Interlacing Diffusions

Abstract

We study in some generality intertwinings between h-transforms of Karlin–McGregor semigroups associated with one dimensional diffusion processes and those of their Siegmund duals. We obtain couplings so that the corresponding processes are interlaced and furthermore give formulae in terms of block determinants for the transition densities of these coupled processes. This allows us to build diffusion processes in the space of Gelfand–Tsetlin patterns so that the evolution of each level is Markovian. We show how known examples naturally fit into this framework and construct new processes related to minors of matrix valued diffusions. We also provide explicit formulae for the transition densities of the particle systems with one-sided collisions at either edge of such patterns.

Appendix

We collect here the proofs of some of the facts regarding conjugate diffusions that were stated and used in previous sections.

We first give the derivation of the table on the boundary behaviour of a diffusion and its conjugate. Keeping with the notation of Sect. 13.2 consider the following quantities with x ∈ I^∘ arbitrary,

$$\displaystyle \begin{aligned} & N(l)=\int_{(l^+,x]}^{}(s(x)-s(y))M(dy)=\int_{(l^+,x]}^{}(s(x)-s(y))m(y)dy ,\\ &\Sigma(l)=\int_{(l^+,x]}^{}(M(x)-M(y))s(dy)=\int_{(l^+,x]}^{}(M(x)-M(y))s'(y)dy. \end{aligned} $$

We then have the following classification of the boundary behaviour at l (see e.g. [29]):

• l is an entrance boundary iff N(l) < ∞,  Σ(l) = ∞.

• l is a exit boundary iff N(l) = ∞,  Σ(l) < ∞.

• l is a natural boundary iff N(l) = ∞,  Σ(l) = ∞.

• l is a regular boundary iff N(l) < ∞,  Σ(l) < ∞.

From the relations $\hat {s}'(x)=m(x)$ and $\hat {m}(x)=s'(x)$ we obtain the following,

$$\displaystyle \begin{aligned} &\hat{N}(l)=\int_{(l^+,x]}^{}(\hat{s}(x)-\hat{s}(y))\hat{m}(y)dy=\Sigma(l),\\ &\hat{\Sigma}(l)=\int_{(l^+,x]}^{}(\hat{M}(x)-\hat{M}(y))\hat{s}'(y)dy=N(l). \end{aligned} $$

These relations immediately give us the table on boundary behaviour, namely: If l is an entrance boundary for X, then it is exit for $\hat {X}$ and vice versa. If l is natural for X, then so it is for its conjugate. If l is regular for X, then so it is for its conjugate. In this instance as already stated in Sect. 13.2 we define the conjugate diffusion $\hat {X}$ to have boundary behaviour dual to that of X, namely if l is reflecting for X then it is absorbing for $\hat {X}$ and vice versa.

Proof (Proof of Lemma 2.1)

There is a total number of 5² boundary behaviours (5 at l and 5 at r) for the L-diffusion (the boundary behaviour of $\hat {L}$ is completely determined from L as explained above) however since the boundary conditions for an entrance and regular reflecting ($\mathcal {D}_sv=0$) and similarly for an exit and regular absorbing boundary ($\mathcal {D}_m\mathcal {D}_sv=0$) are the same we can pair them to reduce to 3² cases (b.c.(l), b.c.(r)) abbreviated as follows:

$$\displaystyle \begin{aligned} & (nat,nat),(ref,ref),(abs,abs),(nat,abs),(ref,abs),\\ & \quad (abs,ref),(abs,nat),(nat,ref),(ref,nat). \end{aligned} $$

We now make some further reductions. Note that for x, y ∈ I^∘,

$$\displaystyle \begin{aligned} \mathsf{P}_t {\mathbf{1}}_{[l,y]}(x)=\hat{\mathsf{P}}_t {\mathbf{1}}_{[x,r]}(y) \iff \mathsf{P}_t {\mathbf{1}}_{[y,r]}(x)=\hat{\mathsf{P}}_t {\mathbf{1}}_{[l,x]}(y). \end{aligned} $$

After swapping x ↔ y this is equivalent to,

$$\displaystyle \begin{aligned} \hat{\mathsf{P}}_t {\mathbf{1}}_{[l,y]}(x)=\mathsf{P}_t {\mathbf{1}}_{[x,r]}(y). \end{aligned} $$

So we have a bijection that swaps boundary conditions with their duals $(\mathsf {b.c.}(l),\mathsf {b.c.}(r))\leftrightarrow (\widehat {\mathsf {b.c.}(l)},\widehat {\mathsf {b.c.}(r)})$. Moreover, if $\mathfrak {h}:(l,r)\to (l,r)$ is any homeomorphism such that $\mathfrak {h}(l)=r,\mathfrak {h}(r)=l$ and writing H_t for the semigroup associated with the $\mathfrak {h}(X)(t)$-diffusion and similarly $\hat {\mathsf {H}}_t$ for the semigroup associated with the $\mathfrak {h}(\hat {X})(t)$-diffusion we see that,

$$\displaystyle \begin{aligned} \mathsf{P}_t {\mathbf{1}}_{[l,y]}(x)=\hat{\mathsf{P}}_t {\mathbf{1}}_{[x,r]}(y) \ \ \forall x,y \in I^\circ \iff \mathsf{H}_t {\mathbf{1}}_{[l,y]}(x)=\hat{\mathsf{H}}_t {\mathbf{1}}_{[x,r]}(y) \ \ \forall x,y \in I^{\circ}. \end{aligned} $$

And we furthermore observe that, the boundary behaviour of the $\mathfrak {h}(X)(t)$-diffusion at l is the boundary behaviour of the L-diffusion at r and its boundary behaviour at r is that of the L-diffusion at l and similarly for $\mathfrak {h}(\hat {X})(t)$. We thus obtain an equivalent problem where now (b.c.(l), b.c.(r)) ↔ (b.c.(r), b.c.(l)). Putting it all together, we reduce to the following 4 cases since all others can be obtained from the transformations above,

$$\displaystyle \begin{aligned} (nat,nat),(ref,nat),(ref,ref),(ref,abs). \end{aligned} $$

The first case is easy since there are no boundary conditions to keep track of and is omitted. The second case is the one originally considered by Siegmund and studied extensively in the literature (see e.g. [21] for a proof). We give the proof for the last two cases.

First, assume l and r are regular reflecting for X and so absorbing for $\hat {X}$. Let $\mathcal {R}_{\lambda }$ and $\hat {\mathcal {R}}_{\lambda }$ be the resolvent operators associated with P_t and $\hat {\mathsf {P}}_t$ then with f being a continuous function with compact support in I^∘ the function $u=\mathcal {R}_{\lambda }f$ solves Poisson’s equation $\mathcal {D}_m\mathcal {D}_su-\lambda u=-f$ with $\mathcal {D}_su(l^+)=0, \mathcal {D}_su(r^-)=0$. Apply $\mathcal {D}_m^{-1}$ defined by $\mathcal {D}_m^{-1}f(y)=\int _{l}^{y}m(z)f(z)dz$ for y ∈ I^∘ to obtain $\mathcal {D}_su-\lambda \mathcal {D}_m^{-1} u=-\mathcal {D}_m^{-1}f$ which can be written as,

$$\displaystyle \begin{aligned} \mathcal{D}_{\hat{m}}\mathcal{D}_{\hat{s}}\mathcal{D}_m^{-1}u-\lambda\mathcal{D}_m^{-1} u=-\mathcal{D}_m^{-1}f. \end{aligned} $$

So $v=\mathcal {D}_m^{-1}u$ solves Poisson’s equation with $g=\mathcal {D}_m^{-1}f$,

$$\displaystyle \begin{aligned} \mathcal{D}_{\hat{m}}\mathcal{D}_{\hat{s}}v-\lambda v=-g, \end{aligned} $$

with the boundary conditions $\mathcal {D}_{\hat {m}}\mathcal {D}_{\hat {s}}v(l^+)=\mathcal {D}_{s}\mathcal {D}_{m}\mathcal {D}_{m}^{-1}u(l^+)=\mathcal {D}_s u(l^+)=0$ and $\mathcal {D}_{\hat {m}}\mathcal {D}_{\hat {s}}v(r^-)=0$. Now in the second case when l is reflecting and r absorbing we would like to check the reflecting boundary condition for $v=\mathcal {D}_m^{-1}u$ at r. Namely, that $(\mathcal {D}_{\hat {s}})v(r^-)=0$ and note that this is equivalent to $(\mathcal {D}_{m})v(r^-)=u(r^-)=0$. This then follows from the fact that (since r is now absorbing for the L-diffusion) $(\mathcal {D}_m\mathcal {D}_s)u(r^-)=0$ and that f is of compact support. The proof proceeds in the same way for both cases, by uniqueness of solutions to Poisson’s equation (see e.g. Section 3.7 of [39]) this implies $v= \hat {\mathcal {R}}_{\lambda }g$ and thus we may rewrite the relationship as,

$$\displaystyle \begin{aligned} \mathcal{D}_m^{-1}\mathcal{R}_{\lambda}f= \hat{\mathcal{R}}_{\lambda}\mathcal{D}_m^{-1}f. \end{aligned} $$

Let now f approximate δ_x with x ∈ I^∘ to obtain with r_λ(x, z) the resolvent density of $\mathcal {R}_{\lambda }$ with respect to the speed measure in I^∘× I^∘,

$$\displaystyle \begin{aligned} \int_{l}^{y}r_{\lambda}(z,x)m(z)dz=m(x)\hat{\mathcal{R}}_{\lambda}{\mathbf{1}}_{[x,r]}(y). \end{aligned} $$

Since r_λ(z, x)m(z) = m(x)r_λ(x, z) we obtain,

$$\displaystyle \begin{aligned} \mathcal{R}_{\lambda}{\mathbf{1}}_{[l,y]}(x)=\hat{\mathcal{R}}_{\lambda}{\mathbf{1}}_{[x,r]}(y), \end{aligned} $$

and the result follows by uniqueness of Laplace transforms.

It is certainly clear to the reader that the proof only works for x, y in the interior I^∘. In fact the lemma is not always true if we allow x, y to take the values l, r. To wit, first assume x = l so that we would like,

$$\displaystyle \begin{aligned} \mathsf{P}_t {\mathbf{1}}_{[l,y]}(l)\overset{?}{=}\hat{\mathsf{P}}_t {\mathbf{1}}_{[l,r]}(y)=1 \ \forall y. \end{aligned} $$

This is true if and only if l is either absorbing, exit or natural for the L-diffusion (where in the case of a natural boundary we understand P_t1_[l,y](l) as lim_x→lP_t1_[l,y](x)). Analogous considerations give the following: The statement of Lemma 13.1 remains true with x = r if r is either a natural, reflecting or entrance boundary point for the L-diffusion. Enforcing the exact same boundary conditions gives that the statement remains true with y taking values on the boundary of I.

Remark 13.15

For the reader who is familiar with the close relationship between duality and intertwining first note that with the L-diffusion satisfying the boundary conditions in the paragraph above and denoting as in Sect. 13.2 by P_t the semigroup associated with an L-diffusion killed (not absorbed) at l our duality relation becomes,

$$\displaystyle \begin{aligned} P_t {\mathbf{1}}_{[x,r]}(y)=\hat{\mathsf{P}}_t {\mathbf{1}}_{[l,y]}(x) . \end{aligned} $$

It is then a simple exercise, see Proposition 5.1 of [16] for the general recipe of how to do this, that this is equivalent to the intertwining relation,

$$\displaystyle \begin{aligned} P_t\Lambda=\Lambda\hat{\mathsf{P}}_t, \end{aligned} $$

where Λ is the unnormalized kernel given by $(\Lambda f)(x)=\int _{l}^{x}\hat {m}(z)f(z)dz$. This is exactly the intertwining relation obtained in (13.26) with n₁ = n₂ = 1.

Entrance Laws

For x ∈ I and $\mathfrak {h}_n$ a positive eigenfunction of $P_t^n$ we would like to compute the following limit that defines our entrance law $\mu _t^x\left (\mathbf {y}\right )$ (with respect to Lebesgue measure) and corresponds to starting the Markov process $P^{n,\mathfrak {h}_n}_t$ from (x, ⋯ , x),

$$\displaystyle \begin{aligned} \mu_t^x\left(\mathbf{y}\right):=\lim_{(x_1,\cdots,x_n)\to x \mathbf{1}}e^{-\lambda t}\frac{\mathfrak{h}_n(y_1,\cdots,y_n)}{\mathfrak{h}_n(x_1,\cdots,x_n)}\det\left(p_t(x_i,y_j)\right)^n_{i,j=1} . \end{aligned} $$

Note that, since as proven in Sect. 13.3.10 all eigenfunctions built from the intertwining kernels are of the form $\det \left (h_i(x_j)\right )^n_{i,j=1}$ we will restrict to computing,

$$\displaystyle \begin{aligned} \mu_t^x\left(\mathbf{y}\right):= e^{-\lambda t}\det\left(h_i(y_j)\right)^n_{i,j=1}\lim_{(x_1,\cdots,x_n)\to x\mathbf{1}}\frac{\det\left(p_t(x_i,y_j)\right)^n_{i,j=1}}{\det\left(h_i(x_j)\right)^n_{i,j=1}}. \end{aligned} $$

If we now assume that p_t(⋅, y) ∈ Cⁿ⁻¹∀t > 0, y ∈ I^∘ and similarly h_i(⋅) ∈ Cⁿ⁻¹ (in fact we only need to require this in a neighbourhood of x) we have,

$$\displaystyle \begin{aligned} \lim_{(x_1,\cdots,x_n)\to x\mathbf{1}}\frac{\det\left(p_t(x_i,y_j)\right)^n_{i,j=1}}{\det\left(h_i(x_j)\right)^n_{i,j=1}}&=\lim_{(x_1,\cdots,x_n)\to x\mathbf{1}}\frac{\det\left(x_j^{i-1}\right)^n_{i,j=1}}{\det\left(h_i(x_j)\right)^n_{i,j=1}}\\ & \quad \times \frac{\det\left(p_t(x_i,y_j)\right)^n_{i,j=1}}{\det\left(x_j^{i-1}\right)^n_{i,j=1}}\\ &=\frac{1}{\det\left(\partial^{i-1}_xh_j(x)\right)^n_{i,j=1}}\\ & \quad \times \det\left(\partial^{i-1}_xp_t(x,y_j)\right)^n_{i,j=1}. \end{aligned} $$

For the fact that the Wronskian, $\det \left (\partial ^{i-1}_xh_j(x)\right )^n_{i,j=1}>0$ and in particular does not vanish see Sect. 13.3.10. Thus,

$$\displaystyle \begin{aligned} \mu_t^x\left(\mathbf{y}\right)=const_{x,t}\times \det\left(h_i(y_j)\right)^n_{i,j=1}\det\left(\partial^{i-1}_xp_t(x,y_j)\right)^n_{i,j=1}, \end{aligned} $$

is given by a biorthogonal ensemble as in (13.29). The following lemma, which is an adaptation of Lemma 3.2 of [47] to our general setting, gives some more explicit information.

Lemma 13.4

Assume that for x′ in a neighbourhood of x there is a convergent Taylor expansion ∀t > 0, y ∈ I^∘,

$$\displaystyle \begin{aligned} \frac{p_t(x',y)}{p_t(x,y)}=f(t,x')\sum_{i=0}^{\infty}\left(x'-x\right)^i\phi_i(t,y), \end{aligned} $$

for some functions f, {ϕ_i}_i≥0 that in particular satisfy f(t, x)ϕ₀(t, y) ≡ 1. Then $\mu _t^x\left (\mathbf {y}\right )$ is given by the biorthogonal ensemble,

$$\displaystyle \begin{aligned} const_{x,t}\times \det\left(h_i(y_j)\right)^n_{i,j=1} \det\left(\phi_{i-1}(t,y_j)\right)^n_{i,j=1} \prod_{i=1}^{n}p_t(x,y_i). \end{aligned} $$

If moreover we assume that we have a factorization ϕ_i(t, y) = yⁱg_i(t) then $\mu _t^x\left (\mathbf {y}\right )$ is given by the polynomial ensemble,

$$\displaystyle \begin{aligned} const^{\prime}_{x,t} \times \det\left(h_i(y_j)\right)^n_{i,j=1}\det\left(y^{i-1}_j\right)^n_{i,j=1} \prod_{i=1}^{n}p_t(x,y_i). \end{aligned} $$

Proof

By expanding the Karlin–McGregor determinant and plugging in the Taylor expansion above we obtain,

$$\displaystyle \begin{aligned} \frac{\det\left(p_t(x_i,y_j)\right)^n_{i,j=1}}{\prod_{i=1}^{n}p_t(x,y_i)}&=\prod_{i=1}^{n}f(t,x_i)\sum_{k_1,\cdots,k_n \ge 0}^{}\prod_{i=1}^{n}\left(x_i-x\right)^{k_i}\\ & \quad \times \sum_{\sigma \in \mathfrak{S}_n}^{} sign(\sigma)\prod_{i=1}^{n}\phi_{k_i}(t,y_{\sigma(i)})\\ &=\prod_{i=1}^{n}f(t,x_i)\sum_{k_1,\cdots,k_n \ge 0}^{}\prod_{i=1}^{n}\left(x_i-x\right)^{k_i}\det\left(\phi_{k_i}(t,y_j)\right)^n_{i,j=1}. \end{aligned} $$

First, note that we can restrict to k₁, ⋯k_n distinct otherwise the determinant vanishes. Moreover, we can in fact restrict the sum over k₁, ⋯ , k_n ≥ 0 to k₁, ⋯ , k_n ordered by replacing k₁, ⋯ , k_n by k_τ(1), ⋯ , k_τ(n) and summing over $\tau \in \mathfrak {S}_n$ to arrive at the following expansion,

$$\displaystyle \begin{aligned} \frac{\det\left(p_t(x_i,y_j)\right)^n_{i,j=1}}{\prod_{i=1}^{n}p_t(x,y_i)}&=\prod_{i=1}^{n}f(t,x_i)\\ & \quad \sum_{0\le k_1<k_2<\cdots<k_n }^{}\det\left(\left(x_j-x\right)^{k_i}\right)^n_{i,j=1}\det\left(\phi_{k_i}(t,y_j)\right)^n_{i,j=1}. \end{aligned} $$

Now, write with k = (0 ≤ k₁ < ⋯ < k_n) ,

$$\displaystyle \begin{aligned} \chi_{\mathbf{k}}(z_1,\cdots,z_n)=\frac{\det\left(z^{k_i}_j\right)^n_{i,j=1}}{\det\left(z^{i-1}_j\right)^n_{i,j=1}}, \end{aligned} $$

for the Schur function and note that $\lim _{(z_1,\cdots ,z_n)\to 0}\chi _{\mathbf {k}}(z_1,\cdots ,z_n)=0$ unless k = (0, ⋯ , n − 1) in which case we have χ_k ≡ 1. We can now finally compute,

$$\displaystyle \begin{aligned} &\lim_{(x_1,\cdots,x_n)\to x\mathbf{1}} \frac{\det\left(p_t(x_i,y_j)\right)^n_{i,j=1}}{\det\left(x_j^{i-1}\right)^n_{i,j=1}}\\ & \quad =\lim_{(x_1,\cdots,x_n)\to x\mathbf{1}} \frac{\det\left(p_t(x_i,y_j)\right)^n_{i,j=1}}{\det\left((x_j-x)^{i-1}\right)^n_{i,j=1}}=\prod_{i=1}^{n}p_t(x,y_i)\\ & \qquad \times \lim_{(x_1,\cdots,x_n)\to x\mathbf{1}} \prod_{i=1}^{n}f(t,x_i)\\ & \qquad \times \sum_{0\le k_1<k_2<\cdots<k_n }^{}\chi_{\mathbf{k}}(x_1-x,\cdots,x_n-x)\det\left(\phi_{k_i}(t,y_j)\right)^n_{i,j=1}\\ & \quad =f^n(t,x)\times \prod_{i=1}^{n}p_t(x,y_i)\det\left(\phi_{i-1}(t,y_j)\right)^n_{i,j=1}. \end{aligned} $$

The first statement of the lemma now follows with,

$$\displaystyle \begin{aligned} const_{x,t}=e^{-\lambda t} f^n(t,x)\frac{1}{\det\left(\partial^{i-1}_xh_j(x)\right)^n_{i,j=1}}. \end{aligned} $$

The fact that when ϕ_i(t, y) = yⁱg_i(t) we obtain a polynomial ensemble is then immediate.