Stochastic Differential Equation using Ito's Lemma

[cdbsmith]

I am new to SDE, and especially Ito's Lemma. I have a question that I simply cannot answer. It attached.

Can someone please help?

 

[Euge]

I am new to SDE, and especially Ito's Lemma. I have a question that I simply cannot answer. It attached.

Can someone please help?

Hi cdbsmith,

Let $u = \log y $, $\mu_t = \alpha - \beta u$, and $\sigma_t = \delta$. Then $du = \mu_t dt + \sigma_t dX_t$. Let $g(u) = e^u$. We have by Ito's lemma

$\displaystyle dg = \left(g'(u)\mu_t + \frac{g"(u)}{2}\sigma_t^2\right) dt + g'(u)\sigma_t\, dX_t $,

$\displaystyle dg = \left[e^u(\alpha -\beta u) + \frac{e^u}{2}\delta^2\right] dt + e^u\delta\, dX_t$,

$\displaystyle \frac{d(e^u)}{e^u} =(\alpha - \beta u + \frac{1}{2}\delta^2) dt + \delta\, dX_t$,

$\displaystyle \frac{dy}{y} = (\alpha - \beta\log y + \frac{1}{2}\delta^2) dt + \delta\, dX_t$.

 

[cdbsmith]

Hi cdbsmith,

Let $u = \log y $, $\mu_t = \alpha - \beta u$, and $\sigma_t = \delta$. Then $du = \mu_t dt + \sigma_t dX_t$. Let $g(u) = e^u$. We have by Ito's lemma

$\displaystyle dg = \left(g'(u)\mu_t + \frac{g"(u)}{2}\sigma_t^2\right) dt + g'(u)\sigma_t\, dX_t $,

$\displaystyle dg = \left[e^u(\alpha -\beta u) + \frac{e^u}{2}\delta^2\right] dt + e^u\delta\, dX_t$,

$\displaystyle \frac{d(e^u)}{e^u} =(\alpha - \beta u + \frac{1}{2}\delta^2) dt + \delta\, dX_t$,

$\displaystyle \frac{dy}{y} = (\alpha - \beta\log y + \frac{1}{2}\delta^2) dt + \delta\, dX_t$.

Thanks, Euge!

But, can you explain to me the steps? Ito's Lemma is confusing for me and I'm having a hard time understanding it.

Thanks again!

 

[Euge]

Thanks, Euge!

But, can you explain to me the steps? Ito's Lemma is confusing for me and I'm having a hard time understanding it.

Thanks again!

Sure, let's start with this. Suppose you have a process $dY_t = \mu_t dt + \sigma_t\, dX_t$ where $X_t$ is a Brownian motion. If $T > 0$ and $f(t, x)$ is a function that is in $C^{1,2}_{t, x}([0, T] ;(0,\infty))$, that is, continuously differentiable with respect to $t$ on $[0, T]$ and continuously twice-differentiable with respect to $x$ on $\Bbb (0,\infty)$, then $f(t, Y_t)$ satisfies the SDE

$\displaystyle df(t, Y_t) = \left(f_t + \mu_t f_x + \frac{\sigma_t^2}{2} f_{xx}\right) dt + \sigma_t f_x \, dX_t$.

This is a version of Ito's lemma which is applicable to several SDE. Now there are more general versions of the lemma which deal with cases where $X_t$ is a semi-martingale, but for your problem the above formula will do.

In your SDE, I let $u = \log y$ so that $y = e^u := g(u)$. Then I can find $dy$ by using Ito's formula with $g$. Note that $g$ is independent of $t$, so $g_t = 0$. That's how I got

$\displaystyle dg = \left(g'(u)\mu_t + \frac{g''(u)}{2}\sigma_t^2\right) dt + g'(u)\sigma_t$.

I hope this helps.

 

[blue_raver22]

Hello there,

I understand that you are new to Stochastic Differential Equations (SDE) and Ito's Lemma. I would be happy to provide some clarification for you.

SDE is a type of differential equation that involves randomness or uncertainty. It is often used in mathematical models to describe systems that are subject to random fluctuations, such as stock prices or weather patterns.

Ito's Lemma is a mathematical tool used to solve SDEs. It allows us to calculate the expected value of a function of a stochastic process, which is a key component in solving SDEs.

Now, for your question, I would need more context to provide a specific answer. However, I can explain the general process of using Ito's Lemma to solve an SDE.

First, we need to define the stochastic process in question, which is typically denoted by dX(t). Then, we apply Ito's Lemma to find the differential of the function we are interested in, denoted by dF(X(t)). This differential can then be used to solve the SDE.

I hope this helps to clarify your understanding of SDE and Ito's Lemma. If you have any further questions, please feel free to ask. Good luck with your studies!