Archive for December 2nd, 2012

On Patch Stationariness II (RWLA,MCT,GT,AM Part X)

December 2nd, 2012 No comments

Claim:  Suppose p(x,y) = f_1(x) \cdot g_1(y) + f_2(x) \cdot g_2(y) and \int p(x,y) dx = 1 .  This last restriction causes freedom of choice of three functions, say the first three, and then g_2(y) = \frac{1 - g_1 (y) \int_0^1 f_1(x) dx}{\int_0^1 f_2(x) dx} = \frac{1 - g_1(y) F_1}{F_2} .  This basic construction of p(x,y) is the premise of all claims regarding patches I've proven here on my blog.  Take an a(x) with \int_0^1 a(x) dx = 1, with form that will be defined in the course of the proof.   Then B = \int a(1-y) g_1(y) dy = \sum_{i=0}^{\infty} a^i(1-y) G_1^{i+1} \vert_0^1 provided the sum converges.  If it does, this gives rise to the "stationary" p_\infty (x,y) = \frac{f_2(x)}{F_2} - \left( \frac{f_2(x) F_1}{F_2} - f_1(x) \right) B (and it is also a patch).

Proof: Since we are looking for stationary p_\infty (x,y), we are looking for  a(x) \star p(x,y) = \lambda a(x) with  \lambda = 1 in direct analogy with discrete Markov chains.  The star operator is explicitly as I've defined it previously implies \int_0^1 a(1 - y) p(x,y) dy = a(x) .  By the definition of  p(x,y) , we have that \int_0^1 a(1 - y) \left( f_1(x) g_1(y) + f_2(x) \left( \frac{1 - g_1(y) F_1}{F_2} \right) \right) dy = a(x) .  Expansion results in a(x) = f_1(x) \int_0^1 a(1-y) g_1(y) dy + \frac{f_2(x)}{F_2} \cdot 1- \frac{F_1}{F_2} f_2(x) \int_0^1 a(1-y) g_1(y) dy , where we have simplified  \int_0^1 a(1-y) dy to 1 because the transformation to the y-axis does not change the integral result.

Next we have the expression

a(x) = \frac{f_2(x)}{F_2} - \left( \frac{F_1 f_2(x)}{F_2} - f_1(x) \right) \int_0^1 a(1-y) g_1(y) dy


a(x) = \frac{f_2(x)}{F_2} - \left( \frac{F_1 f_2(x)}{F_2} - f_1(x) \right) B and derivatives

a^i(x) = \frac{f_2^i(x)}{F_2} - \left( \frac{F_1 f_2^i(x)}{F_2} - f_1^i(x) \right) B

We want to obtain B, so that the expression  \int_0^1 a(1-y) g_1(y) dy can be clearly defined.  We use the tabular method to simplify the integration by parts.

 \begin{array}{ccccc} \vert & Derivatives & \vert & Integrals & \vert \\ \vert & a(1-y) & \vert & g_1(y) & \vert \\ \vert & -a'(1-y) & \vert & G_1^1(y) & \vert \\ \vert & a''(1-y) & \vert & G_1^2(y) & \vert \\ \vert & \vdots & \vert & \vdots & \vert \end{array}

and obtain

 B = a(1-y) G_1^1(y) + a^{'}(1-y) G_1^2(y) + \ldots \vert_0^1 = \sum_{i=0}^\infty a^i (1-y) G_1^{i+1} \vert_0^1

Lastly the steady-state "stationary" patch is exactly this a(x), so that p_\infty(x,y) = a(x) as we desired to show.

Notice that \int_0^1 p_\infty(x,y) dx = 1 by hypothesis, since it is equal to a(x).

Existence.  Lastly, notice that p_\infty (x,y) exists provided the sum converges.  In previous exercises, I have shown that such converges in two specific examples: when derivatives of a(x) vanish (when a(x) is a finite polynomial) and when derivatives of a(x) are periodic but vanish for the periodic terms.