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I quote a paper from Delbaen and Shirakawa (2002). I will write in italics my observations/questions.

Starting from a stochastic differential equation of the form:

$$dr_t=\alpha\left(r_{\mu}-r_t\right)dt+\beta\sqrt{\left(r_t-r_m\right)\left(r_M-r_t\right)}dW_t\tag{1}$$ with $\left\{W_t\right\}_{t\geq0}$ a standard Wiener process in the filtered probability space $\left(\Omega,\mathcal{F},\left\{\mathcal{F}_n\right\},\mathbb{P}\right)$. We assume $\alpha,\beta>0$ and $r_m<r_{\mu}<r_M$, which guarantee the existence of stationary distribution.

(1. Why do the assumption $\alpha,\beta>0$ and $r_m<r_{\mu}<r_M$ guarantee stationarity of distribution? What is exactly meant here?)

$\alpha$ represents the speed of reversion to the longrun mean $r_{\mu}$.
Then, for diffusion of $(1)$, set $\sigma(x)=\beta\sqrt{\left(r_t-r_m\right)\left(r_M-r_t\right)}$. For any $x,y\in[r_m,r_M]$, it holds that: $$|\sigma^2(x)-\sigma^2(y)|=\beta|r_m+r_M-(x+y)||x-y|\leq\beta(r_M-r_m)|x-y|\tag{2}$$ This means that the diffusion coefficient function $\sigma(x)$ is Holder $\frac{1}{2}$ continous.

(2. Why does $(2)$ translate into the fact that $\sigma(x)$ is Holder $\frac{1}{2}$ continous? Doesn't $(2)$ denote instead that $\sigma^2(x)$ is Holder continous (with coefficient $1$)?)

Since $\mu(x)=\alpha(r_{\mu}-x)$ is Lipschitz continuous, the pathwise uniqueness of the stochastic differential equation is guaranteed by the general uniqueness theorem.

(3. Could you please give me some reference for such a theorem? In this specific case, does this theorem rely on Lipschitz condition on $\mu(x)$ combined with Holder continuity of $\sigma(x)$? If so, why do such two conditions guarantee the uniqueness of the path of SDE $(1)$?)

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    $\begingroup$ Can you give us a link or a complete reference to the paper? Also, you seem to have misspelled the name of the first author. $\endgroup$ Nov 11, 2020 at 19:41
  • $\begingroup$ Of course. I have just edited the question @IosifPinelis $\endgroup$ Nov 11, 2020 at 20:10

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Concerning(3): If I am not mistaken, this type of condition is sometimes referred to as Yamada-Watanabe condition for pathwise uniqueness. Your particular case is just a special case of Thm 2.1 in these lecture notes: https://fam.tuwien.ac.at/~schmock/notes/Yamada-Watanabe.pdf

EDIT: If the drift coefficient $b$ is Lipschitz continuous and the diffusion coefficient $\sigma$ is $\frac{1}{2}$-Hölder, then they satisfy an inequality like $$ |b(x)-b(y)| + |\sigma(x)-\sigma(y)|^2 \leq K |x-y| $$ for some $K>0$. Now you can use this to show pathwise uniqueness for the SDE $$ dX_t = b(X_t)dt + \sigma(X_t)dB_t. $$ Suppose you have two strong solutions $X^1$ and $X^2$ for this SDE. Then you can apply Itô's formula to the process $\Delta_t:= X^1_t -X^2_t$ and the function $$ \phi_{\varepsilon}(x) = (\varepsilon + x^2)^{\frac{1}{2}}. $$ If you then take the limit $\varepsilon \to 0$ you can show via Gronwall's inequality that indeed $\mathbb{E}[|X^1_t - X^2_t|] = 0$ for all $t\geq 0$. The strategy for the proof in the more general case in the lecture notes I linked above is essentially the same, just a bit more technical.

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  • $\begingroup$ First of all, thank you a lot. Starting from the reference you give (TH. 2.1 in those lecture notes), do you think that in the paper from Delbaen and Shirakawa (2002) the part concerning the Osgood conditions (to be met by $\rho$ and $G$) are simply skipped/taken for granted? $\endgroup$ Nov 12, 2020 at 8:25
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    $\begingroup$ The case of a Lipschitz continuous drift coefficient and a 1/2-Hölder continuous diffusion coefficient is very common, so showing that the Osgood conditions hold is skipped most of the time and also not too hard to prove. You can also derive the uniqueness more elementary in your situation, I have added some information to my post. $\endgroup$
    – jonask
    Nov 12, 2020 at 8:54
  • $\begingroup$ I cannot see the info you have added yet. What would be the other way (more elementary) to show uniqueness? OK, so in general, also in my it would not be that hard to show Osgood condition fulfillment $\endgroup$ Nov 12, 2020 at 9:05
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    $\begingroup$ Took a bit longer to add the addtional information, but now it should be there $\endgroup$
    – jonask
    Nov 12, 2020 at 9:11
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    $\begingroup$ I cannot yet comment because my reputation is too low, but I think the reference given in the paper ("Stochastic processes with applications" by B. Bhattacharya and E. Waymire is quite good. I think they explain the link between hitting probabilities and the associated PDE boundary value problem quite well. Maybe it helps if you think about the hitting probabilities for a Brownian motion first because there you can use some elementary calculations to derive it and compare your results to Proposition 9.1 on page 415 to get some intuition. $\endgroup$
    – jonask
    Nov 12, 2020 at 9:54
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Your equation (1) is a Jacobi diffusion equation. In particular, it matches equation (2.1) in this paper if you take there $X_t=r_t$, $\sigma=\beta$, $\beta=\alpha$, $m=(r_M+r_m)/2$, $z=(r_M-r_m)/2$, and $\gamma=\dfrac{r_\mu-m}z$. As explained in the linked paper, if the distribution of $X_0$ (that is, of your $r_0$) is the shifted and rescaled Beta-distribution on the interval $(m−z,m+z)=(r_m,r_M)$ given by formula (2.2) in the linked paper, then the distribution of the diffusion process is stationary. This answers your questions 1 and 3.

Concerning your question 2, for $C:=\beta(r_M-r_m)$ you have $$|\sigma(x)-\sigma(y)| =\frac{|\sigma^2(x)-\sigma^2(y)|}{\sigma(x)+\sigma(y)} \\ \le\min\Big(\frac{C|x-y|}{\sigma(x)+\sigma(y)},\sigma(x)+\sigma(y)\Big) \\ \le\sqrt{C|x-y|},\tag{*}$$ so that $\sigma$ is Hölder $1/2$-continuous. (The latter displayed inequality is a special case of the inequality $\min(\frac cu,u)\le\sqrt c$ for real $c,u>0$.)

Responses to the comments by the OP:

in Delbaen and Shirakawa (2002) it is not specified whether r0 follows a beta-distribution, but it is simply stated that "$\alpha,\beta>0$ and $r_m<r_\mu<r_M$ guarantee the existence of stationary distribution."

The logic here is quite simple: We can choose the distribution of $r_0$ to be the shifted and rescaled Beta-distribution on the interval $(m−z,m+z)=(r_m,r_M)$ given by formula (2.2) in the linked paper, and then the distribution of the diffusion process will be stationary. So, a stationary distribution exists.

As for the pathwise uniqueness, it was addressed in answer by jonask.


Concerning the answer to question 2: Inequalities $|\sigma^2(x)-\sigma^2(y)|\le\sigma(x)+\sigma(y)$ and even $\sigma^2(x)-\sigma^2(y)\le\sigma(x)+\sigma(y)$ will be of course false in general, and I stated nothing of that sort. Rather, part of what display (*) says is that $$\frac{|\sigma^2(x)-\sigma^2(y)|}{\sigma(x)+\sigma(y)}\le\sigma(x)+\sigma(y),$$ and this is very simple algebra.

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  • $\begingroup$ [First part] First of all, thank you a lot for all your time and precious help. $$$$ As to answer to question 1.: in Delbaen and Shirakawa (2002) it is not specified whether r0 follows a beta-distribution, but it is simply stated that "$\alpha,\beta>0$ and $r_m<r_{\mu}<r_M$ guarantee the existence of stationary distribution". Why is that true? Additionally, why does this answer to question $(3)$ about uniqueness as well? $\endgroup$ Nov 12, 2020 at 9:26
  • $\begingroup$ [Second part] As to answer to question 2.: could you please help me understand why the inequality: $$\frac{|\sigma^2(x)-\sigma^2(y)|}{\sigma(x)+\sigma(y)} \le\min\Big(\frac{C|x-y|}{\sigma(x)+\sigma(y)},\sigma(x)+\sigma(y)\Big)\tag{1}$$ hold true? I guess that, given that $|\sigma^2(x)-\sigma^2(y)|\leq C|x-y|$ (simple algebra. I would define $C=\beta^{\color{red}{2}}(r_M-r_m)$), one can show as well that $\sigma^2(x)-\sigma^2(y)\leq\sigma(x)+\sigma(y)$, hence explaining inequality $(1)$ above. But how is that possible to show that $|\sigma^2(x)-\sigma^2(y)|\leq\sigma(x)+\sigma(y)$? $\endgroup$ Nov 12, 2020 at 9:26
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    $\begingroup$ @Strictly_increasing : In my answer, I have added responses to your comments. $\endgroup$ Nov 12, 2020 at 15:33

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