I'm getting into asset pricing and was looking at Ito's Lemma, but cannot understand a few steps that are given.

Ito's Lemma states that given

$$dx_t = \mu dt + \sigma dz_t \\ y_t = f(t, x_t)$$


$$(1) \quad dy_t = \frac{\partial f}{\partial t} dt + \frac{\partial f}{\partial x} dx_t + \frac{1}{2} \frac{\partial^2 f}{\partial x^2} dx^2_t \\$$

I understand this part using chain rule and a second order Taylor expansion, of the second equation. I don't understand why then the following holds:

$$(*) \quad dy_t = \left[\frac{\partial f}{\partial t} + \frac{\partial f}{\partial x} \mu + \frac{1}{2} \frac{\partial^2 f}{\partial x^2} \sigma^2 \right] dt + \left[ \frac{\partial f}{\partial x} \sigma \right] dz_t \\$$

When I substitute in $dx_t$ into $(1)$ and use the fact that $dz^2_t = dt$, it isn't enough to arrive at $(*)$. I think $dx^2_t$ can literally be interpreted as $(dx_t)^2$, but if there is a better way of handling that term, any guidance there would be appreciated.

  • 2
    $\begingroup$ I suggest a powerful reading in Hull's "Options and Derivatives" where you can find an interesting appendix to prove Ito's lemma $\endgroup$ – Alexis L. Jul 1 '16 at 15:02
  • $\begingroup$ Strange that no-one answered this for forever. On math.stackexchange, I'm sure you could have gotten help quickly! This isn't economics specific. $\endgroup$ – Matthew Gunn Oct 7 '16 at 2:18

$$dx_t = \mu dt + \sigma dz_t \\ y_t = f(t, x_t)$$

A key idea here is that $\left( dx_t \right)^2=\left( \ldots \right)dt^2 + \left(\ldots\right) dzdt + \sigma^2 dz_t^2 = \sigma^2 dt$. The loose reasoning is that $\left( dz_t\right)^2 = dt$ and all the other terms (i.e. $dt^2$ and $dz\, dt$) are infinitely smaller than $dt$.

Horribly loose intuition for $dz_t^2 = dt$ is that $dz_t$ is normally distributed with variance $dt$, and hence the expectation of the square of $dz_t$ is $dt$.

In any case, we then have:

\begin{align*} \quad dy_t &= \frac{\partial f}{\partial t} dt + \frac{\partial f}{\partial x} dx_t + \frac{1}{2} \frac{\partial^2 f}{\partial x^2} dx^2_t\\ &= \frac{\partial f}{\partial t} dt + \frac{\partial f}{\partial x} \left( \mu dt + \sigma dz_t \right)+ \frac{1}{2} \frac{\partial^2 f}{\partial x^2} \sigma^2 dt \\ &= \left(\frac{\partial f}{\partial t} + \frac{\partial f}{\partial x}\mu + \frac{1}{2} \frac{\partial^2 f}{\partial x^2} \sigma^2 \right) dt + \left( \frac{\partial f}{\partial x}\right)\sigma dz_t \end{align*}

Which is Ito's lemma.

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Maybe your Itô formula is just wrong. Look at:


On page 6 you will find Itôs Lemma. You don't have to interpret $dx_t^2$ as $(dx_t)^2$. The $dx_t^2$ part in your formula is in fact $(dx_t)^2$.

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