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At time $t$, there are a measure of particle whose distribution is described by the p.d.f. $g(x,t)$. Each particle follows a geometric Brownian motion with zero drift, $\frac{dx^{i}}{x^{i}}=\sigma dB^{i}$, where $dB^i$ is independent across $i$. By the law of large number, the diffusion terms cancel each other out in the distribution. Their distribution then follows the Kolmogorov Forward Equation (KFE): $$\frac{d}{dt}g(x,t)=\frac{d^{2}}{dx^{2}}\left(g(x,t)\frac{1}{2}σ^{2}x^{2}\right)$$

Denote $X_t=\mathbb \int x^i_t di = \int g(x,t)xdx$ to be the mean of the particles. How does $X_t$ evolve?

An intuitive but probably not really rigorous derivation would be:

$$\begin{aligned}dX_{t} &=d\int x_{t}^{i}di\\ &=\int dx_{t}^{i}di\\ &=\int\left(\sigma x_{i}dB^{i}\right)di\\ &=0\\ \end{aligned} $$

The last step uses the argument that "the diffusion terms cancel each other out", but it doesn't seem very rigorous to me. What's the $di$ measure and can we interchange its integral with the derivative?

Another approach is to use the Kolmogorov forward equation: $$\begin{aligned}\frac{dX_{t}}{dt} =\frac{1}{2}\sigma^{2}\int_{0}^{\infty}x\frac{d^{2}}{dx^{2}}\left(g(x,t)x^{2}\right)dx\\ =\frac{1}{2}\sigma^{2}\int_{0}^{\infty}2g(x,t)x+4g_{x}(x,t)x^{2}+x^{3}g_{xx}(x,t)dx\\ =\frac{1}{2}\sigma^{2}\left(x^{2}g(x,t)+x^{3}g_{x}(x,t)\right)|_{0}^{\infty}\end{aligned}$$

Now we are left with a bunch of higher-order terms and limits. Assume the first moment exists for $g(x,t)$, is the integral always zero, or will it eventually be distribution-dependent?

P.S. the derivation using the $di$ measure proves a stronger property than the KFE approach above. It essentially says for any diffusion processes, not only the geometric brownian motion, they will cancel each other out for the first moment. Is there any way to establish that as well using KFE? i.e.

$$\int\frac{d^{2}}{dx^{2}}\left(g(x,t)σ^{2}(x)\right)dx=0$$

Zhiyu Fu
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  • This $\int x^{i}{t},di$ thing looks a bit weird to me. Maybe write instead $\frac{1}{N}\sum{i} x^{i}_{t}$ with the understanding that $N\to\infty$? – Ali Aug 28 '21 at 00:01
  • yeah I also don't understand it very well...but economists use it all the time – Zhiyu Fu Aug 28 '21 at 02:46
  • though now I believe (but fail to prove) with some regularity conditions, the last integral is always evaluated to be zero. – Zhiyu Fu Aug 28 '21 at 02:50
  • Sorry I don't think I can provide any actual help. A minor niggle though: I think there's an $x$ missing in each of your $\int \frac{d^{2}}{dx^{2}}\cdots$ expressions, i.e. should be $\int x\frac{d^{2}}{dx^{2}}\cdots$. – Ali Aug 28 '21 at 06:01
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    The formulas your are juggeling around with are neither rigours nor intuitive, there are simply wrong. It's nonsense. Besides, the mean of geometric BM is an old hat, and you could have found solutions to your questions immediately without any problem using google. – Tobsn Aug 28 '21 at 08:08
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    @Tobsn That is clearly unfriendly language according to the code of conduct (the very first example given!), and to a new user. Moreover you are wrong, it is not nonsense. We have already established that economists are responsible for much of the strangeness. – Ali Aug 28 '21 at 23:11
  • @Ali It's admittedly harsh, but I don't see where it's unfriendly. Besides, personally I think an unfriendly but honest feedback is better than a friendly but dishonest one. And yes I've seen the comment 'economists use it all the time'. But we've not been given any reference so far. Apart, even if some economists were using it, that's not making it less nonsensical. – Tobsn Aug 29 '21 at 07:47
  • @Tobsn "the mean of geometric BM is an old hat" true, but "the mean of a population of independent BM" is not easily found using google. I wouldn't have written a long question waiting indefinitely for answers or harsh comments if I haven't spent an afternoon googling. Or let me reply in the same straightforward and honest way: You can make more contribution to the community if you don't assume people haven't done their homework. – Zhiyu Fu Aug 31 '21 at 21:41
  • @Ali thanks for pointing it out! corrected. – Zhiyu Fu Aug 31 '21 at 21:48
  • We still got no reference. – Tobsn Sep 01 '21 at 20:40
  • https://www.bis.org/publ/work939.htm Section 3.1 paragraph 2 by Ricardo Reis, a well-respected macroeconomist at LSE. – Zhiyu Fu Dec 27 '21 at 20:06

1 Answers1

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I will just focus on the quantity

$$X_{t}:=\int_{[0,1]}x^{i}_{t}di,$$

(the [0,1] is the choice from "The constraint on public debt when r < g but g < m" as mentioned in the comments).

First, we have to make sense of this integral. We start with the discrete Riemann-integral

$$X^{N}_{t}:=\frac{1}{N}\sum_{i=1}^{N}x_{t}^{i},$$

where $x_{t}^{i}$ are N-iid geometric Brownian motions with $x_{0}^{i}=1$ and so $X^{N}_{0}=1$. By taking Ito derivative, we have

$$dX^{N}_{t}:=\frac{1}{N}\sum_{i=1}^{N}dx_{t}^{i}=\frac{1}{N}\sum_{i=1}^{N}\sigma x_{t}^{i}dB_{t}^{i},$$

$$X^{N}_{t}:=X^{N}_{0}+\frac{1}{N}\sum_{i=1}^{N}\left(\int^{t}_{0}\sigma x_{s}^{i}dB_{s}^{i}\right).$$

The Ito integral term has zero expectation as explained here Dealing with a term coming from Ito formula.

A sufficient condition for the integral $\int_0^t f(\omega, s)\, dB_s$ to be a martingale on $[0,T]$ is that

  1. $f(\omega,s)$ is adapted, measurable in s, and
  2. $\mathbb{E}\left(\int_0^T f^2(\omega,s)\,ds\right) < \infty$. In this case, indeed, $\mathsf{E} \left(\int_0^T f(\omega,s)\, dB_s\right)=0$.

Therefore, $E[X^{N}_{t}]=1$. So by the strong law of large numbers

$$X^{N}_{t}\to 1~~a.s..$$

Thomas Kojar
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