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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 $X_t$. =(

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1 Answer 1

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  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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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

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