Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I state my problem in a few lines then describe what I have already done.

I have a quite simple stochastic differential equation (SDE):

$dx=-2x \, dt+\sqrt{1-x^2} \, dW$ with $W$ a brownian.

I want to compute $\displaystyle{\lim_{t\to 0}}~\mathbb{E}\left[B_t\tanh\left(A_t\frac{x(t)-x(0)}{t}\right)|x(0)\right]$ and can't manage to do it.

I want to describe a given phenomenon obeying my SDE, thus the factors B_t and A_t will depend on $t$. This is to ensure that as I decrease the time increment $t$ through which I approximate my continuous phenomenom by a "discrete" growth rate, as one cannot differentiate a Brownian, I will converge towards a given value. It is equivalent to the normalisation having to be applied to a random walk diffusion coefficient when one wants to converge to the "underlying" brownian. EDIT : $A_t\propto \sqrt{t}$ and $B_t \propto \frac{1}{\sqrt{t}}

This is my problem, any suggestions are welcome, below i expand on where I am, and how I approach the problem :

Let $\phi(x(t),t)$ be a twice differentiable function, Ito's lemma yields that if $\phi(x(t),t)$ is solution of : $$(1)~\frac{\partial \phi }{\partial t}-2x\frac{\partial \phi }{\partial x}+\frac{1-x^2}{2}\frac{\partial^2 \phi}{\partial x^2}=0\textrm{, with }\phi(x,0)=\Phi(x(0))\textrm{ as initial condition,}$$ then $\phi(x(t),t)=E[\Phi(x(0))|x(t)=x]$.

Noting that the partial differential equation (1) can be rewritten as $\partial_t\phi=M[\phi]$ with $M[.]$ a linear operator, any function of the form : $$\phi(x,t)=f(x)+\displaystyle{\sum_{n=1}^\infty}\frac{t^n}{n!}M^n[f]$$ is a solution to the PDE (1), satisfying the initial condition $\phi(x,0)=f(x)$.

Taking $\Phi(x(0))=x(0)$, for example, and with $t>0$, yields $$\mathbb{E}[x(t)|x(0)]=x(0)e^{-2t}$$ thus $\displaystyle{\lim_{t\to 0}}~\mathbb{E}\left[\frac{x(t)-x(0)}{t}|x(0)\right]=-2x(0)$

My problem is that $B_t~tanh\left(A_t\frac{x(t)-x(0)}{t}\right)$ is undefined at $t=0$, thus I cannot use the same approach.

I tried to compute the characteristic function as i can compute all the moments, then Fourier-transform it to get the distribution, but it doesn't seem to give any meaningful result, which I guess comes from the fact that the distribution tends towards pathological functions (Dirac distributions) as $t\to 0$ thus I i am not allowed to switch the limit and the expectation operator anymore.

I can expand if need be, but i think I'll already be lucky if anyone read this far :).

Thanks in advance


I also tried to Taylor expand the $tanh$ (not bothering about radius of convergence to start with) apply the expectation operator, which I can compute when applied to any power of $x$. By $b_{2n}$ I denote the coefficient of the Taylor expansion of $tanh$. \begin{eqnarray} B_t\tanh\left(A_t\frac{x(t)-x(0)}{t}\right)&=&B_t\tanh\left(A_t\frac{\Delta x}{t}\right)\\ &=&B_t\displaystyle{\sum_{n=1}^\infty}b_{2n}\frac{A_t^{2n-1}}{t^{2n-1}}\Delta x^{2n-1}\\ &=&B_t\displaystyle{\sum_{n=1}^\infty}b_{2n}\frac{A_t^{2n-1}}{t^{2n-1}} \displaystyle{\sum_{k=0}^{2n-1}}(-1)^k\binom{2n-1}{k}x(0)^k x(t)^{2n-1-k} \end{eqnarray}

For simplicity I denote $x(0)=x$ and $x(t)=x_t$, and I apply the expectation operator : $$B_t\displaystyle{\sum_{n=1}^\infty}b_{2n}\frac{A_t^{2n-1}}{t^{2n-1}} \displaystyle{\sum_{k=0}^{2n-1}}(-1)^k\binom{2n-1}{k}x^k \mathbb{E}[x(t)^{2n-1-k}|x]$$ \begin{eqnarray} =&B\displaystyle{\sum_{n=1}^\infty}b_{2n}\frac{A_t^{2n-1}}{t^{2n-1}}\displaystyle{\sum_{k=0}^{2n-1}}(-1)^k\binom{2n-1}{k}x^k \displaystyle{\sum_{m=0}^{\infty}}\frac{t^m}{m!}M^m[x^{2n-1-k}]\\ \end{eqnarray} Moreover, computations give that : $$\forall(n,k)\in\mathbb{N^2},~M^n(x^k)=\displaystyle{\sum_{i=0}^{\lfloor k/2\rfloor}}\alpha_{i,n}(k)x^{2i+\delta_k}$$ with $\delta_k=0$ if $k$ is even, $\delta_k=1$ if not, $$\alpha_{i,n+1}(k)=\alpha_{i,n}(k)(2i^2+3i+\delta_k(2i+1))-\alpha_{i+1,n}(k)(2i^2+3i+\delta_k(2i+1))$$ and initial condition : $\alpha_{i,0}(k)=\delta_{i,\lfloor k/2 \rfloor}$, with $\delta_{i,j}$ Kronecker's symbol. We can rearrange the sum :

\begin{eqnarray} S&=&B\displaystyle{\sum_{n=1}^\infty}b_{2n}\frac{A_t^{2n-1}}{t^{2n-1}}\displaystyle{\sum_{m=0}^{\infty}}\frac{t^m}{m!}\displaystyle{\sum_{k=0}^{2n-1}}(-1)^k\binom{2n-1}{k}x^k \displaystyle{\sum_{i=0}^{\lfloor k/2\rfloor}}\alpha_{i,m}(2n-1-k)x^{2i+\delta_{k+1}}\\ &=&B\displaystyle{\sum_{n=1}^\infty}b_{2n}\frac{A_t^{2n-1}}{t^{2n-1}}\displaystyle{\sum_{m=0}^{\infty}}\frac{t^m}{m!}\displaystyle{\sum_{k=0}^{2n-1}}(-1)^k\binom{2n-1}{k}\displaystyle{\sum_{i=\lfloor k/2\rfloor}^{n-1}} x^{2i+1} \alpha_{i,m}(2n-1-k) \end{eqnarray}

In addition $x^{2i+1}\alpha_{i,m}(k)$ is a polynomial in $k$ of degree $d\leq 2m$, therefore its sum over $i$ has a maximal degree of $2m$ too. This implies that $$\forall m< n, ~\displaystyle{\sum_{k=0}^{2n-1}}(-1)^k\binom{2n-1}{k}\displaystyle{\sum_{i=\lfloor k/2\rfloor}^{n-1}} x^{2i+1} \alpha_{i,m}(2n-1-k)=0$$ We are left with : \begin{eqnarray} S&=&B_t\displaystyle{\sum_{n=1}^\infty}b_{2n}\frac{A_t^{2n-1}}{t^{2n-1}}\displaystyle{\sum_{m=n}^{\infty}}\frac{t^m}{m!}\displaystyle{\sum_{i=0}^{n-1}} x^{2i+1}\displaystyle{\sum_{k=0}^{2i+1}}(-1)^k\binom{2n-1}{k} \alpha_{i-\lfloor k/2\rfloor,m}(2n-1-k) \end{eqnarray}

If we take $A\propto t$ and $B\propto \frac{1}{t}$, as $t\to0$, only the term $n=1$ is kept, therefore we obtain a result linear in $x$, which is neglecting all the terms in $x$ coming from the term $M^n(x^k)$. If $A\propto \sqrt{t}$ and $B\propto \frac{1}{\sqrt{t}}$ we keep all the terms $m=n$. This expansion reduces to $$S=\displaystyle{\sum_{k=0}^{\infty}}(-1)^{k+1} a_k C^{2k+1}x(1-x^2)^k$$ where $\lfloor n/2 \rfloor ! \leq a_k \leq n!$ numerically (I calculated the first 100 terms...). Thus this expansion diverges.

This seems normal to me (now) because $tanh$ has a finite radius of convergence, and as $t\to 0$ when taking the expected value, $\frac{\Delta x}{\sqrt{t}}$ gets larger and larger for some values of $x_t$, those divergences are taken care of by the distribution of $x_t$ which converges to $0$, but the taylor expansion assumes that the argument of $tanh$ is bounded, which is not the case.

share|cite|improve this question
you must want the $|x(0)$ outside of the parens, also , I would like the $\frac 1t$ outside the parens beacause for small t (say $x(0) = 0$ $x(t)$ doesn't know it's not a brownian motion, so $\frac {x(t)} t$ is quite large. – mike Jan 22 '13 at 19:40
Thank you, I edited the typos. As to your suggestion, I agree, this corresponds to taking $A\propto t$ and $B\propto \frac{1}{t}$. – alexis Jan 23 '13 at 13:41
This modifies the specification of the problem, but not its computability so far as i can tell : the $\lim$ and the $\mathbb{E}$ operator still do not commute. – alexis Jan 23 '13 at 14:08
If you move the $\frac 1t$ outside you are computing the generator. With the $\frac 1t$ inside $tanh() \approx 1 $ or $-1$ depending on whether $x(t) > 0$ or not. That is because $\frac {x(t)} t $ is large. – mike Jan 23 '13 at 14:48
Could you expand/link as to what generator means in this context? As to $\frac{x(t)}{t}$ being large, actually I do not think it is a sufficient argument to consider $A\propto t$ and $B\propto \frac{1}{t}$. I might be wrong, but as the expectation is time dependent (i.e. the distribution of $x(t)$ depends on $t$ given the value of $x(0)$), the pathologies you describe could be "taken care of" by the distribution converging towards Dirac-like functions as $\frac{x(t)}{t}$ diverges, could they not? It seems to me that you implicitly switch the limit and the expectation, which I think is illicit – alexis Jan 23 '13 at 15:25

One could begin with the decomposition $$ \frac{x(t)-x(0)}t=a(t)x(0)+b(t)y(0)W(t)+b(t)z(t), $$ where $$ a(t)=\frac{\mathrm e^{-2t}-1}t,\quad b(t)=\frac{\mathrm e^{-2t}}t,\quad y(s)=\mathrm e^{2s}\sqrt{1-x(s)^2} $$ and $$ z(t)=\int_0^t(y(s)-y(0))\mathrm dW(s). $$ When $t\to0$, $a(t)\to-2$, $b(t)\sim1/t$, $W(t)=\sqrt{t}V$ where $V$ is standard normal, and $z(t)$ is centered normal with variance $O(t^2)$, hence $$ \frac{x(t)-x(0)}t=y(0)\frac1{\sqrt{t}}V+U(t), $$ where $U(t)$ is a family of random variables bounded in $L^2$. Now, for every function $u$ such that $|u(t)|\ll1/\sqrt{t}$ when $t\to0$, $$ \tanh\left(A\left(y(0)\frac1{\sqrt{t}}V+u(t)\right)\right)\to\mathrm{sign}(AV). $$ If made rigorous, all this would yield $$ \mathbb E\left(\tanh\left(A\frac{x(t)-x(0)}t\right)\right)\to\mathbb E\left(\mathrm{sign}(AV)\right)=0. $$ Edit: It appears that $A=A_t$ should depend on $t$, for example with $A_t\propto\sqrt{t}$. If $A_t=a\sqrt{t}+o(\sqrt{t})$, the above suggests that $$ \mathbb E\left(\tanh\left(A_t\frac{x(t)-x(0)}t\right)\right)\to\mathbb E\left(\tanh\left(a\sqrt{1-x(0)^2}V\right)\right)=0. $$ Edit 2: The decomposition we started with follows from Itô's formula since $$ \mathrm d(\mathrm e^{2t}x(t))=2\mathrm e^{2t}x(t)\mathrm dt+\mathrm e^{2t}\mathrm dx(t)=y(t)\mathrm dW(t), $$ hence $$ \mathrm e^{2t}x(t)=x(0)+\int_0^ty(s)\mathrm dW(s), $$ which is equivalent to $$ x(t)=\mathrm e^{-2t}x(0)+\mathrm e^{-2t}\int_0^ty(s)\mathrm dW(s), $$ that is, $$ x(t)-x(0)=(\mathrm e^{-2t}-1)x(0)+\mathrm e^{-2t}z(t)+\mathrm e^{-2t}y(0)W(t). $$

share|cite|improve this answer
Thank you for your answer. I can rewrite your decomposition as follows: $$\frac{x(t)-x(0)}{t}=\frac{E(x(t))-x(0)}{t}+\frac{E(x(t))}{x(0)t}\left(y(0)W(t)‌​+ \int_0^t(y(s)-y(0))dW(s)\right)$$ I do not understand where it comes from. Could you give me a track to follow? Taking this formula as given for now, I am not sure I agree with your result. Firstly, in your last line you intervert the limit and the expectation operator, which I don't think is legit as the dominated convergence theorem would require to be able to find a function g dominating the distribution f(x,t) of x at t. – alexis Feb 16 '13 at 13:32
Secondly, as I was telling Mike a few weeks ago, the A and B factors depend on t. The idea being that were I to approximate the stochastic process by a random walk of time step t I would like to converge towards the underlying stochastic process, so I have to consider $A\propto\sqrt{t}$. In this case, I guess that instead of converging towards $sign(AV)$ the tangent converges towards an indefinite quantity, hence my problem. – alexis Feb 16 '13 at 13:38
The decomposition is straightforward from Itô's formula. Regarding your "Firstly", the domination is obvious since $|\tanh|\leqslant1$ uniformly. – Did Feb 16 '13 at 13:39
Regarding your "Secondly", I did miss that $A$ and $B$ depend on $t$ (say, you should really write them $A_t$ and $B_t$, no?). Then the way $A_t$ depends on $t$ seems to be crucial--but I see nothing about that in your question. – Did Feb 16 '13 at 13:42
OK, I have a suggestion, and I would like you to consider it seriously: please spend more time trying to understand what is written and working on it before jumping to the conclusion that something must be wrong. (Anyway I cannot continue forever to debunk the issues you have with what I write which seem to come partly from the fact that you read too quickly.) For instance... if one writes down explicitely $U(t)$ (which I omitted since I thought my post was already pretty explicit), one sees indeed a contribution $-2x(0)$. This is $O(1)$ hence $\ll1/\sqrt{t}$. .../... – Did Feb 17 '13 at 7:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.