SERVING THE QUANTITATIVE FINANCE COMMUNITY

• 1
• 2

Alan
Posts: 9214
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

### Re: Singular and degenerate elliptic partial differential equation

Well, to be frank, I don't think it was your conjecture when you started this thread, or you would have written a different question.

My gut feel is that a putative proof has nothing to do with series expansions/Frobenius stuff, etc. Ignoring the singular nature of those delta functions, the conjectured solution is separable $p(x,y) = f(x) g(y)$. Maybe that is something to explore.

lovenatalya
Topic Author
Posts: 152
Joined: December 10th, 2013, 5:54 pm

### Re: Singular and degenerate elliptic partial differential equation

Alan wrote:
Well, to be frank, I don't think it was your conjecture when you started this thread, or you would have written a different question.

My gut feel is that a putative proof has nothing to do with series expansions/Frobenius stuff, etc. Ignoring the singular nature of those delta functions, the conjectured solution is separable $p(x,y) = f(x) g(y)$. Maybe that is something to explore.

Well, the A=0 in the first comment is the stationary equation after stripping out the x function. I asked to prove the poles and singularities of the solution. Once poles or singularities are proved, the solution will be unintegrable. Then the Dirac delta function will be the only choice. That is why I did not conjecture the delta functions directly, as it will be a direct consequence.
For the joint bivariate stationary equation, the best you can do is to assume $p(x,y) = e^{-x^2}g(x,y)$. $\mathcal A g(x,y)=0$ in my very first post, with $q=-2x$, is precisely the stationary equation after separating out $e^{-x^2}$. Separation of variables was the first thing came to my mind when I looked at this problem. But it does not look like you could do that. I do not think you can separate out y function from x function completely since you have x appearing in both the first and second order partial over y and with different powers which prevents you from dividing them out and moving x completely to the x partial terms.

What to do?

Alan
Posts: 9214
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

### Re: Singular and degenerate elliptic partial differential equation

Another thing to try might be conditioning on the X(t) path.

Related to that, I suggest you revisit the 1D case, where $X_t \equiv x_0$, a fixed real number and the process is

$dY_t = x_0 (1-Y_t) Y_t dt + x_0 (1-Y_t) Y_t dB_t$.

So, the ultimate terminal density ( admitting distributions) is $p(y; x_0)$.  Now,  what exactly do you think  $p(y; x_0)$ is?  Based on some things you wrote before, I suspect we disagree on the answer to that. Also, do you have a proof for your answer to this case?

I revisit that because I think you have to totally nail down that case before conditioning on an X-path.

Then, the next case (under a conditioning approach) would be: stick with the 1D case, but now take $X_t \equiv x(t)$, a given deterministic function of time.

So, now the 1D problem is:

$dY_t = x(t) (1-Y_t) Y_t dt + x(t) (1-Y_t) Y_t dB_t$,

and the terminal density is some $p(y; x(\cdot))$.

Now, can you conjecture  $p(y; x(\cdot))$?  Suppose $x(t) = \sin t$, or something else that oscillates at arbitrary large t?

I don't know the answer -- I am just making some suggestions about an approach.

lovenatalya
Topic Author
Posts: 152
Joined: December 10th, 2013, 5:54 pm

### Re: Singular and degenerate elliptic partial differential equation

@Alan:

I like your idea. I actually thought of this conditioning approach too before, but was scared off by having to consider all possible $x$ and integrate. I then sought to look for other more general and slicker qualitative methods which would give me the singularities without calculation. I will revisit this idea and report back.

Yes, I can solve the stationary density as a function of $x$ for the first SDE exactly.

Why do we need to look at the $x(t)$ as a function of time? Since we are looking at the terminal density, we just need to integrate the density over the terminal distribution of $x$, i.e., $\int_x dx\, e^{-x^2} p(x,y)$, right?

Alan
Posts: 9214
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

### Re: Singular and degenerate elliptic partial differential equation

To answer your last question, I need the answer to mine: what's $p(y;x_0)$ exactly?

lovenatalya
Topic Author
Posts: 152
Joined: December 10th, 2013, 5:54 pm

### Re: Singular and degenerate elliptic partial differential equation

Here is the answer to Alan's question.

The stationary probability density $p(x,y)$ conditioned on $x$ is
$$0 = -\frac{\partial}{\partial y}(xy(1-y)p(x,y))+\frac12\frac{\partial^2}{\partial y^2}\big((xy(1-y))^2p(x,y)\big)$$
Integrate over $y$ once,
$$c(x) = -\frac1xy(1-y)p+\frac12\frac{\partial}{\partial y}\big((y(1-y))^2p\big)$$
or
$$\frac{\partial}{\partial y}p+2\frac{1-\frac1x-2y}{y(1-y)}p-\frac{2c(x)}{(y(1-y))^2}=0.$$
Solve again
$$p(x,y) = y^{-2+\frac2x}(1-y)^{-2-\frac2x}\left[2c(x)\int_{\frac12}^yds\, s^{1-\frac2x}(1-s)^{1+\frac2x}+\frac1{16}p\Big(x,y=\frac12\Big)\right].$$
The first term is holomorphic on the domain $y\in[0,1]$ while the second term requires $x\in(-2,2)$ for $p(x,y)$ to be integrable over $y$, so long as $p\big(x,y=\frac12\big)\not = 0$.

Please check my calculation. If it is correct, the $y$-integrability region is confined to a strip in the plane. Shall we conclude the Dirac delta function resides only outside this strip or the whole plane?

Alan
Posts: 9214
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

### Re: Singular and degenerate elliptic partial differential equation

No, I get something different for the integrand. Anyway, take x = 1, which is certainly in (-2,2). It's easy to see that the second term is not integrable as y->1. Bottom line: IMO, this kind of analysis unfortunately goes nowhere. If done carefully (and I have not made a mistake), I conclude the two integration constants are identically 0 for all x. So p(x,y) = 0 in the interior of (0,1), leaving only the possibility of Dirac masses at the endpoints.

I had thought we were on agreement on that point and the issue was going to be over an expression of the Dirac terms. So, it doesn't make any sense at this point to elaborate on why I suggested looking at x(t) until we have agreement on the behavior for constant x.

lovenatalya
Topic Author
Posts: 152
Joined: December 10th, 2013, 5:54 pm

### Re: Singular and degenerate elliptic partial differential equation

@Alan:
You are right. I made some careless algebraic mistakes. Here is the corrected solution.

$$p(x,y) = y^{-2+\frac2x}(1-y)^{-2-\frac2x}\left[2c(x)\int_{\frac12}^yds\, s^{-\frac2x}(1-s)^{\frac2x}+\frac1{16}p_2\Big(x,y=\frac12\Big)\right].$$
Consider the first term. Consider the neighbourhood of $y=0$. The analysis of the neighbourhood of $y=1$ will be similar. The highest order singularities is a first order pole when $x\not=2$. When $x=2$, the highest order singularity is $\frac{\ln y}{y}$. So the first term is never integrable at either end of the $y$ interval.

Consider the second term. In the neighbourhood of $y=0$, it requires $x\in(0,2)$ for $p_2(x,y)$ to be integrable over $y$. In the neighbourhood of $y=1$, it requires $x\in(-2,0)$. So the second term is never integrable over the whole $y$ interval for $x\not=0$.

Do we agree? If so, we have almost reached the conclusion. But while it is intuitive, how do we argue rigorously, that the joint $(x,y)$ stationary PDE $\mathcal A p(x,y)=0$ drops the terms of $x$ partial derivatives once it is conditioned on $x$? I tried integrating the joint $(x,y)$ stationary PDE over $x$ the $x$ in the $y$ partial terms frustrated the effort.

Alan
Posts: 9214
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

### Re: Singular and degenerate elliptic partial differential equation

Yes, we now agree that neither term in what you just posted for p(x,y) is integrable. Beyond that, you are trying to jump ahead way too fast.

The next issue to come to agreement on is, when x is a constant, then what is $p(x,y)$, admitting distributions? Run some simulations (of the 1D process) for various constant x values (positive and negative), see what you get, make a precise conjecture, and I will say if I agree with it or not. (Clearly, it involves delta functions). That should be the next step.

lovenatalya
Topic Author
Posts: 152
Joined: December 10th, 2013, 5:54 pm

### Re: Singular and degenerate elliptic partial differential equation

From the previous solution of $p(x,y)$ for fixed $x$ we already know it is unintegrable for any $x$. Do we agree that the only choice is $p(x,y)=c_1(x)\delta(y)+c_2(x)\delta(1-y)$ for some positive $c_1(x)$ and $c_2(x)$? If so, are you asking to see what exactly $c_1$ and $c_2$ are? Do we really need to know their exact forms to proceed to the joint stationary PDE in the very first post?
Last edited by lovenatalya on December 13th, 2017, 1:43 am

Alan
Posts: 9214
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

### Re: Singular and degenerate elliptic partial differential equation

yes to all 3 questions.

We already have a final (conjectured) answer -- the question is how to prove it.

My suggestion is perhaps a conditioning argument will work -- i.e. you condition on the x(t) paths (in the original bivariate process). To get the conditioning argument to work, you need to know $p(y|x(\cdot))$, the univariate terminal density conditional on a deterministic path x(t) for the 1D y-process. This would be the terminal y-density for the process $dY_t = x(t) Y_t (1-Y_t) (dt + dB_t)$, where $x(\cdot)$ is some generic, real-valued, continuous function of time. I am trying to work us toward agreement on $p(y|x(\cdot))$. So the answer to all questions is yes.

lovenatalya
Topic Author
Posts: 152
Joined: December 10th, 2013, 5:54 pm

### Re: Singular and degenerate elliptic partial differential equation

@Alan et al:

I have some simulation results and theoretical explorations to be posted tomorrow.