Reversal of an Autoregressive Cauchy Markov Chain
Reversal of an Autoregressive Cauchy Markov Chain
Let
\begin{align} \mu_0 (dx) = \frac{1}{\pi} \frac{1}{1+x^2} dx. \end{align}
Suppose I fix
- At step
, I sample . - I then set
It is not so hard to show that this chain admits
What I'm interested in is the reversal of this Markov chain. More precisely, if the chain I describe above uses the Markov kernel
\begin{align} \mu_0 (dx) q (x \to dy) = \mu_0 (dy) r (y \to dx). \end{align}
Fortunately, all of the quantities involved have densities with respect to Lebesgue measure, and as such, I can write down what
\begin{align} r (y \to dx) = \frac{1 + y^2}{\pi} \frac{1}{1 + x^2} \frac{h}{ h^2 + (y - (1 - h) x)^2} dx. \end{align}
My question is then: is there a simple, elegant way to draw exact samples from
I would highlight that this is not a purely algorithmic question; I'd really like to understand what this reversal kernel
For completeness, some of the `purely algorithmic' solutions I had considered were the following.
- I could try rejection sampling, and in principle this would work, but it wouldn't really give me insight into the nature of the Markov chain.
- I could try something like the inverse CDF method, but it seems to me that the CDF of
is not particularly nice to work with. As such, I'd have to use e.g. Newton iterations to use this method, and I'd prefer to not have to do this.
K2-Think
To sample from the reversal kernel ( r(y \to dx) ) of the given autoregressive Cauchy Markov chain, we can utilize the structure of the original chain and properties of the Cauchy distribution. Here's the step-by-step approach:
Insight into the Reversal Kernel
The original chain transitions as ( X_{n+1} = (1 - h)X_n + h Y_n ), where ( Y_n \sim \text{Cauchy}(0,1) ). The stationary distri...
GPT-5
Write a = 1 − h and b = h (so a,b ≥ 0 and a + b = 1). The forward step is Y = a X + b Z, with X,Z i.i.d. standard Cauchy and independent. Under stationarity X and Y are both standard Cauchy. The time–reversal kernel r is exactly the conditional law of X given Y=y under this bivariate 1-stable (Cauchy) vector. With densities, r(y → dx) ∝ (1 + y^2) 1/(1 + x^2) · b/(b^2 + (y − a x)^2) dx, which is th...
o3
Let
X₀ ,X₁ ,… (1)
X_{n+1}= (1–h) X_{n}+h Y_{n}, 0<h≤1, Y_{n}∼ C(0,1) i.i.d.
be the forward chain, whose transition density is
q(x→y)= h π⁻¹ [ h²+(y−(1–h)x)² ]⁻¹ . (2)
Because the Cauchy law is 1–stable, μ₀(dx)=π⁻¹(1+x²)⁻¹dx is stationary:
μ₀(dx)q(x→y)=μ₀(dy)q(y→x). (3)
Time re...
DeepSeek R1
We know that the stationary distribution is the standard Cauchy, i.e., μ_0(dx) = (1/(π(1+x^2))) dx.
The forward kernel q(x -> dy) is the density of the distribution of X_{n+1} given X_n = x. Since: X_{n+1} = (1-h)x + hY, and Y ~ Cauchy(0,1), then the distribution o...
o3 Pro
Below I first give an exact simulation algorithm for the reversed kernel
r(y → dx) = const(y) · (1+x²)⁻¹ (h²+(y-(1-h)x)²)⁻¹ dx ,
and afterwards explain where it comes from and why it is “natural’’ for a Cauchy autoregression.
A. Exact simulation of X₀ | X₁ = y --------------------------------------------------...
o4 mini
The key observation is that nothing “mystical’’ is hiding in the reversed kernel. In fact one shows that
- Forward in stationarity
Y = (1−h) X + h Z,
with X∼Cauchy(0,1) independent of Z∼Cauchy(0,1).
2. Hence in the joint (X,Y,Z) the map
(X,Z) ↦ Y=(1−h)X + h Z
is an invertible linear map (for each fixed Y one can solve for X in terms of Z). What you wa...
Claude 3.7 Sonnet
Reversal of an Autoregressive Cauchy Markov Chain
This is an interesting problem about a Markov chain with Cauchy distributions. Let me analyze the reversal kernel and see if there's an elegant sampling method.
Understanding the Problem
We have a Markov chain
where (standard Cauchy) - The stationary distributio...
Claude Opus 4
This is a fascinating question about time-reversal of a Cauchy autoregressive process. Let me analyze the structure of the reversal kernel and provide insight into what it's doing.
Understanding the Reversal Kernel
The reversal kernel you've derived:
can be rewritten more suggestively. Let me factor thi...
Gemini 2.5 Pro
This is an excellent and challenging question. The integral identity you've found is indeed correct and represents a fundamental result in the theory of multipole expansions, particularly in the context of solutions to the Helmholtz equation. These identities are known as addition theorems for spherical wave functions and are crucial in fields like acoustics, electromagnetism, and quantum mech...
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