$ d\big( \frac{S_t}{B_t} \big)d\tilde{W_t} $$

$$ d\tilde{W_t} = \frac{\mu -r}{\sigma} dt +dW_t $$

by Girsanov’s Theorem that $\{a_t \}_{t\in [0,T]}$ be and adapted $\{ \mathcal{F} \}_{t\in [0,T]}$ Ito Process, so that $\mathbb{Q}$ is an equivalent measure on $\mathbb{P}$ such that $\tilde{W_t}$ is a Brownian Motion, $\tilde{W_t} = W_t +\int_0^t a_u du$ .

$$ dS_t = rS_t \ d_t +\sigma S_t \ d\tilde{W_t} $$

Implication

The implication is that,

We can construct a relicating portfolio $(S_t, B_t)$ with value process $\{ V_t \}_{t\in[0,T]}$ . Under $\mathbb{Q}$ the discounted value process exists that is a martingale $\{ e^{-rt}V_t \}$ .

$$ e^{-rt} C_t =\mathbb{E}_{\mathbb{Q}}[ e^{-rT} C_T | \mathcal{F}_t ] $$

$$ =\mathbb{E}_{\mathbb{Q}}[ e^{-rT} V_T | \mathcal{F}_t ] = e^{-rt} V_t $$

Risk Neutral Expectation Pricing Formula

We therefore get,

$$ \frac{C_t}{B_t} = \mathbb{E}_{\mathbb{Q}}\bigg[ \frac{C_T}{B_t} \bigg| \mathcal{F}_t \bigg] $$

$$ B_T=1,\quad and \quad B_t=exp\{ \int_0^t r_s ds \} $$

Tensorflow – Keras – beginner-Example

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Simulating Geometric Brownian Motion (GBM)

Simulating-Geometric-Brownian-Motion-GBM Download

Here is a realisation of the previous blog, A bit Stochastic Calculus.

Implications

1. Stochastic Integral is different from Ordinary Integral, so as to differentiation. The Stochastic Differential Equation, S.D.E. we normally use to mimic the movement of a stock is as the following,

$$ d S=\mu S d + \sigma S dW$$

, where $dW$ follows a Brownian Motion.

The Brownian Motion also has the following critical properties:

Martingale: $E(S_{t+k}|F_t)=E(S_t|F_t)$ for k>1.
Quadratic Variation: $\sum_{i=1}^n (S_i-S_{i-1})^2 \to t$
Normality: An increment of $S_{t+dt}$ is normally distributed, with mean zero and variance $t_i – t_{i-1}$

2. What does Ito’s Lemma tell us?

If the stock price, $S_t$ , follows a stochastic process, then the financial contracts $F(S_t)$ , as a function of the underlying stock price, follow another stochastic process.

3. Taylor Expansion helps

See notes about Paul Willmott’s book’s paragraph.

4. Different Forms of Stochastic Processes could be applied.

Example 1: $dS=\mu \ dt + \sigma \ dX$

Example 2: $dS=\mu S \ dt +\sigma S \ dX$

For this, $dV = \frac{\partial V}{\partial t}dt+\frac{\partial V}{\partial S}dS+\frac{1}{2}\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2}dt$ , which is also called the Geometric Brownian Motion (GBM). If in a specific form of value function $V=F(S)=ln(S)$ , then $dF = (\mu -\frac{1}{2}\sigma^2)dt + \sigma \ dX$ . See derivation in attached notes.

Example 3: A mean-reverting random walk. $dS=(v-\mu S)dt +\sigma dX$

Example 4: Another mean-reverting r.w. $dS=(v-\mu S)dt +\sigma S^{1/2}dX$

The stochastic term is altered compared with example 3. Now if S ever gets close to zero the randomness decreases,

A bit Stochastic Calculus

A-bit-about-Stochastic-Calculus Download

See the HTML file for full detals.

Key Takeaways

Property of $\{W_t\}$ :

$ W_t – W_s \sim N(0, t-s) $
$(W_t – W_{t-1})$ and $W_{t-i} – W_{t-i+1}$ are uncorrelated. So, $\int_0^t dW_u = \sum^t dW_u = W_t$

For,

$$ dS_t = \mu S_t \ dt +\sigma S_t \ dW_t $$

Why the Geometric Brownian Motion of $\{S_t\}$ is designed in that form?

The answer might be,

$$ dS_t /S_t = \mu \ dt +\sigma \ dW_t $$

$$\int_0^t dS_t /S_t = \int_0^t \mu \ dt + \int_0^t \sigma \ dW_t $$

$$ log(S_t) = \mu \ t +\sigma \ W_t $$

Taking the first difference is similar to differentiation. ( $d(log(S_T)) = log(S_t/S_t-1) = log(1+r_t) \ approx r_t$ ).

$$ r_t = \mu + \sigma \ \Delta W_t $$

The return, $\{r_t\}$ , is equal to a mean, $\mu$ , plus a stochastic term. That is a random walk.

We apply a transformation, if $S_t$ follows a Geometric Brownian Motion, then $f(S_t)$ follows another,

In calculating $d f(S_t)$ , we would get, (by Taylor Expansion)

A Special Form of $f(\cdot)$ is $f(S) = log(S)$ ,

$$ d\ log(S_t) = \bigg( \mu – \frac{1}{2}\sigma^2 \bigg)dt + \sigma \ dW_t $$

Integrate the above equation from time 0 to time t, then we would get,

$$ log(S_t) = log(S_0) + (\mu – \frac{1}{2}\sigma^2)t + \sigma W_t $$

Brownian Motion to Normal Distribution

Codes are shown in the HTML file.

Brownian-Motion-Normal-Dist Download

Markov Property

The Markov Property states that the expected value of the random variable $S_i$ conditional upon all of the past events only depends on the previous value $S_{i−1}$. The implication of the Markov Property is that the

Martingale Property

best expectation of future price, $(S_{t+k}$, is the current price, $S_t$. The current price contains all the information until now.

$$ \mathbb{E}(S_{t+k}|F_t)=S_t, \text{ for k>=0} $$

That is called the Martingale Property.

Quadratic Variation

The Quadratic Variation is defined as,

$$ \sum_{i=1}^k (S_i – S_{i-1})^2 $$

In the coin toss case, each movement must be either “1” or “-1”, so $S_i – S_{i-1} = \pm 1$. Thus, $(S_i – S_{i-1})^2 =1$ for all “i”. Therefore, the Quadratic Variation equals “k”.

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