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 $X_t$. =(