LECTURE 11

Two notions of uniqueness for stochastic differential equations. 1. Pathwise uniqueness of the stochastic differential equation dX = bdt + σdB, X(0) = x means that if X_{1}(t) and X_{2}(t) are two solutions with the same Brownian motion, then X_{1}(t) = X_{2}(t) for all t > 0 almost surely. 2. Uniqueness in the sense of law means that if X_{1}(t) is a solution of dX_{1}(t) = b(t, X_{1}(t))dt + σ(t, X_{1}(t))dB_{1}(t), X_{1}(0) = x and X_{2}(t) is a solution of dX_{2}(t) = b(t, X_{2}(t))dt + σ(t, X_{2}(t))dB_{2}(t), X_{2}(0) = x, then X_{1} and X_{2} have the same law - that is the collection of probabilities P (X(·) ∈ A) where A is a (measurable) subset of the space of continuous functions.

Pathwise uniqueness implies uniqueness in law but not the other way around.

Markov property of solutions of stochastic differential equations. Let X(t) be the unique (in law) solution after time s of the stochastic differential equation

dX(t) = b(t, X(t))dt + σ(t, X(t))dB(t), t > s X(s) = x.

One denotes the expectation with respect to this process as E_{s,x}[·]. Sometimes it is useful to note explicitly the dependence on the initial conditions and people write things like X_{s,x}(t).

The solutions are Markov processes, i.e. for any nice function f(x), and any t ≥ s,

E_{0,x}[f(X(t)) | F(s)] = E_{s,X(s)}[f(X(t))].

Proof. This follows from the uniqueness which guarantees that for t > s, X_{s,X(s)}(t) = X_{0,x}(t) because otherwise they would represent two distinct solutions to the sde starting at X(s) at time s.

Transition probabilities. Since X(t) is a Markov process, its law is determined by the transition probabilities,

p(s, x, t, A) = P (X(t) ∈ A | X(s) = x), t > s.

Another way to write this is

E_{s,x}[f(X(t))] = ∫ f(y)p(s, x, t, dy)

for any function f.

Recall that transition probabilities always satisfy the Chapman-Kolmogorov equations p(s, x, t, A) = ∫ p(s, x, u, dy)p(u, y, t, A) for any s < u < t, and that all finite dimensional distributions of the process can be written in terms of them,

P (X(t_{1}) ∈ A_{1}, …, X(t_{n}) ∈ A_{n}) = ∫_{A_{n}} ⋯ ∫_{A_{1}} p(0, x, t_{1}, dx_{1}) ⋯ p(t_{n−1}, x_{n−1}, t_{n}, dx_{n}).

Except in degenerate cases, the transition probabilities of a diffusion process will have densities

p(s, x, t, A) = ∫_{A} p(s, x, t, y)dy.

Note the abuse of notation.

Backward and forward equations. Suppose that X(t) is the unique (in law) solution of a stochastic differential equation dX(t) = b(t, X(t))dt + σ(t, X(t))dB(t), X(0) = x. We know that the solution is a Markov process. How do we find the transition probabilities? Recall our old friend the operator

L_{t}f = \frac{1}{2} ∑_{i,j=1}^{d} a_{i,j}(t, x)\frac{∂^{2}f}{∂x_{i}∂x_{j}} + ∑_{i=1}^{d} b_{i}(t, x)\frac{∂f}{∂x_{i}}

where a = σσ^{T}. Its adjoint is

L_{t}^{∗}f(x) = \frac{1}{2} ∑_{i,j=1}^{d} \frac{∂^{2}}{∂x_{i}∂x_{j}} (a_{i,j}(t, x)f(x)) − ∑_{i=1}^{d} \frac{∂}{∂x_{i}} (b_{i}(t, x)f(x)).

Proposition. The transition probabilities satisfy the forward or Fokker-Planck equations

\frac{∂}{∂t}p(s, x, t, y) = L_{t}^{∗}p(s, x, t, y)

where the operator L_{t}^{∗} is acting on the y variable. x and s are just fixed parameters playing no part in the partial differential equation. They only appear in the initial condition

lim_{t↓s} p(s, x, t, y) = δ(y − x).

Proof. Let f(t, x) be any smooth function. By Ito’s formula

E_{s,x}[f(T, X(T))] − f(s, x) = E_{s,x} [∫_{s}^{T} \{\frac{∂f}{∂t} + L_{t}f\} (t, X(t))dt] .

In terms of the transition probabilities we can write this as

∫ f(T, y)p(s, x, T, y)dy − f(s, x) = ∫_{s}^{T} ∫ \{\frac{∂f}{∂t} + L_{t}f\} (t, y)p(s, x, t, y)dydt.

Integrating by parts, once in the t variable and twice in the y variable we get

∫_{s}^{T} ∫ f(t, y)\{−\frac{∂}{∂t} + L_{t}^{∗}\}p(s, x, t, y)dydt = 0.

If ∫_{s}^{T} ∫ f(t, y)g(t, y)dydt = 0 for every smooth function, then g(t, y) = 0 for s ≤ t ≤ T. This gives the forward equation.

Proposition. The transition probabilities satisfy the backward or Kolmogorov equations

−\frac{∂}{∂s}p(s, x, t, y) = L_{s}p(s, x, t, y).

Here the operator L_{s} acts on the variable x and the t and y are parameters appearing only in the final condition

lim_{s↑t} p(s, x, t, y) = δ(y − x).

Proof. Let f(x) be any smooth function u(s, x) be the unique solution of −\frac{∂}{∂s}u = L_{s}u for 0 ≤ s < t with u(t, x) = f(x). By Ito’s formula, u(s, X(s)) is a martingale up to time t. Taking expectation

E_{s,x}[u(s, X(s))] = E_{s,x}[u(t, X(t))].

By the final condition, the right hand side is ∫ f(z)p(s, x, t, z)dz. The left hand side is just u(s, x). So

u(s, x) = ∫ f(z)p(s, x, t, z)dz.

Letting f_{n}(z) be a sequence of smooth functions tending to δ(y − z) we get in the limit that p satisfy the backward equations.

Examples.

1. Brownian motion in 1 dimension. The operator L = \frac{1}{2} \frac{∂^{2}}{∂x^{2}} . The forward equation is \frac{∂}{∂t}p(s, x, t, y) = \frac{1}{2} \frac{∂^{2}}{∂y^{2}} p(s, x, t, y) with initial condition lim_{t↓s} p(s, x, t, y) = δ(y − x)

and the backward equation is −\frac{∂}{∂s}p(s, x, t, y) = \frac{1}{2} \frac{∂^{2}}{∂x^{2}} p(s, x, t, y) with final condition lim_{s↑t} p(s, x, t, y) = δ(y − x). Both are of course satisfied by the heat kernel

p(s, x, t, y) = \frac{1}{\sqrt{2π(t−s)}}e^{− \frac{(x−y)^{2}}{2(t−s)}} .

2. Brownian motion in d dimensions. The operator L = \frac{1}{2}Δ = \frac{1}{2} ∑_{i=1}^{d} \frac{∂^{2}}{∂x_{i}^{2}} .

The forward equation is \frac{∂}{∂t}p(s, x, t, y) = \frac{1}{2} ∑_{i=1}^{d} \frac{∂^{2}}{∂y_{i}^{2}} p(s, x, t, y) with initial condi-

tion lim_{t↓s} p(s, x, t, y) = δ(y − x) and the backward equation is −\frac{∂}{∂s}p(s, x, t, y) = \frac{1}{2} ∑_{i=1}^{d} \frac{∂^{2}}{∂x_{i}^{2}} p(s, x, t, y) with final condition lim_{s↑t} p(s, x, t, y) = δ(y − x). Both are

satisfied by the heat kernel p(s, x, t, y) = \frac{1}{(2π(t−s))^{d/2}} e^{− \frac{|x−y|^{2}}{2(t−s)}} .

3. Brownian motion with constant drift b and constant volatility σ^{2}. Consider the constant coefficient stochastic differential equation

dX(t) = bdt + σdB(t), X(0) = x.

The solution is X(t) = bt + σB(t) so it is easy to check that

p(s, x, t, y) = \frac{1}{\sqrt{2πσ^{2}(t − s)}}e^{− \frac{(y−x−b(t−s))^{2}}{2σ^{2}(t−s)}} .

You can also check that it satisfies the forward equation \frac{∂}{∂t}p(s, x, t, y) = \frac{σ^{2}}{2} \frac{∂^{2}}{∂y^{2}} p(s, x, t, y)−

b\frac{∂}{∂y} p(s, x, t, y) and the backward equation −\frac{∂}{∂s}p(s, x, t, y) = \frac{σ^{2}}{2} \frac{∂^{2}}{∂x^{2}} p(s, x, t, y)+b\frac{∂}{∂x}p(s, x, t, y).

4. Ornstein-Uhlenbeck process. For the stochastic differential equation dX(t) = −ρX(t)dt + dB(t) the corresponding Markov generator is L = \frac{1}{2} \frac{∂^{2}}{∂x^{2}} − ρx\frac{∂}{∂x}. The forward equation is \frac{∂}{∂t}p(s, x, t, y) = \frac{1}{2} \frac{∂^{2}}{∂y^{2}} p(s, x, t, y)+ρ\frac{∂}{∂y} (yp(s, x, t, y)). The backward

equation is −\frac{∂}{∂s}p(s, x, t, y) = \frac{1}{2} \frac{∂^{2}}{∂x^{2}} p(s, x, t, y) + ρx\frac{∂}{∂x}p(s, x, t, y). Since the process is time-homogeneous (i.e. b and σ do not depend on time) the solution must be of the form p(s, x, t, y) = p(t − s, x, y). Letting v(t, x) = p(t, xe^{ρt}, y) one gets the equation e^{2ρt} \frac{∂v}{∂t} = \frac{1}{2} \frac{∂^{2}v}{∂x^{2}} which can be transformed into the standard heat equation by τ = (2ρ)^{−1}(1 − e^{−2ρt}). Solving and undoing all these transformations one gets Mehler’s formula,

p(t, x, y) = \frac{e^{− \frac{ρ(y−xe^{−ρt})^{2}}{(1−e^{−2ρt})}}}{\sqrt{2π (\frac{1−e^{−2ρt}}{2ρ} )}}.

5. Geometric Brownian motion. The stochastic differential equation is dX(t) = rX(t)dt + σX(t)dB(t). The Markov generator is L = \frac{1}{2}σ^{2}x^{2} \frac{∂^{2}}{∂x^{2}} + rx\frac{∂}{∂x}. Since the process is time-homogeneous the transition probabilities have the form p(s, x, t, y) = p(t − s, x, y) and the backward equation becomes

\frac{∂}{∂t}p(t, x, y) = \frac{1}{2}σ^{2}x^{2} \frac{∂^{2}}{∂x^{2}} p(t, x, y) + rx\frac{∂}{∂x}p(t, x, y), p(0, x, y) = δ(y − x).

If you like, you can try to solve it. But in this case we actually know the solution of the stochastic differential equation is

X(t) = xe^{σB(t)+(r− \frac{1}{2} σ^{2})t}.

So

P_{x}(X(t) ∈ A) = P (xe^{σB(t)+(r− \frac{1}{2} σ^{2})t} ∈ A).

In other words, for fixed x and t, p(t, x, y) is the density function of ϕ(B(t)) where ϕ(y) = xe^{σy+(r− \frac{1}{2} σ^{2})t}. Recall that if X is a random variable with density f the density of ϕ(X) is given by (f ∘ ϕ^{−1})/ϕ′. Since ϕ^{−1}(y) = σ^{−1} (log \frac{y}{x} − (r − \frac{1}{2}σ^{2})t)) we have

p(t, x, y) = \frac{1}{σy\sqrt{2πt}}e^{−^{\frac{(log \frac{y}{x} −(r− \frac{1}{2} σ^{2})t))^{2}}{2tσ^{2}}}} .

You can check it satisfies the backward equation if you like long calculations.

6. Black-Scholes. The price at time t of a European call option with maturity T and strike price K is

E[e^{−r(T −t)}(X(T) − K)_{+} | F(t)] = e^{−r(T −t)}E_{t,X(t)}[(X(T) − K)_{+}]

where X(t) is a geometric Brownian motion dX(t) = rX(t)dt + σX(t)dB(t) where r is the riskless interest rate and σ is the stock volatility. Now

E_{t,x}[((X(T) − K)_{+}] = ∫ (y − K)_{+}p(T − t, x, y)dy

where p(t, x, y) are the transition probabilities of the geometric Brownian motion given above. It is not hard to write down the answer in terms of the error function

Φ(x) = ∫_{−∞}^{x} \frac{e^{− \frac{y^{2}}{2}}}{\sqrt{2π}} dy.

We get

E[e^{−r(T −t)}(X(T) − K)_{+} | F(t)] =X(t)Φ (\frac{log \frac{X(t)}{K} + (r + \frac{1}{2}σ^{2})(T − t)}{σ\sqrt{T − t}})

− e^{−r(T −t)}KΦ (\frac{log \frac{X(t)}{K} + (r − \frac{1}{2}σ^{2})(T − t)}{σ\sqrt{T − t}}) .

Diffusion process as probability measure on C([0, ∞)). You can either think of Brownian motion as a collection of random variables B(t), t ≥ 0 on some probability space (Ω, F, P), or as a probability measure P on the space C([0, ∞)) of continuous functions. If A is a (measurable) subset of continuous functions, then P (A) is just the probability that a Brownian path falls in that subset.

The same holds for any diffusion. If X(t) is the solution of the stochastic differential equation dX(t) = b(t, X(t))dt + σ(t, X(t))dB(t), X(0) = x then we can let P_{x}^{a,b} denote the probability measure on the space of continuous functions with

P_{x}^{a,b}(A) = P rob(X(·) ∈ A).

Here a = σσ^{T} and we label it by a and not σ because it really only depends on a. Also we use the notation X(·) to indicate that we are thinking of the whole function as opposed to X(t) which is just the value of that function at t.

Relation of P_{x}^{a,b} for different values of x. The measure P_{x}^{a,b} is of course concentrated on paths starting at x, so if x_{1} ≠ x_{2} then P_{x_{1}}^{a,b} ⊥ P_{x_{2}}^{a,b} meaning they are concentrated on completely different subsets of C([0, ∞)) .

Relation of P_{x}^{a,b} for different values of a. Just as with Brownian motion, it is possible to show that the quadratic variation

lim_{n→∞} ∑_{i=0}^{⌊2^{n}T ⌋} |X(\frac{i + 1}{2^{n}}) − X(\frac{i}{2^{n}})|^{2} = ∫_{0}^{T} a(t, X(t))dt

almost surely. Hence if a_{1} ≠ a_{2}, P_{x}^{a_{1},b} ⊥ P_{x}^{a_{2},b}.

Sometimes it may be useful to know how different they are. Suppose we solve the two constant coefficient sde’s dX_{1}(t) = bdt + σ_{1}dB(t) and dX_{2}(t) = bdt + σ_{2}dB(t). We ask for the relative probability that X_{i}(t_{1}) = x_{1}, … X_{i}(t_{n}) = x_{n}. From the formula we worked out we have

\frac{P (X_{1}(t_{1}) ∈ dx_{1}, …, X_{1}(t_{n}) ∈ dx_{n})}{P (X_{2}(t_{1}) ∈ dx_{1}, …, X_{2}(t_{n}) ∈ dx_{n})} ; = (\frac{σ_{2}^{2}}{σ_{1}^{2}} )^{n/2} exp \{− (\frac{1}{σ_{1}^{2}} − \frac{1}{σ_{2}^{2}} ) ∑_{i=0}^{n−1} \frac{(x_{i+1} − x_{i} − b(t_{i+1} − t_{i}))^{2}}{2(t_{i+1} − t_{i})} \}

which tells you that the relative probability at n times is roughly of size e^{cn}.

Relation of P_{x}^{a,b} for different values of b (Cameron-Martin-Girsanov formula). Let’s try the same thing with dX_{1}(t) = b_{1}dt+σdB(t) and dX_{2}(t) = b_{2}dt+σdB(t). Then \frac{P (X_{1}(t_{1})∈dx_{1},...,X_{1}(t_{n})∈dx_{n})}{P (X_{2}(t_{1})∈dx_{1},...,X_{2}(t_{n})∈dx_{n})} is given by

exp \{− ∑_{i=0}^{n−1} \frac{(x_{i+1} − x_{i} − b_{1}(t_{i+1} − t_{i}))^{2} − (x_{i+1} − x_{i} − b_{2}(t_{i+1} − t_{i}))^{2}}{2σ^{2}(t_{i+1} − t_{i})} \} .

Rewriting the term in the exponential we get

− ∑_{i=0}^{n−1} σ^{−1}(b_{2} − b_{1})σ^{−1}(x_{i+1} − x_{i} − b_{2}(t_{i+1} − t_{i})) − \frac{1}{2} ∑_{i=0}^{n−1}(σ^{−1}(b_{2} − b_{1}))^{2}(t_{i+1} − t_{i}).

Under P_{x}^{a,b_{2}}, σ^{−1}(X(t) − b_{2}t) is a Brownian motion, so if we take limits, then under this measure we get

\frac{dP_{x}^{a,b_{1}}}{dP_{x}^{a,b_{2}}} = exp \{− ∫_{0}^{t} σ^{−1}(b_{2} − b_{1})dB(s) − \frac{1}{2} ∫_{0}^{t}(σ^{−1}(b_{2} − b_{1}))^{2}ds\} .

This is the Cameron-Martin-Girsanov formula. In particular the formula tells us that as long as this thing is integrable with respect to P_{x}^{a,b_{2}}, then P_{x}^{a,b_{1}} << P_{x}^{a,b_{2}}, i.e. if P_{x}^{a,b_{2}}(A) = 0 then P_{x}^{a,b_{1}}(A) = 0 as well.

Problems.

1. i. Show that X(t) = (1 − t) ∫_{0}^{t} \frac{1}{1−s}dB(s) is the solution of

dX(t) = −\frac{X(t)}{1 − t}dt + dB(t),  0 ≤ t < 1,  X(0) = 0.

ii. Show that lim_{t↑1} X(t) = 0 almost surely. X(t) is called the Brownian bridge. iii. Show that X(t) is Gaussian and calculate its mean and covariance. iii. For fixed constants a and b solve the stochastic differential equation

dX(t) = \frac{b − X(t)}{1 − t}dt + dB(t),  0 ≤ t < 1  X(0) = a.

2. i. Solve the stochastic differential equation dX(t) = λX(t)dB(t), X(0) = 1. ii. For n = 0, 1, … define

H_{n}(t, x) = (−1)^{n} \frac{t^{n}}{n!} e^{\frac{x^{2}}{2t}} \frac{d^{n}}{dx^{n}} e^{− \frac{x^{2}}{2t}} .

These are called the Hermite polynomials. iii. Show that

e^{λx− \frac{1}{2} λ^{2}t} = ∑_{n=0}^{∞} λ^{n}H_{n}(t, x).

iv. Show that the Hermite polynomials play the role for stochastic calculus that x^{n}/n! play for ordinary calculus:

dH_{n+1}(t, B(t)) = H_{n}(t, B(t))dt.

3. Let B(t) be Brownian motion. Use Ito’s formula to find an expression for E[B^{n}(t)] in terms of E[B^{n−2}(t)]. Use induction to compute E[B^{n}(t)] for every positive integer n.

4. Let

C_{K}(s, ω, t) = E[(X(t) − K)_{+} | F(s)]

where X(t) is the solution of a stochastic differential equation dX(t) = b(t, X(t))dt + σ(t, X(t))dB(t).

i. Show that C_{K}(s, ω, t) = C_{K}(s, X(s)), i.e. it only depends on ω through X(s). ii. Show that \frac{∂^{2}C_{K}(s,x,t)}{∂K^{2}} = p(s, x, t, K) where p are the transition probabilities for

X(t).