Re: Correlation P<-->(x,y) = P<-->(x,0) + P<-->(0,y) + 2F(x,y) - 1

From: Osher Doctorow (mdoctorow_at_comcast.net)
Date: 08/15/04 

Date: Sun, 15 Aug 2004 00:40:45 +0000 (UTC)

On 13 Aug 04 23:27:11 -0400 (EDT), Osher Doctorow wrote:
>Theorem 1. P<-->(x,y) = P<-->(x,0) + P<-->(0,y) + 2F(x,y) - 1
>It is of some interest to notice that (1) has the functional form:
>
>5) f(x,y) = f(x,0) + f(0,y) + h(x,y)
>
>with h(x,y) = 2F(x,y) - 1. Conditions under which P<-->(x,y) is
>the unique solution and h(x,y) = 2F(x,y) - 1 are also interesting.

The Method of Characteristics in partial differential equations
(PDEs) seems to hint at a factorization:

1) (Dx - k)(Dy - k) = Dxy - kDx - kDy = 0

Since P<-->(x,y) = P-->(x,y) + P<--(x,y) - 1 = 2F(x,y) - FX(x) -
                                               -FY(y) + 1

we have:

2) Dxy[P<-->(x,y)] = 2Dxy[F(x,y)]
3) Dx[P<-->(x,y)] = 2Dx[F(x,y)] - fX(x)
4) Dy[P<-->(x,y)] = 2Dy[F(x,y)] - fY(y)

so that (1) becomes:

5) 2Dxy[F(x,y)] -2kDx[F(x,y)] + kfX(x) -2kDy[F(x,y)] + kfY(y) = 0

or in other words:

6) 2(1 - k)Dxy[F(x,y)] + k[fX(x) + fY(y) - 2Dx[F(x,y)] - 2Dy[F(x,y)]
   = 0

But Dxy[F(x,y)] = f(x,y), so (6) reduces to:

7) 2(1 - k)f(x,y) + k[fX(x) + fY(y)] - 2Dx[F(x,y)] - 2Dy[F(x,y)] = 0

We could solve this for f(x,y) if we had some handle on Dx[F(x,y)]
and Dy[F(x,y)].

Osher Doctorow

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