# what's the difference between RDE and SDE?

what's the difference between random differential equation and stochastic differential equation? does stochastic differential equations include random differential equation?

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never heard about RDE, could you please out a link on example of such notion? – Ilya Oct 13 '11 at 6:19
sciencedirect.com/science/book/9780126548501 Random Differential Equations in Science and Engineering, a book about random differential equation – Jim Oct 13 '11 at 6:39
They need 30 min in my library to find that book. I will take a look at it and then write you unless anybody other will answer. – Ilya Oct 13 '11 at 8:05
thank you so much – Jim Oct 13 '11 at 8:34
you are welcome – Ilya Oct 13 '11 at 10:13

In the book you've cited the RDE refers to the equation of the form $$\dot X = f(X,Y,t),\quad X(t_0) = X_0.\quad(\star)$$ where $X_0$ is a r.v. and $Y$ is a stochastic process.

There see Section 5.4, p. 134 where it is written:

Accordingly, we follow Sisky [Stochastic Differential Equation, 1967] and classify $(\star)$ into three basic types:

1. Random Initial Conditions.
2. Random inhomogeneous parts.
3. Random coefficients.

The first type is the most simple, i.e. it is just an ODE with random initial conditions, others are more complicated and refer to SDEs.

I may miss some points - but my impression is the following. The RDE $(\star)$ has to be formalized to have nice properties such as existence and uniqueness of the solution. It seems that it more refers to the model of some process which then have to be studied using probability theory.

One of the way to formalize such equation was given by Ito through the notion of Ito integral. In that case $f(x,y,t) = a(t,x)+b(t,x)y$ the process $Y$ assumed to be Gaussian white noise and $(\star)$ reduces to $$dX_t = a(t,X_t)\,dt+b(t,X_t)dB_t\quad (*)$$ where $B_t$ is a Brownian motion.

Some comments now: to be precise, $\dot{B_t}\neq Y_t$ if $Y_t$ is a white noise. Moreover, $\dot B_t$ does not exist at all if you refer it to the derivative w.r.t. time $t$. The derivative $\dot X_t$ doe not exist as well. Differentials in $(*)$ are not differentials like in analysis but rather Ito differentials and to be very precise, you should write $(*)$ in the integral form: $$X_t-X_0 = \int\limits_0^ta(s,X_s)\,ds+\int\limits_0^tb(s,X_s)dB_s\quad (**)$$ because only Ito integral has a precise definition, not Ito differential (correct me, please, if I'm wrong). That's why $(\star)$ is quite imprecise unless you say that $\dot X$ is not a usual derivative and define then what $\dot X$ mean in the concrete equation.

Ito integral is very popular, say, in financial mathematics. However, it is just one way to formalize $(\star)$. Another way was given by Stratonovich with Stratonovich integral. That is another kind of stochastic integral popular in physical problems like Langevin equation.

The process driven Ito integral can be extended from Brownian motion to more general semimartingales because it saves a lot of nice properties of such processes. As I understood, such an extension exists for Stratonovich integral as well though on wiki its formula is given through the Ito integral.

To finish this answer, I would say that SDE is a way to formalize RDE. I strongly recommend you to read the book by B. Oksendal to better understand these things. As far as I remember, there in the very first chapter he gives models for real-life processes in the form of RDE $(\star)$. However, he proves that the notion of 'derivative' should be changed in order to formalize these equations.

Edited: Ok, with regards to your comment. I would say that in SDE the notion of integral and differential is generalized rather than that of derivative. There are infinitesimal generators of Markov processes (as for Fokker-Planck equation in the book you've cited) but they refer to the path derivatives of functions of $X$, say $f(X_t)$.

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if there exists a s.p., denoted as $\dot{X}(t)$, such that $\lim\limits_{\tau\rightarrow 0}E(\frac{X(t+\tau)-X(t)}{\tau}-\dot{X}(t))^2=0$ then $\dot{X}(t)$ is called the derivative of $X(t)$ So I think the derivative in RDE has its own meaning, different from SDE. Is my opinion right? Can you explain this to me? – Jim Oct 13 '11 at 11:17
I've edited the answer. – Ilya Oct 13 '11 at 11:46
in Section 4.4.1, p.95, property(2) said that if the derivative exists then it's unique. and I think your example is not correct. – Jim Oct 13 '11 at 12:11
@Jim: sorry, of course - I've lost one term in my calculations. Let me fix it. – Ilya Oct 13 '11 at 12:32