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

<p>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 <em>infinitely</em> smaller than $dt$.</p>

<p>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 $dt$.</p>

<p>In any case, we then have:</p>

<p>\begin{align*}
\quad dy_t &amp;= \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\\
&amp;= \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 \\
&amp;= \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*}</p>

<p>Which is Ito's lemma.</p>

CC BY-SA 3.0
2016-10-07T02:16:06.667Z
2016-10-07T02:30:27.217Z
2016-10-07T02:30:27.217Z
Matthew Gunn
<p>Maybe your Itô formula is just wrong. Look at: </p>

<p><a href="http://www.columbia.edu/~mh2078/stochastic_calculus.pdf" rel="nofollow">http://www.columbia.edu/~mh2078/stochastic_calculus.pdf</a></p>

<p>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 $(dx_t)^2$.</p>

CC BY-SA 3.0
2016-07-08T18:16:19.053Z
2016-07-08T18:16:19.053Z
Pablo E
<p>I'm getting into asset pricing and was looking at Ito's Lemma, but cannot understand a few steps that are given.</p>

<p>Ito's Lemma states that given</p>

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

<p>then</p>

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

<p>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:</p>

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

<p>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.</p>

CC BY-SA 3.0
2016-07-01T15:02:03.793Z
I suggest a powerful reading in Hull's "Options and Derivatives" where you can find an interesting appendix to prove Ito's lemma
Alexis L.
CC BY-SA 3.0
2016-10-07T02:18:07.743Z
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.
Matthew Gunn
<asset-pricing><stochastic-calculus>
Ito's Lemma derivation
Kitsune Cavalry
# Ito's Lemma derivation

(Asked by: username_0 on 2016-06-30)

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

$$dx\_t = \mu dt + \sigma dz\_t \\
y\_t = f(t, x\_t)$$

then

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

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.

**Tags:** asset-pricing, stochastic-calculus
> **Comments:**
> **username_1**: I suggest a powerful reading in Hull's "Options and Derivatives" where you can find an interesting appendix to prove Ito's lemma  
> **username_2**: 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.  

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## 2 answers

### Answer by username_2 on 2016-10-07. Score: 2 | Accepted answer

$$dx\_t = \mu dt + \sigma dz\_t \\
y\_t = f(t, x\_t)$$

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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### Answer by username_3 on 2016-07-08. Score: -1

Maybe your Itô formula is just wrong. Look at:

<http://www.columbia.edu/~mh2078/stochastic_calculus.pdf>

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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