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I'm trying to solve stochastic differential equation: $$d X(t)=X(t) d t+d W(t)$$

Here, W(t) denotes a Brownian Motion. Comparing with Itô's lemma (with $f\left(W_{t}, t\right) = X(t)$), $$ d X(t)=\partial_{w} X(t) d W_{t}+\frac{1}{2} \partial_{w}^{2} X(t) d t+\partial_{t} X(t) d t $$ I get two equations

a)

$$ \partial_{w} X(t) = 1 $$ $$ \partial_{w} X(t) = 1 \Rightarrow X(t)= W + C(t) $$

b) $$ \frac{1}{2} \partial_{w}^{2} X(t) +\partial_{t} X(t) = X(t) $$ substituting result from part (a) we have $$ \frac{1}{2} \partial_{w}^{2} (W + C(t)) +\partial_{t} (W + C(t)) = W + C(t) $$ $$ C'(t) = W + C(t) $$ which can be solved as $$ C(t) = C_0e^t-W $$ The issue starts when I try to substitute this back $$ X(t)= W + C_0e^t-W $$ $$ X(0)= C_0e^0 $$ $$ X_0= C_0 $$ $$ X(t)= X_0e^t $$ Now equation (a) does not hold anymore, because $$\partial_{w} X(t)=\partial_{w} X_{0} e^{t}= 0$$ Anyone can point me what am I missing or what I'm solving wrong? Thanks.

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After your clarifying comment, I see the following problem with your approach: You assume that the solution can be written in the form $f(t,W_t)$, but for this particular SDE this is not the case. So you need to use a slightly different method.

You can still use the approach of matching coefficients, but you should start with $$X_t = \alpha(t)Y(t)=\alpha(t)\left[X_0 + \int_0^t \beta(s)dW_s\right],$$ where $\alpha$ and $\beta$ are deterministic functions of time.

Now, applying Ito's Lemma we obtain $$ dX_t=\alpha'(t)Y(t)dt+\alpha(t)\beta(t)dW(t)$$

Matching the coefficients we have $$ \alpha'(t)Y(y)=X(t) \\ \alpha(t)\beta(t)=1$$ From the first equation, we conclude that $\alpha(t)=e^t$, which implies that $\beta(t)=e^{-t}$. Thus, $X_t$ can be written as $$ X_t = e^t X_0 + e^t \int_0^t e^{-s}dW_s$$ As you can see X_t is not in the form of $f(t,W_t)$.


Alternatively, this equation can be solved by applying Ito's Lemma to $e^{-t}X_t$, where $f(t,x)=e^{-t}x$. Then write the resulting equation in the integral form and multiply both sides of it by $e^t$.

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  • $\begingroup$ I would disagree with the first part of your answer, you have just used altered notations. Itô's lemma says: $$ d f\left(W_{t}, t\right)=\partial_{w} f\left(W_{t}, t\right) d W_{t}+\frac{1}{2} \partial_{w}^{2} f\left(W_{t}, t\right) d t+\partial_{t} f\left(W_{t}, t\right) d t $$ and $f\left(W_{t}, t\right) = X(t)$. I assume you mean I should write $X(W_t, t)$, but that shouldn't be an issue and all the textbooks that I see write it in the same manner. $\endgroup$ – Blade Apr 24 at 2:29
  • $\begingroup$ I updated the answer based on your clarification. But to be honest, I have never seen notation $\partial_w X_t$ used for $\partial_w f(W_t,t)$ in this context. $\endgroup$ – Mdoc Apr 24 at 5:17
  • $\begingroup$ Because it is a wrong notation if $\partial_\omega$ stands for $\dfrac{\partial}{\partial W_t}$. That's why Ito Lemma exists. $\endgroup$ – QFi Apr 24 at 6:08
  • $\begingroup$ @QFi You can find same notation used in this document from some lecture at NYU: math.nyu.edu/faculty/goodman/teaching/StochCalc2012/notes/… look at eq.3 $\endgroup$ – Blade Apr 24 at 14:36
  • $\begingroup$ @Mdoc Thanks for the update, is $$ X_{t}=\alpha(t) Y(t)=\alpha(t)\left[X_{0}+\int_{0}^{t} \beta(s) d W_{s}\right] $$ a general rule or formula based off of the form of SDE? $\endgroup$ – Blade Apr 25 at 1:22

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