MATH-7770
STOCHASTIC DIFFERENTIAL EQUATIONS SPRING 2012
Lecture 25. More on Poisson kernels. Diffusion processes as Markov processes.
Let me return to Poisson kernels.
Suppose 𝝃𝑡 is a time-homogeneous diffusion process corresponding to the linear operator 𝐿; and 𝐺 is a region in ℝ𝑟 such that almost surely 𝜏 𝒙 = min{𝑡 ≥ 0 : 𝝃𝑡𝒙 ∈
/ 𝐺} < ∞.
∙ the region 𝐺 is bounded; and
∙ the boundary ∂𝐺 is smooth; and
(
)
∙ the operator 𝐿 is ∑
strictly elliptic; i. e. the diffusion matrix 𝑎𝑖𝑗 (𝒙) is positive definite
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𝑟
for all 𝒙 ∈ 𝐺 ∪ ∂𝐺:
𝑖, 𝑗 𝑎𝑖𝑗 (𝒙) ⋅ 𝜆𝑖 𝜆𝑗 ≥ 𝐶 ⋅ ∣𝝀∣ for every 𝝀 ∈ ℝ , where 𝐶 is a positive
constant; –
then, under some conditions, a Poisson kernel 𝑝(𝒙, 𝒚), 𝒙 ∈ 𝐺, 𝒚 ∈ ∂𝐺, exists.
This is not a precise formulation: We haven’t specified how smooth the boundary
should be: once continuously differentiable? twice continuously differentiable? five times?
And we didn’t specify what supplementary conditions we impose: is it enough that the
coefficients 𝑎𝑖𝑗 (𝒙) and the drift coefficients 𝑏𝑖 (𝒙) should satisfy a Lipschitz condition?
or some weaker assumptions are enough? or we should require three-times continuous
differentiability of 𝑎𝑖𝑗 and twice continuous differentiability of 𝑏𝑖 ? I won’t give a precise
formulation here: better you look up a book on (elliptic) partial differential equation and
see what is written on the subject there.
It is clear, though, that whatever the precise conditions should be, they should be
satisfied for (half) the Laplace operator (𝑎𝑖𝑗 (𝒙) ≡ 𝛿𝑖𝑗 , 𝒃(𝒙) ≡ 0) and 𝐺 being a circle (or
a sphere in dimension > 2).
For 12 Δ and 𝐺 being the unit-radius circle the Poisson kernel is easiest written in polar
coordinates: for 𝒙 ∈ 𝐺 having the polar coordinates (𝑟, 𝜃) and 𝒚 ∈ ∂𝐺 with coordinates
(1, 𝜑) we have
(2𝜋)−1 (1 − 𝑟2 )
𝑝(𝒙, 𝒚) =
.
(25.1)
1 − 2𝑟 cos(𝜃 − 𝜑) + 𝑟2
I won’t check it, differentiating twice in the Cartesian coordinates 𝑥1 , 𝑥2 and checking that
the measure on ∂𝐺 with the density 𝑝(𝒙, ∙) converges weakly to 𝛿𝒙0 , the unit measure
concentrated at the point 𝒙0 = (1, 𝜃0 ), as 𝑟 → 1− , 𝜃 → 𝜃0 , and am not going to give you
this as a problem. Better you check it yourself for your own satisfaction (or look it up in
some book). At least try to draw the graph of (25.1) as a function of 𝜑 for some (𝑟, 𝜃)
with 𝑟 close to 1.
If the region 𝐺 is not bounded; or the boundary is not smooth; or the operator 𝐿 is
no strictly elliptic (is degenarate elliptic, with (𝑎𝑖𝑗 ) being only nonnegative definite), it’s
possible that(the Poisson
kernel still exists, and it’s possible that it doesn’t exist. Example:
)
(
)
1 0
𝑎𝑖𝑗 (𝒙) ≡
, 𝒃(𝒙) ≡ 0: that is, the first component of our diffusion process is a
0 0
one-dimensional Wiener process, while the second component remains constant (𝑑𝜉𝑡2 ≡ 0).
Draw a picture of the region 𝐺 and see that the distribution of the exit point 𝝃𝜏𝒙𝒙 is
𝐺
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concentrated at two points, so it has no density with respect to the curve length ℓ(𝑑𝒚), no
Poisson kernel.
( ) On the other hand, it turns out that if the diffusion matrix is the same,
0
while 𝒃 ≡
, a Poisson kernel exists and can be written explicitly for some nice regions
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(probably we return to this later).
The only example we had before (25.1) was one with an unbounded region 𝐺 (the
upper half-plane) and the Cauchy distribution.
Now we return to the question of diffusion processes being Markov processes.
We have proved that the diffusion process 𝝃𝑡𝑠0 , 𝒙0 = 𝝃𝑡𝑠0 , 𝒙0 (𝜔) starting from the
point 𝒙0 at time 𝑡0 is, for fixed 𝑡0 and 𝑠, measurable in (𝒙0 , 𝜔) with respect to the
𝜎-algebra ℬ 𝑟 × ℱ𝑠 .
Before we go further, let us reformulate what we have found out in a different language
(in the spirit of Problems 1* – 6* ):
For every pair of time moments 𝑡0 ≤ 𝑠 there exists a functional 𝒇 (𝑡0 , 𝒙0 , 𝑠;
𝒙𝑢 , 𝑡0 ≤ 𝑢 ≤ 𝑠) (𝒙0 is not the value of the function 𝒙∙ at the point 𝑢 = 0: I hope
no misunderstanding will arise), depending on 𝒙0 ∈ ℝ𝑟 and on a function 𝒙𝑢 , 𝑡0 ≤ 𝑢 ≤ 𝑠,
belonging to the space C[𝑡0 , 𝑠] of continuous functions on this interval, the functional
being measurable with respect to ℬ 𝑟 × ℬ [𝑡0 , 𝑠] (C[𝑡0 , 𝑠]), such that
𝝃𝑡𝑠0 , 𝒙0 = 𝒇 (𝑡0 , 𝒙0 , 𝑠; 𝑾𝑢 − 𝑾𝑡0 , 𝑡0 ≤ 𝑢 ≤ 𝑠)
(25.2)
almost surely. (It’s clear why we can take 𝑾𝑢 − 𝑾𝑡0 rather than just 𝑾𝑢 in (25.2):
all stochastic integrals depend only on increments of the Wiener process, and they don’t
change if we subtract 𝑾𝑡0 from all its values.)
How can we construct the functional 𝒇 so that (25.2) holds?
Even if we had solved problems 1* – 6* , it seems unlikely that we could use it here:
the measurable space in these problems was, at most, the 𝑟-dimensional space, or its Borel
subset, with the 𝜎-algebra of its Borel subsets; and here we have an infinite-dimensional
space C = C[𝑡0 , 𝑠]. So: how?
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