LECTURE 8

Today we are going to have an informal lecture on the connection between probability and partial differential equations. It will also give us a chance to introduce some of the most basic pde’s.

Dirichlet problem. Let D be a domain in R^{d}, i.e. a bounded connected open set with a smooth boundary ∂D. You are presented with a continuous function f defined on the boundary ∂D and you are asked to come up with a function u satisfying

\{ \begin{matrix}Δu = 0 & in D, \\ u = f & on ∂D. \end{matrix}

Here the Laplacian Δ is the operator

Δu = \frac{∂^{2}u}{∂x_{1}^{2}} + ⋯ + \frac{∂^{2}u}{∂x_{d}^{2}}.

Discrete version of the Dirichlet problem. As a first try we approximate the problem on a lattice, with lattice width ϵ taken pretty small. Let D_{ϵ} be the approximation of D and ∂D_{ϵ} the boundary. The discrete Laplacian is given by

Δ_{ϵ}u(x) = ϵ^{−2} ∑_{y∼x}[u(y) − u(x)]

where y ∼ x means that y is a nearest neighbour lattice site to x. We have a function f defined on the lattice sites in the boundary and the discrete Dirichlet problem is to come up with a function u defined on D_{ϵ} satisfying

\{ \begin{matrix}Δ_{ϵ}u = f & in D_{ϵ}, \\ u = f & on ∂D_{ϵ}. \end{matrix}

Solution by random walks. Let X_{n} be a symmetric simple random walk starting at x ∈ D_{ϵ}. In other words

X_{n} = Y_{1} + ⋯ + Y_{n} + x

where each Y_{i} is chosen with uniform distribution from the 2d basic vectors in the coordinate directions (we use e to denote such basic lattice vectors.) Let τ be the first time that the random walk is in the boundary ∂D_{ϵ}. Then X_{τ} is the first site in the boundary that the walk gets to. We claim the answer to our problem is

u(x) = E_{x}[f(X_{τ})].

The subscript x tells you that the walk starts at x.

We have two things to check. First that Δ_{ϵ}u = 0 inside. By double conditioning we have

u(x) = E_{x}[E_{x}[f(X_{τ}) | Y_{1}]].

This latter can be written as

∑_{e} E_{x}[f(X_{τ}) | Y_{1} = e]P (Y_{1} = e).

The last term P (Y_{1} = e) = 1/2d. Note that

E_{x}[f(X_{τ}) | Y_{1} = e] = E_{x+e}[f(X_{τ})] = u(x + e).

So we have that

u(x) = Av_{y∼x}u(y)

and therefore Δ_{ϵ}u = 0.

The second thing we have to check is that u = f on ∂D_{ϵ}. This is obvious since if x ∈ ∂D_{ϵ}, then τ = 0 and X_{τ} = x, so there is no probability and u(x) = E_{x}[f(X_{τ})] = f(x).

Solution of the Dirichlet problem by Brownian motion. Let W_{t} be a Brownian motion starting at x ∈ D. Let τ be the exit time

τ = inf{t ≥ 0 : W_{t} ∉ D}.

Note that W_{τ} is the point in the boundary where the Brownian motion exits. We claim the solution of the Dirichlet problem is

u(x) = E_{x}[f(W_{τ})].

We need the mean value property of harmonic functions which says that

Δu = 0 in D ⇔ Av_{|y−x|=ϵ}u(y) = u(x) for any ϵ with {|y − x| ≤ ϵ} ⊂ D.

The first thing to show is that our u satisfies the mean value property, and is therefore harmonic. Let x be a point in D and ϵ be small enough that the ball of radius ϵ around x is contained in D as well. Let σ denote the hitting time of the surface of the ball of radius ϵ by a Brownian motion starting at x. Then W_{σ} is the point where the Brownian motion first exits the ball of radius ϵ. By double conditioning we have

u(x) = E_{x}[E_{x}[f(W_{τ}) | W_{σ}]].

This can be rewritten as

∫_{|y−x|=ϵ} E_{x}[f(W_{τ}) | W_{σ} = y]P (W_{σ} ∈ dy).

The distribution of W_{σ} is uniform on the sphere of radius ϵ, by the spherical symmetry of Brownian motion. By the strong Markov property the integrand is

E_{x}[f(W_{τ}) | W_{σ} = y] = E_{y}[f(W_{τ})] = u(y).

Therefore we have proved that u(x) is equal to its average over the sphere of radius ϵ. This proves the mean value property.

The second thing to show is that if x_{n} is a sequence of points in D with x_{n} → y where y is a point in the boundary then u(x_{n}) → f(y). Let’s write it out. We need

E_{x_{n}}[f(W_{τ})] → f(y).

What it means is that the distribution of the exit point W_{τ} is becoming concentrated near y. Note that this is non-trivial, and in fact if the boundary is terrible enough (eg. a very sharp inward pointing spike) may not even be correct. The proof is a little tricky so we will leave it for later.

2nd order linear partial differential operators. Let σ be a d×d matrix, a = σ^{T} σ, and b a d-vector. Let Y be a random variable with mean 0 and covariance I. Then Z = ϵ^{1/2}σY + ϵb has mean ϵb and covariance ϵa. Let u be a smooth function on R^{d}. By Taylor’s theorem we have

u(x + Z) − u(x) = ∑_{i=1}^{d} Z_{i} \frac{∂u}{∂x_{i}} (x) + \frac{1}{2} ∑_{i,j=1}^{d} Z_{i}Z_{j} \frac{∂^{2}u}{∂x_{i}∂x_{j}} (x) + O(|Z|^{3}).

So

lim_{ϵ→0} \frac{1}{ϵ}E[u(x + Z) − u(x)] = Lu(x)

where

Lu = \frac{1}{2} ∑_{i,j=1}^{d} a_{ij} \frac{∂^{2}u}{∂x_{i}∂x_{j}} + ∑_{i=1}^{d} b_{i} \frac{∂u}{∂x_{i}} .

Generalized Dirichlet problem. Suppose we want to solve

\{ \begin{matrix}Lu = 0 & in D, \\ u = f & on ∂D \end{matrix}

where

Lu = \frac{1}{2} ∑_{i,j=1}^{d} a_{ij}(x)\frac{∂^{2}u}{∂x_{i}∂x_{j}} + ∑_{i=1}^{d} b_{i}(x)\frac{∂u}{∂x_{i}} .

We claim the solution has the following representation: Let X_{t} be the solution of the stochastic differential equation

dX_{t} = σ(X_{t})dW_{t} + b(X_{t})dt, X_{0} = x.

Let τ be the exit time from the region D. Then X_{τ} is the exit point on the boundary ∂D. The formula for u is

u(x) = E_{x}[f(X_{τ})].

To show it we cannot use the same conditioning argument as before because we lost the symmetry. Suppose u is the solution of the generalized Dirichlet problem. We claim that u(X_{t}) is a martingale (note we stop the process when it hits the boundary). To show this we apply Ito’s formula to u(X_{t}). We get

du(X) = ∑_{i=1}^{d} \frac{∂u}{∂x_{i}} dX_{i} + \frac{1}{2} ∑_{i,j=1}^{d} \frac{∂^{2}u}{∂x_{i}∂x_{j}} dX_{i}dX_{j}.

To simplify this, plug in the sde. Note that

dX_{i}dX_{j} = ∑_{ℓ,ℓ′} σ_{iℓ}σ_{jℓ′}dW_{ℓ}dW_{ℓ′} = ∑_{ℓ} σ_{iℓ}σ_{jℓ}dt = a_{ij}dt.

You get

du(X) = Lu(X)dt + σ(X)∇u(X) · dW.

In particular, if Lu = 0, then u(X_{t}) − u(X_{0}) = ∫_{0}^{t} σ(X_{s})∇u(X_{s}) · dW_{s} which is always a martingale. By the optional stopping theorem

E_{x}[u(X_{τ})] = u(x).

But note that X_{τ} ∈ ∂D, so u(X_{τ}) = f(X_{τ}) so we are done.

Poisson equation. Suppose we want instead

\{ \begin{matrix}Lu = f & in D, \\ u = 0 & on ∂D \end{matrix}

where f is some given function defined in D. Let X_{t} and τ be the same as above. By Ito’s formula we have du(X) = f(X)dt + dM where M is a martingale. By the optional stopping time theorem

E_{x}[u(X_{τ}) − u(X_{0}) − ∫_{0}^{τ} f(X_{s})ds] = 0.

But X_{τ} ∈ ∂D so u(X_{τ}) = 0. Also u(X_{0}) = u(x) so we have the representation formula

u(x) = E_{x}[∫_{0}^{τ} f(X_{s})ds].

Heat equation. The heat equation on R^{d} is

\{ \begin{matrix}\frac{∂u}{∂t} = \frac{1}{2}Δu,  t > 0 \\ u(0, x) = f(x). \end{matrix}

This is one of those special cases where the solution can be written explicitly in terms of functions we have names for. The answer is

u(t, x) = ∫ f(y)p(0, x, t, y)dy

where

p(0, x, t, y) = \frac{e^{− \frac{|y−x|^{2}}{2t}}}{(2πt)^{d/2}} .

Since these are the transition probabilities for Brownian motion we have automatically the representation formula

u(t, x) = E_{x}[f(W_{t})].

Diffusion equation. In an inhomogeneous medium with convection the heat equation becomes

\{ \begin{matrix}\frac{∂u}{∂t} = Lu,  t > 0 \\ u(0, x) = f(x), \end{matrix}

where

Lu = \frac{1}{2} ∑_{i,j=1}^{d} a_{ij}(x, t)\frac{∂^{2}u}{∂x_{i}∂x_{j}} + ∑_{i=1}^{d} b_{i}(x, t)\frac{∂u}{∂x_{i}} .

Let X_{t} be the solution of the stochastic differential equation

dX_{t} = σ(X_{t}, t)dW_{t} + b(X_{t}, t)dt, X_{0} = x.

By Ito’s formula

du(T − t, X_{t}) = \{−\frac{∂u}{∂t} + Lu\} (T − t, X_{t})dt + σ(X_{t}, t)∇u(X_{t}, t)dW_{t}.

So if u solves the diffusion equation then u(T − t, X_{t}) is a martingale. Therefore

E_{x}[u(T − T, X_{T})] = u(T, x).

But u(0, X_{T}) = f(X_{T}) so we have the formula

u(T, x) = E_{x}[f(X_{T})].

Diffusion equation with killing (interest). Let r be a constant. We want to solve

\{ \begin{matrix}\frac{∂u}{∂t} = \frac{1}{2}Lu − ru,  t > 0 \\ u(0, x) = f(x). \end{matrix}

We can rewrite it as \frac{∂e^{rt}u}{∂t} = \frac{1}{2}Δe^{rt}u so the answer is just

u(t, x) = E_{x}[e^{−rt}f(X_{t})].

This is used a lot in finance. For example in the Black-Scholes analysis we want to evaluate

u(t, x) = E_{x}[e^{−rt}(X_{t} − a)_{+}]

where r is the interest rate, a is the strike price, and X_{t} is the solution of the sde

dX_{t} = rX_{t}dt + σX_{t}dW_{t}.

Now we know that u is the solution of the pde

\{ \begin{matrix}\frac{∂u}{∂t} = \frac{1}{2}σ^{2}x^{2} \frac{∂^{2}u}{∂x^{2}} + rx\frac{∂u}{∂x} − ru \\ u(0, x) = (x − a)_{+}. \end{matrix}

As chance would have it, the explicit answer can be written down in terms of the error function Φ(x) = ∫_{−∞}^{x} \frac{e^{−y^{2}/2}}{\sqrt{2π}} dy. The answer is

u(t, x) = xΦ (\frac{1}{σ\sqrt{t}} \{log(x/a) + (r + σ^{2}/2)t\})+ae^{−rt}Φ (\frac{1}{σ\sqrt{t}} \{log(x/a) + (r − σ^{2}/2)t\}) .

Feynman-Kac formula. Suppose V (x) is a function and we want to solve

\{ \begin{matrix}\frac{∂u}{∂t} = \frac{1}{2}Δu + V (x)u,  t > 0 \\ u(0, x) = f(x). \end{matrix}

It’s like the heat equation with killing, but the rate is no longer a constant. The Feynman-Kac formula says the answer has the representation

u(t, x) = E_{x}[e^{∫_{0}^{t} V (W_{s})ds}f(W_{t})].

To see why this is so, using Ito’s formula we have

\begin{aligned}d \{u(T − t, W_{t})e^{∫_{0}^{t} V (W_{s})ds}\} = & \{−\frac{∂u}{∂t} + \frac{1}{2}Δu + V u\} (T − t, X_{t})e^{∫_{0}^{t} V (W_{s})ds}dt \\ & + σ(X_{t}, t)∇u(X_{t}, t)e^{∫_{0}^{t} V (W_{s})ds}dW_{t}. \end{aligned}

So u(T − t, W_{t})e^{∫_{0}^{t} V (W_{s})ds} is a martingale. Therefore

E_{x}[u(T − T, W_{t})e^{∫_{0}^{T} V (W_{s})ds}] = u(T, x).

Since u(0, W_{t}) = f(W_{t}) this is the Feynman-Kac formula.

Question. Note that in the above calculation there is no extra Ito term corresponding

to the exponential e^{∫_{0}^{t} V (W_{s})ds}. Why not?