LECTURE 15

Deficiencies of the Black-Scholes model. Two of the most well known deficiencies of the Black-Scholes model are 1. Failure to predict the volatility smile, 2. Empirical evidence questioning the log-normality of asset prices. Several refinements have been proposed, many of them using jump processes which are generalizations of the Poisson process.

Characteristic functions of compound Poisson distributions. The Poisson distribution is p(n) = \frac{λ^{n}}{n!} e^{−λ}, n = 0, 1, . . .. Its characteristic function is ∑_{n=0}^{∞} e^{izn}p(n) =

∑_{n=0}^{∞} \frac{(λe^{iz})^{n}}{n!} e^{−λ} = e^{λ(e^{iz}−1)}.

Let N have a Poisson distribution and X_{1}, X_{2}, … be i.i.d. r.v.’s. Then Y = X_{1} + ⋯ + X_{N} has a compound Poisson distribution. Suppose that ϕ(z) is the characteristic function of X_{i}. Then the characteristic function of Y is E[e^{izY}] = E[e^{iz(X_{1}+···+X_{N})}] = ∑_{n=0}^{∞} E[e^{iz(X_{1}+···+X_{n})}]P (N = n) = ∑_{n=0}^{∞} ϕ^{n}(z)p(n) = e^{λ(ϕ(z)−1)}. Let μ be the distribution of the X_{i}. Then ϕ(z) = ∫ e^{izy}μ(dy). If ν = λμ then −λ(ϕ(z) − 1) = ∫_{R−{0}}(1 − e^{izy})ν(dy). Let η = ∫ \frac{y}{1+y^{2}} ν(dy). Then we can write the characteristic function of a compound Poisson distribution as

exp (−ψ(z))  where  ψ(z) = −iηz + ∫_{R−\{0\}}(1 − e^{izy} + \frac{y}{1 + |y|^{2}} )ν(dy)

where ν is a finite measure on R − {0}.

Skewness, kurtosis, fat tails. If X is a random variable with mean μ and variance σ^{2} then the skewness and excess kurtosis are given by

χ = \frac{E[(X − μ)^{3}]}{σ^{3}} κ = \frac{E[(X − μ)^{4}]}{σ^{4}} − 3.

A kurtosis of larger than 3 says that the distribution is in some sense more spread out than a normal distribution.

Consider a compound Poisson distribution where the underlying distribution is normal: Let N be a Poisson random variable of mean 1 and X_{1}, X_{2}, … be N (0, 1). Let X = X_{1} + ⋯ + X_{N}. One can check μ = 0 and σ = 1. The skewness is 0 by symmetry. The excess kurtosis is

∑_{n=0}^{∞} E[(X_{1} + ⋯ + X_{n})^{4}]p(n) = ∑_{n=0}^{∞}(3n + n(n − 1))p(n) − 3 = 1.

So it is more spread out than a normal distribution.

Infinitely divisible laws. An infinitely divisible distribution F is one such that for each n = 1, 2, … it can be written as F = F_{n}^{n∗} for some distribution F_{n}. In other words if a random variable X has an infinitely divisible law, then for each n there are random variables X_{1}, …, X_{n} which are i.i.d. and X = X_{1} + ⋯ + X_{n}.

Examples of infinitely divisible distributions are the normal distribution and the compound Poisson distribution, and in fact every infinitely divisible distribution is the limit of compound Poissons.

The reason they are important is that any distribution which arises as the limit of a sequence sums of i.i.d. random variables will be infinitely divisible.

Levy-Khinchine formula. A Levy measure is a measure on R − {0} such that ∫_{R−{0}}(y^{2} ∧ 1)ν(dy) < ∞. A distribution is infinitely divisible if and only if there are numbers η and σ and a Levy measure ν such that the characterstic function is given by

exp (−ψ(z))  where  ψ(z) = −iηz + \frac{σ^{2}z^{2}}{2} + ∫_{R−\{0\}}(1 − e^{izy} + \frac{y}{1 + |y|^{2}} )ν(dy).

A special class of examples are those whose characteristic functions take the form exp{−imz +k|z|^{α}} with 0 < α ≤ 2. α = 2 is the Normal distribution. α = 1 is the Cauchy distribution.

Levy processes. One characterisation of Brownian motion is that it is the unique process with stationary, independent increments and continuous sample paths. Now we drop the assumption of continuous sample paths. A Levy process X(t) is a process with stationary independent increments, i.e., if 0 = t_{0} < t_{1} < ⋯ < t_{n} then P (X(t_{j})−X(t_{j−1}) ∈ A_{j}, j = 1, …, n) = ∏_{j+1}^{n} P (X(t_{j}) − X(t_{j−1}) ∈ A_{j}).

Examples of Levy processes are Brownian motion and compound Poisson processes. Since the process is the sum of i.i.d. increments, i.e. X(t) = (X(t/n) − X(0)) +

(X(2t/n) − X(t/n)) + ⋯ + (X(t) − X(n − 1)t/n)), it has a fortiori infinitely divisible marginal distributions. Conversely there is a Levy process corresponding to each infinitely divisible distribution.

Cauchy process. Let X_{1}, X_{2}, … be independent Cauchy random variables and let S_{n} = X_{1} + ⋯ + X_{n}. Notice that the characteristic function of \frac{S_{n}}{n} is ϕ^{n}(\frac{z}{n}) = (e^{k| \frac{z}{n} |})^{n} = ϕ(z). In other words, \frac{S_{n}}{n} has the same distribution as each X_{i}. The limit Y (t) = lim_{n→∞} \frac{S_{⌊nt⌋}}{n} is called the Cauchy process.

Markov property. Because of the independent increments property, Levy processes are Markov processes. For any reasonable function f,

f(t, X(t)) − f(0, X(0)) − ∫_{0}^{t}[\frac{∂}{∂s} + L]f(X(s))ds

is a martingale where

Lf(x) = \frac{1}{2} ∑_{i,j=1}^{d} a_{ij} \frac{∂^{2}f}{∂x_{i}∂x_{j}} + ∑_{i=1}^{d} η_{i} \frac{∂f}{∂x_{i}} + ∫ (f(x + y) − f(x) + \frac{y · ∇f}{1 + |y|^{2}} )ν(dy)

and a = σσ^{T}. All the connections of diffusions with partial differential equations have exact analogies here with L replacing the partial differential operators there. So the resulting equations are partial differential-difference equations (which are unfortunately rather hard to solve).

A model for stock prices with jumps. Let N(t) be a Poisson process with intensity λ. We modify the geometric Brownian motion model for prices by adding, at exponential times, downward jumps by a factor J by considering

dS = μSdt + σSdB + (J − 1)SdN.

As usual this means

S(T) − S(0) = ∫_{0}^{T} μS(t)dt + ∫_{0}^{T} σS(t)dB(t) + ∫_{0}^{T} (J − 1)S(t)dN(t).

Note that N(t) is a function of bounded variation and the last integral can be defined without resorting to all sorts of stochastic mumbo jumbo. In fact N(t) just has jumps of size 1 at times 0 < t_{1} < t_{2} < ⋯ < t_{n} < T and the last integral is really just

∫_{0}^{T} (J − 1)S(t)dN(t) = ∑_{i=1}^{n}(J − 1)S((t_{i})_{−})

where S(t_{−}) = lim_{u↑t} S(u) is the left hand limit.

If you choose J = 0.95 then what you get is geometric Brownian motion plus price drops of 5% at times t_{1}, …, t_{n}, which are random with independent exponential times between them of average length 1/λ.

You could also have the J be random, modelling random sizes of drops in the market. Choose initially J_{1}, J_{2}, … independent out of some distribution F and define the last object to be

∑_{i=1}^{n}(J_{i} − 1)S((t_{i})_{−}).

Of course the resulting process is a combination of geometric Brownian motion and a compound Poisson process. The corresponding Markov generator is

Lf(x) = μx\frac{∂f}{∂x} + \frac{1}{2}σ^{2} \frac{∂^{2}f}{∂x^{2}} + λ ∫ (f(yx) − f(x))dF (y).

We want to see what happens when we try to hedge in the presence of jumps. Suppose we have a portfolio consisting of one (European) option and −Δ of the stock. Let V (S(t), t) be the value of the option at time t. The portfolio is worth Π = V (S, t)−δS. Differentiating in time and using the self-financing condition

dΠ = (\frac{∂V}{∂t} + \frac{1}{2}σ^{2}S^{2} \frac{∂^{2}V}{∂S^{2}} ) dt + (\frac{∂V}{∂S} − Δ) dS + (V (JS) − V (S) − Δ(J − 1)S) dN.

Remember that N jumps at discrete times t_{1}, t_{2}, . . .. If we are not at one of those times then we think we are in the Black-Scholes world and we want to choose \frac{∂V}{∂S} = Δ so that the ‘risky’ term drops out and we obtain a deterministic partial differential equation for the option price by setting dΠ equal to the risk free return. On the other hand at those discrete times we still have the jumps. So we obtain

dΠ = (\frac{∂V}{∂t} + \frac{1}{2}σ^{2}S^{2} \frac{∂^{2}V}{∂S^{2}} ) dt + (V (JS) − V (S) − (J − 1)S \frac{∂V}{∂S} ) dN.

The risk cannot be fully hedged so what is to be done? The simplest thing is just to take expectation and set is equal to the risk free return on the portfolio. This is justified by saying that the diversifiable risk should not be rewarded so the jumps should not be priced into the option. I would be very grateful if someone could tell me what that means. You get

\frac{∂V}{∂t} + \frac{1}{2}σ^{2}S^{2} \frac{∂^{2}V}{∂S^{2}} − rS \frac{∂V}{∂S} − rV + λ ∫ (V (yS) − V (S) − (y − 1)S \frac{∂V}{∂S} ) dF (y) = 0.

If F is a lognormal distribution then there is an exact solution given by

V (S, t) = ∑_{n=1}^{∞} \frac{(λ̃(T − t))^{n}}{n!} e^{−_{λ}(T −t)}V_{BS}(S, t, σ_{n}, r_{n})

where λ̃ = λ(1 + k), σ_{n} = σ^{2}(1 + \frac{n(1+k)^{2}}{T −t}), r_{n} = r − λk + \frac{n log(1+k)}{T −t} and k = ∫ (y − 1)dF (y).

Drawbacks of Levy processes as models for stock movements. 1. They cannot be hedged; 2. Too many parameters to calibrate; 3. Difficult to simulate; 4. No exact solutions.

Continuous space random walk models. In practice one has discrete times 0, \frac{1}{n}, \frac{2}{n}, … when trading happens and therefore sophisticated models like Levy processes are not necessary. Instead we choose a jump law μ(dy) which is a probability measure on R, we choose independent random variables X_{1}, X_{2}, … with common distribution μ and we let X(t) = n^{−1/2}(X_{1} + ⋯ + X_{⌊nt}). So this process waits time \frac{1}{n} then takes a jump of size \frac{1}{\sqrt{n}}× a random variable with distribution μ, then waits another time \frac{1}{n} and takes another such jump, etc. The process can be specified by saying that for any nice f,

f(X(t)) − f(X(0)) − ∑_{i=0}^{⌊nt⌋−1} Lf(X(i/n))

is a martingale where

Lf(x) = ∫ (f(x + y) − f(x))μ(dy).

Use of continuous space random walk models to explain the volatility smile. Suppose we use a continuous space random walk to model our stock price. We have S_{\frac{j+1}{n}} = S_{\frac{j}{n}} exp\{\frac{r}{n} + σX_{j}\} where X_{1}, X_{2}, … are independent with distribution μ. For risk neutrality we insist that E[X_{j}] = ∫ xμ(dx) = 0 to match the interest rate. We choose some distribution for μ which should have the following properties: 1. It should have fatter tails than the normal distribution, 2. It should be negatively skewed. We have already discussed the empirical justification for 1. The negative skewness is because it is observed that large negative jumps in the market are less rare than large positive jumps. We already decided the distribution should have mean zero. The variance can be set to 1 without loss of generality because of the multiplicative parameter σ. In order to calibrate the model one would like the distribution to depend on two extra parameters, and for these we may as well choose the skewness χ and the excess kurtosis κ. So we have a family of distributions μ_{χ,κ} depending on real parameters χ and κ and μ_{0,0} is the normal distribution.

The price of a call option with strike price K and expiration date T at time t < T is V_{χ,κ}(S(t), T, t, K, σ, r) = E[e^{−r(T −t)}(S_{T} − K)_{+} | F_{t}].

It depends on σ, χ and κ through the law of S_{T}.

We also have the Black-Scholes solution

V_{0,0}(S, T, t, K, σ, r) = SΦ (\frac{log(S/K) + (r + \frac{1}{2}σ^{2})(T − t)}{σ\sqrt{T − t}})−Ke^{−r(T −t)}Φ (\frac{log(S/K) + (r − \frac{1}{2}σ^{2})(T − t)}{σ\sqrt{T − t}})

where Φ(x) = (2π)^{−1/2} ∫_{−∞}^{x} e^{−y^{2}/2}dy.

Observing the market we can determine the true price of the call, C(K, T) as a function of K and T, as well as S and r. Solving for σ it terms of K and T, i.e. defining implicitly

V_{0,0}(S, T, t, K, σ(K, T), r) = C(K, T)

gives the implied volatility σ(K, T).

We could also define σ̄_{σ,χ,κ}(K, T) implicitly by

V_{0,0}(S, T, t, K, σ̄_{σ,χ,κ}(K, T), r) = V_{χ,κ}(S, T, t, K, σ, r).

Note that σ̄_{σ,0,0}(K, T) = σ.

The idea then is to choose σ, χ, κ in order to minimize |σ̄_{σ,χ,κ}(K, T) − σ(K, T)|. If we graph σ̄_{σ,χ,κ}(K, T) as a function of K for fixed T we see the ‘smile’ which just

means deviation from the constant line σ. As T gets larger the smile fades away - the curve becomes more and more a constant. The reason is that the central limit theorem takes over, and no matter what kernel we started with, S_{T} becomes closer and closer to lognormal. Qualitatively, χ is responsible for the slope of the curve and κ is responsible for the curvature. For example, the explanation for the smile is that a higher volatility is required to produce the same weight in the tails for the normal model as for the model with higher kurtosis.

Jump process heteroskedactic model. Another explanation for the volatility smile is that the volatility σ should itself be a random process. One way to model this is to have σ(t) hop between two values σ_{1} and σ_{2} as a Poisson process which is independent of the Brownian motion used to drive the process. In other words we have

dS(t) = μS(t)dt + σ(t)S(t)dB(t), dσ(t) = (1_{σ(t_{−})=σ_{1}} − 1_{σ(t_{−})=σ_{2}})(σ_{2} − σ_{1})dN(t). For each fixed realization σ(·, ω) we know the price is given by

V (σ(·)) = SΦ (\frac{log(S/K) + (r + \frac{1}{2}σ^{2})(T − t)}{\sqrt{∫_{t}^{T} σ^{2}(u)du}})−Ke^{−r(T −t)}Φ (\frac{log(S/K) + (r − \frac{1}{2}σ^{2})(T − t)}{\sqrt{∫_{t}^{T} σ^{2}(u)du}}) .

We look for E[V (σ(·))] where the expectation is over the random σ(·). We could simulate several different σ(·) and take the average giving us a good approximation.

Problem. (Law of rare events) Let A_{k,n}, 1 ≤ k ≤ n be events in a probability space (Ω, F, P) and suppose that for each n, A_{1,n}, …, A_{n,n} are independent. Suppose also that

lim_{n→∞} max_{1≤k≤n} P (A_{k,n}) = 0  and  lim_{n→∞} ∑_{k=1}^{n} P (A_{k,n}) = λ

where 0 < λ < ∞. Prove that Y_{n} = ∑_{k=1}^{n} 1_{A_{k,n}} converges in law to a standard Poisson random variable with mean λ.