LECTURE 12

Review of Radon-Nikodym. Let P and P̃ be two probability measures on (Ω, F).

• P̃ << P if P̃(A) = 0 whenever P (A) = 0. We say P̃ is absolutely continuous with respect to P.

• If there exist disjoint sets A and B such that P (A) = 1 and P̃(B) = 1 then we say P and P̃ are mutually singular and write P ⊥ P̃.

• If P̃ << P and P << P̃ then we say P and P̃ are equivalent.

Radon-Nikodym Theorem. If P̃ << P then there exists a non-negative random variable Z such that for any bounded random variable X,

Ẽ[X] = E[XZ]

where Ẽ is expectation with respect to P̃ and E is expectation with respect to P. Z is often denoted \frac{d_{P}}{dP}. Example. Suppose that P is a probability measure on the real line R such that

P (A) = 0 whenever the Lebesgue measure of A is zero. Then by the Radon-Nikodym theorem there exists a probability density function f(x) such that

P (A) = ∫_{A} f(x)dx.

(Note that this does not follow directly from the way we stated the theorem. What has to be done?)

Cameron-Martin-Girsanov formula. Let B(t), 0 ≤ t ≤ T be a Brownian motion on a probability space (Ω, F, P) with filtration F(t), 0 ≤ t ≤ T. Let θ(t), 0 ≤ t ≤ T be a

process adapted to F(t), 0 ≤ t ≤ T and satisfying E[e^{\frac{1}{2} ∫_{0}^{T} θ^{2}(t)dt}] < ∞. Define

B̃(t) = ∫_{0}^{t} θ(u)du + B(t)

and

Z(t) = e^{− ∫_{0}^{t} θ(u)dB(u)− \frac{1}{2} ∫_{0}^{t} θ^{2}(u)du}.

Define a new probability measure P̃ by P̃(A) = ∫_{A} Z(T)dP forall A ∈ F, or equivalently, for every bounded random variable X,

Ẽ[X] = E[XZ(T)].

Then under P̃, B̃(t), 0 ≤ t ≤ T is a Brownian motion.

Bayes rule. If X is F(t)-measurable then for 0 ≤ s < t,

Ẽ[X | F(s)] = \frac{1}{Z(s)}E[XZ(t) | F(s)].

Proof. Recall that there are two requirements from the definition of conditional expectation:

1. Ẽ[X | F(s)] should be measurable with respect to F(s).

2. We should have ∫_{A} Ẽ[X | F(s)]dP̃ = ∫_{A} XdP̃ for any A ∈ F(s). So we just have to check that \frac{1}{Z(s)}E[XZ(t) | F(s)] works in 1. and 2. It is clearly

F(s)-measurable, by definition of conditional expectation so 1. is easy. To check 2. let A ∈ F(s). We can write

∫_{A} \frac{1}{Z(s)}E[XZ(t) | F(s)]dP̃ = Ẽ[1_{A} \frac{1}{Z(s)}E[XZ(t) | F(s)]].

Now by definition, Ẽ[X] = E[XZ(s)] for any F(s)-measurable X, so

Ẽ[1_{A} \frac{1}{Z(s)}E[XZ(t) | F(s)]] = E[1_{A}E[XZ(t) | F(s)]].

But since 1_{A} is measurable with respect to F(s), it can be taken into the inside expectation. But then we just have a double expectation which can be rewritten as E[1_{A}XZ(t)] which by definition is just Ẽ[1_{A}X] = ∫_{A} XdP̃ as desired.

Risk-neutral valuation. We have a stock and a risk-free bond. The stock price S(t) is assumed to satisfy

dS(t) = S(t) {μ(t)dt + σ(t)dB(t)}  (1)

where B(t) is a Brownian motion on (Ω, F, P) with filtration F(t). μ(t) and σ(t) and the interest rate r(t) are all adapted to F(t). The investor starts with X(0) and at time t has Δ(t)S(t) in stock and X(t) − Δ(t)S(t) in bonds. Δ(t) is also assumed to be adapted to F(t). The wealth process X(t) satisfies

dX(t) = Δ(t)dS(t) + r(t)(X(t) − Δ(t)S(t))dt.  (2)

The value of a dollar at time t in present dollars is \frac{1}{β(t)} = e^{− ∫_{0}^{t} r(u)du} so all assets should really be discounted by this factor.

A risk-neutral measure is a probability measure P̃ equivalent to P under which all tradable assets are martingales after discounting.

If there exists a risk-neutral measure then the value at time s of a hedgable contingent claim paying V (T) at time T must be

β(s)Ẽ[β(T)^{−1}V (T) | F(s)]

for if Δ(t) is the hedge then we have X(T) = V (T) and β^{−1}(t)X(t) is a martingale under P̃ so β^{−1}(s)X(s) = Ẽ[β^{−1}(T)V (T) | F(s)].

To find the risk-neutral measure we rewrite the equation for S(t) as

d \{\frac{S(t)}{β(t)}\} = \frac{σ(t)S(t)}{β(t)} \{θ(t)dt + dB(t)\}

where

θ(t) = \frac{μ(t) − r(t)}{σ(t)} = the market price of risk.  (3)

Let

B̃(t) = ∫_{0}^{t} θ(u)du + B(t)  (4)

so that the equation becomes

d \{\frac{S(t)}{β(t)}\} = \frac{σ(t)S(t)}{β(t)}dB̃(t).

Define P̃(A) = ∫_{A} Z(T)dP where Z(T) = e^{− ∫_{0}^{T} θ(u)dB(u)− \frac{1}{2} ∫_{0}^{T} θ^{2}(u)du} as before. Then by Cameron-Martin-Girsanov, B̃(t) is a Brownian motion under P̃ and hence, since it is a stochastic integral, β^{−1}(t)S(t) is a martingale with respect to P̃. Note that d{β^{−1}X(t)} = Δ(t)d{β^{−1}S(t)} so it is automatically a martingale as well, so all tradable assets are martingales after discounting, as desired.

Example (European call, time and price dependent volatility). Let V (T) = (S(T) − K)_{+}, and σ(t, ω) = σ(t, S(t)), μ(t, ω) = μ(t, S(t)). The value at time 0 is

V (0) = Ẽ[e^{− ∫_{0}^{T} r(u)du}(S(T) − K)_{+}]

where Ẽ is the risk-neutral measure. In other words, S(t) evolves according to the stochastic differential equation d{β^{−1}(t)S(t)} = β^{−1}(t)σ(t)S(t)dB̃(t) where B̃(t) is a Brownian motion under P̃. This is of course equivalent to

dS(t) = S(t){r(t)dt + σ(t)dB̃(t)}.

By the Feynman-Kac formula we know that

u(T, x) = Ẽ_{x}[e^{− ∫_{0}^{T} r(u)du}(S(T) − K)_{+}]

is the solution of the equation

\frac{∂u}{∂T} = \frac{1}{2}x^{2}σ^{2}(T, x)\frac{∂^{2}u}{∂x^{2}} + xr(T)\frac{∂u}{∂x} − r(T)u, u(0, x) = (x − K)_{+}

and our answer V (0) = u(T, S(0)).

Only in special cases (eg. Black-Scholes) is the equation explicitly solvable. Otherwise you may have to use numerical methods.

Martingale representation theorem. Let B(t) be a Brownian motion with filtration F(t) and suppose that X(t) is a martingale with respect to F(t). Then there exists an adapted process δ(t) such that

X(t) = X(0) + ∫_{0}^{t} δ(u)dB(u).

In particular it follows that the process X(t) has continuous sample paths.

We now know that stochastic integrals with respect to Brownian motion are martingales with respect to the Brownian filtration, and vice-versa, martingales with respect to the Brownian filtration arise as stochastic integrals.

Application to risk-neutral valuation. Suppose we have a stock price satisfying equation 1. We define the risk-neutral measure as in equations 3 and 4. Let V (T) be measurable with respect to F(T). We want to use the martingale representation theorem to construct a hedge. For 0 ≤ t ≤ T let

M(t) = Ẽ[β^{−1}(T)V (T) | F(t)].

Then M(t) is a martingale under P̃ with respect to the filtation F(t). By the martingale representation theorem there exists an adapted process γ(t) so that

M(t) = M(0) + ∫_{0}^{t} γ(u)dB̃(u).

Choose Δ(u) to satisfy Δ(u)β^{−1}(u)σ(u)S(u) = γ(u) and let X(t) = β(t)M(t). Then X(t) satisfies an equation of the form 2 and X(T) = V (T), so it is a hedge.

Problems.

1. i. If P and P̃ are equivalent and \frac{d_{P}}{dP} = Z show that \frac{dP}{d_{P}} = \frac{1}{Z}.

Let P_{x}^{a,b} denote the probability measure on C([0, T]) corresponding to the solution of the stochastic differential equation

dX(t) = σ(t, X(t))dB(t) + b(t, X(t))dt, X(0) = x

where a = σσ^{T}.

ii. Let b_{1} ≠ b_{2}. Write expressions for \frac{dP_{x}^{a,b_{1}}}{dP_{x}^{a,b_{2}}} and \frac{dP_{x}^{a,b_{2}}}{dP_{x}^{a,b_{1}}} using the Cameron-Martin-

Girsanov formula. iii. Resolve the apparent contradiction between the answer to ii and i.

2. Let a stock evolve according to dS = μSdt + σSdB and a bond evolve as dβ = rβdt where σ, μ and r are adapted to the Brownian filtration. Let V (T) be a random variable measurable with respect to F(T) representing the payoff at time T of a contingent claim. As we know the value at time t is given by

V (t) = β(t)Ẽ[β^{−1}(T)V (T) | F(t)]

where P̃ is the risk-neutral measure. Show that this can be rewritten in the form

ζ^{−1}(t)E[ζ(T)V (T) | F(t)]

and find an explicit expression for the process ζ(t).

3. Let α(x) = (α_{1}(x), …, α_{n}(x)) be a smooth function from R^{n} to R^{n}. Consider the partial differential equation for x ∈ R^{n}, and t > 0,

\frac{∂u}{∂t} = \frac{1}{2} ∑_{i=1}^{n} \frac{∂^{2}u}{∂x_{i}^{2}} + ∑_{i=1}^{n} α_{i}(x)\frac{∂u}{∂x_{i}} , u(0, x) = f(x).

i. Use the Girsanov theorem to show that the solution is

u(t, x) = E_{x} [e^{∫_{0}^{t} α(B(s))dB(s)− \frac{1}{2} ∫_{0}^{t} |α(B(s))|^{2}ds}f(B(t))] .

ii. Suppose that α(x) = ∇γ(x) for some function γ : R^{n} → R. Use Ito’s formula to show that in this case

u(t, x) = e^{−γ(x)}E_{x} [e^{γ(B(t))− \frac{1}{2} ∫_{0}^{t}(∇γ^{2}(B(s))+Δγ(B(s)))ds}f(B(t))] .

iii. Use the Feynman-Kac formula to show that v(t, x) = e^{γ(x)}u(t, x) is the solution

of

\frac{∂v}{∂t} = \frac{1}{2}Δv − \frac{1}{2}(∇γ^{2} + Δγ)v.