By Joseph L. Doob

From the experiences: "Here is a momumental paintings by way of Doob, one of many masters, within which half 1 develops the aptitude conception linked to Laplace's equation and the warmth equation, and half 2 develops these components (martingales and Brownian movement) of stochastic approach thought that are heavily regarding half 1". --G.E.H. Reuter in brief e-book studies (1985)

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**Additional info for Classical potential theory and its probabilistic counterpart**

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EZ g(EZ). 2 284 Evolving Set Methods It is fairly easy to translate these to mixing time bounds. 6 it is appropriate to let f (z) = 1−z z for L bounds. 5) τ2 ( ) ≤ 2x(1 − x)(1 − C√z(1−z) (x)) π∗ 1 1+ 2 /4 dx 4π∗ x(1 − x)(1 − C√ (x)) z(1−z) 1+3π∗ 1 with the ﬁrst integral requiring x 1 − C√z(1−z) 1+x to be convex. 2 r By making the change of variables x = 1+r and applying a few pessimistic approximations one obtains a result more strongly resembling spectral proﬁle bounds: 1 1 log √ √ 1 − C z(1−z) π∗ 1/ 2 dr τ2 ( ) ≤ √ 2r(1 − C z(1−z) (r)) π∗ 4/ 2 dr √ r(1 − C (r)) 4π∗ z(1−z) For total variation distance related results are in terms of Cz(1−z) (r), and Cz log(1/z) (r) for relative entropy.

PΛ(S, y) = ΛK(S, Proof. PΛ(S, y) = z∈S ˆ y) = ΛK(S, S y Q(S, y) π(z) P(z, y) = π(S) π(S) ˆ S ) π(y) = π(y) K(S, π(S ) π(S) The ﬁnal equality Q(S, y)/π(y). is because S K(S, S ) = S y y K(S, S Q(S, y) π(S) ) = P rob(y ∈ S ) = With duality it becomes easy to write the n step transitions in terms ˆ of the walk K. ˆ n . 4. Let E {x} then ˆ n πS (y) , Pn (x, y) = E n where πS (y) = set S by π. 1S (y)π(y) π(S) denotes the probability distribution induced on Proof. ˆ n )({x}, y) = E ˆ n πS (y) Pn (x, y) = (Pn Λ)({x}, y) = (ΛK n The ﬁnal equality is because Λ(S, y) = πS (y).

3. 15. λP ≥ 1 λˆ , MA P ρP ≥ 1 ρˆ , MA P ΛP (r) ≥ 1 Λ ˆ (r). MA P The log-Sobolev and spectral proﬁle mixing time bounds of P are ˆ thus at worst a factor M A times larger than those of P. ˆ along with the If the distribution π = π ˆ then a Nash inequality for P, 1 relation EP (f, f ) ≥ A EPˆ (f, f ), immediately yields a Nash inequality for P. It is not immediately clear how to compare Nash inequality bounds if π = π ˆ . 5) gives for P. EP (f, f ) to EPˆ (f, f ) and Varπ (f ) to Varπˆ (f ) in the original proofs of the mixing times.