1) I want to know the mechanism how to: show that the process $X_t$ solves this SDE 2) know if my friends though, mine though below are correct/incorrect.

I have the general linear stochastic diferential equation (SDE) with initial condition $X_0$, constants: $c,\sigma$ , I need to show that the process $X_t$ solves this SDE.

SDE:$$dX_t = c X_t dt + \sigma dW_t , t \in [0,T]$$ Stoch. Process:$$X_t = X_0 e^{ct}+\sigma e^{ct} \int_0^t e^{-cs}dW_s.$$

From simple logic, I think I need to insert $X_t$ to SDE( take differential of $X_t$ and receive exactly SDE I have above). Is this mathematically correct way of showing that some solution solve SDE(or simple DE)?

My friend found in the web completely awkward solution to me:

1) Take differential by ?product formula?: $$d(e^{-ct}X_t) = -c e^{-ct} X_t + e^{-ct} dX_t$$ 2) Substitute our SDE $dX_t = c X_t dt + \sigma dW_t , t \in [0,T]$ to the second part of equation.

Finally integrate, and you get what you want: $$X_t = ...$$

My question: I think it is not correct proof, moreover, how come can I understand what should I differentiate ( I mean $d(e^{-ct}X_t)$ is not obvious). Is it? I think it somehow diminished proof.

I think that real proof is:

1) by ITOs formula: $$df(t,W_t) = \dot f_t dt + \dot f_{W_t} dW_t + \frac{1}{2}\ddot f_{W_t,W_t} dt,$$ get $$d X_t = ..$$ 2) see that it looks like $dX_t = c X_t dt + \sigma dW_t , t \in [0,T]$.

Here is how far I got, and what obstacles I have:

If I assume I can put $\frac{d}{dt}$ inside of integral and easily differentiate exponent $e^{-cs}$, I get:

$$dX_t = \frac{\partial }{\partial t} \left(X_0 e^{ct}+\sigma e^{ct} \int_0^t e^{-cs}dW_s \right) dt + \frac{\partial }{\partial W_t} \left(X_0 e^{ct}+\sigma e^{ct} \int_0^t e^{-cs}dW_s \right) dW_t + \frac{1}{2}\frac{\partial }{\partial W_t} \frac{\partial }{\partial W_t} \left(X_0 e^{ct}+\sigma e^{ct} \int_0^t e^{-cs}dW_s \right) dt, $$ or $$dX_t = A dt + B dW_t + \frac{1}{2}C dt,$$ where:

$$ A= X_0 c e^{ct} + \left ( \frac{\partial }{\partial t} \sigma e^{ct} \int_0^t e^{-cs}dW_s \right) =$$

$$= X_0 c e^{ct} + \left ( \sigma ce^{ct} \int_0^t e^{-cs}dW_s + \sigma e^{ct} \int_0^t -c e^{-cs}dW_s \right).$$

Is it correct? Next,

$$B = 0 + \text{this is not easy for me to digest} = $$ $$ = 0 + \sigma e^{ct} e^{-ct} = \sigma$$

$$C = 0 + 0 \text{, because there is no } W_t$$ finally: $$dX_t = \left ( X_0 c e^{ct} + \sigma ce^{ct} \int_0^t e^{-cs}dW_s + \sigma e^{ct} \int_0^t -c e^{-cs}dW_s \right) dt + \sigma dW_t + \frac{1}{2}0$$ $$dX_t = X_0 c e^{ct} dt+ \sigma dW_t $$ But here we see $X_0e^{ct}$, not $X_t$. =(


1 Answer 1

  1. Your friends solution is correct. If $(X_t)_{t \geq 0}$ is a one-dimensional Itô process, then Itô's formula states $$ df(t,X_t)= \partial_x f(t,X_t) \, dX_t + \left(\frac{1}{2} \partial_x^2 f(t,X_t) \right) d\langle X \rangle_t + \partial_t f(t,X_t) \, dt. \tag{1}$$ Your friend used this identity for $f(t,x) := x e^{-ct}$.

  2. Your attempt:

$\frac{\partial}{\partial W_t} \left( X_0 e^{ct} + \sigma e^{ct} \int_0^t e^{-cs} \, dW_s \right)$

It this how you are taught to write down Itô's formula? In my oppinion, that's not a good way to write it this way. The problem is that you cannot apply Itô's formula this way. Itô's formula gives you the differential for $f(t,W_t)$ for (nice) functions $f$. But here, you want to calculate the differential of the expression

$$\int_0^t e^{-cs} \, dW_s,$$

i.e. we need a function $f$ such that

$$f(t,W_t) \stackrel{!??!}{=} \int_0^t e^{-cs} \, dW_s.$$

... tell me: How do you choose $f$? Before you have not chosen such a function $f$, you cannot apply Itô's formula this way. What you are doing is treating it as a constant and that's simply not correct.

In order to solve this SDE (or check that the given process is a solution to the SDE) you really have to use Itô's formula for Itô proceses, i.e. $(1)$.

Remark The solution your friend suggested applies Itô's formula to the process

$$e^{-ct} X_t \tag{1}$$

and, at the first glance, it is not obvious how to come up with this particular process. The idea is the following: Instead of considering the SDE

$$dX_t = c X_t \, dt + \sigma \, dW_t$$

we consider the corresponding ODE

$$dx_t = cx_t \, dt$$

(i.e. we just we leave away the stochastic part). It is well-known that the unique solution to this ordinary differential equation is given by

$$x_t = C e^{ct}$$

where $C \in \mathbb{R}$. So far, $C$ is some "deterministic" constant. Now, however, we return to our stochastic setting and allow $C$ to depend on $\omega$ (this is the counterpart of the variation of constants-approach for SDEs). So, by the previous identity, our new auxilary process $C$ is given by

$$C_t = e^{-ct} X_t$$

... and this is exactly the process from $(1)$.

There are a lot of examples where this approach [i.e. first solve the corresponding ODE and then make a "stochastic" variation of constants] works, ee e.g. this question. However, I don't know any statements for which types of SDEs this approach works and for which it doesn't.

  • $\begingroup$ 2 more questions: 1) How should I find these $f(t,x)$ for different problems? Is it some kind of guessing? 2) What is the exact name of this technic: when we firstly write a stochastic process in implicit way with constant( in our case $e^{-ct}$), then obtain known stochastic process using SDE. Is it a method/some rule? Remark: Do we have something similar in ordinary calculus while solving ODEs? $\endgroup$
    – Ievgenii
    Nov 28, 2015 at 9:22
  • 1
    $\begingroup$ @Ievgenii 1. Rule of thumb: Either it is obvious how to choose it (e.g $e^{t} W_t$ or $\sin(W_t)$ or ....) or there doesn't exist such a function $f$. Usually, the second case occurs if the process is a function of the sample path $\{W_s; s \leq t\}$; for example if we cannot expect that there exists a nice function $f$ such that $$f(t,W_t) = \int_0^t W_s \,ds$$ because the right-hand side depends not only on $W_t$, but also on $\{W_s; s \leq t\}$. 2. See my edited answer. $\endgroup$
    – saz
    Nov 28, 2015 at 10:25
  • $\begingroup$ thank you for really nice explanation. As far as I can see reading other posts here, you are one of few who have knowledge in Stochastics, could you please recommend something except Shreve ? ( I am already renting this book from library) $\endgroup$
    – Ievgenii
    Nov 28, 2015 at 11:04
  • 1
    $\begingroup$ @Ievgenii I personally like the book by René Schilling very much (Brownian Motion - An Introduction to Stochastic Processes - René Schilling & Lothar Partzsch); it contains a lot of material on Brownian motion, but also on stochastic integration and SDEs. $\endgroup$
    – saz
    Nov 28, 2015 at 11:15
  • $\begingroup$ yeah, I read this somewhere in your posts, and checked - it is not in the library. =) Thank you $\endgroup$
    – Ievgenii
    Nov 28, 2015 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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