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Explaining Ito's Lemma

5

4

$begingroup$

Find $$int_{0}^{T}W(t)dW(t)$$ using Ito's Lemma.

Now, I know that the answer to that question is: $int_{0}^{T}W(t)dW(t)= frac{W^2(T)}{2}-frac{T}{2}$
but can somebody explain the idea behind the Ito's Lemma by giving a formal mathematical proof of the above?

I would be grateful if someone could post any "interesting" (but not so hard) application of Ito's Lemma when it comes to the Brownian Motion.

real-analysis stochastic-processes stochastic-calculus brownian-motion martingales

share|cite|improve this question

edited Mar 1 at 15:25

GNUSupporter 8964民主女神 地下教會

14.1k82651

asked Feb 26 '17 at 9:45

user415301

$endgroup$

add a comment | 

5

4

$begingroup$

Find $$int_{0}^{T}W(t)dW(t)$$ using Ito's Lemma.

Now, I know that the answer to that question is: $int_{0}^{T}W(t)dW(t)= frac{W^2(T)}{2}-frac{T}{2}$
but can somebody explain the idea behind the Ito's Lemma by giving a formal mathematical proof of the above?

I would be grateful if someone could post any "interesting" (but not so hard) application of Ito's Lemma when it comes to the Brownian Motion.

real-analysis stochastic-processes stochastic-calculus brownian-motion martingales

share|cite|improve this question

edited Mar 1 at 15:25

GNUSupporter 8964民主女神 地下教會

14.1k82651

asked Feb 26 '17 at 9:45

user415301

$endgroup$

add a comment | 

$begingroup$

Find $$int_{0}^{T}W(t)dW(t)$$ using Ito's Lemma.

Now, I know that the answer to that question is: $int_{0}^{T}W(t)dW(t)= frac{W^2(T)}{2}-frac{T}{2}$
but can somebody explain the idea behind the Ito's Lemma by giving a formal mathematical proof of the above?

I would be grateful if someone could post any "interesting" (but not so hard) application of Ito's Lemma when it comes to the Brownian Motion.

real-analysis stochastic-processes stochastic-calculus brownian-motion martingales

share|cite|improve this question

edited Mar 1 at 15:25

GNUSupporter 8964民主女神 地下教會

14.1k82651

asked Feb 26 '17 at 9:45

user415301

$endgroup$

share|cite|improve this question

edited Mar 1 at 15:25

GNUSupporter 8964民主女神 地下教會

14.1k82651

asked Feb 26 '17 at 9:45

user415301

share|cite|improve this question

edited Mar 1 at 15:25

GNUSupporter 8964民主女神 地下教會

14.1k82651

asked Feb 26 '17 at 9:45

user415301

edited Mar 1 at 15:25

GNUSupporter 8964民主女神 地下教會

14.1k82651

edited Mar 1 at 15:25

GNUSupporter 8964民主女神 地下教會

14.1k82651

1 Answer
1

active

oldest

votes

3

$begingroup$

Roughly, let $f :(t,x) rightarrow f(t,x)$ from $mathbb{R^2}$ to $mathbb{R}$, a function smooth enough, classic differentiation gives $$Delta f(t,x)=frac{partial f}{partial t}Delta t+frac{partial f}{partial x}Delta x+frac{1}{2}frac{partial^2 f}{partial x^2}Delta x^2+frac{1}{2}frac{partial^2 f}{partial t^2}Delta t^2+mathcal{O(Delta x^2)}+mathcal{O(Delta t^2)}$$

Using infinitesimal $Delta x$ and $Delta t$, second order terms vanish, and we write
$$d f(t,x)=frac{partial f}{partial t}d t+frac{partial f}{partial x}d x$$

However, if $x$ refers now to a Brownian motion , moving to infinitesimal $Delta x$ and $Delta t$, $Delta t^2$ still vanishes, but $Delta x^2$ becomes $Delta t$ because of the non-nil quadratic variation of the Brownian motion. That is why you may hear that a Brownian motion is in terms of $sqrt{t}$.

Henceforth, the formula differs from the classic one as $Delta x^2$ is not negligeable anymore , and we have

$$d f(t,x)=frac{partial f}{partial t}d t+frac{partial f}{partial x}d x+frac{1}{2}frac{partial^2 f}{partial x^2}dx^2$$

This is the Ito-formula, and $x$ can be any Ito-process, a Brownian motion being one of them.

Now regarding your exercise :

Let $W$ a Brownian motion , and $f(t,x)=x^2$. You would agree that $W$ is a Brownian motion, hence an Ito process. Furthermore $f$ is twice differentiable.
You can apply the Ito's lemma, and use that $frac{partial f}{partial t}=0$, $frac{partial f}{partial x}=2x$ and $frac{partial^2 f}{partial x^2}=2$.
Here $x$ refers to the Brownian motion $W$

Thus,

$$d f(t,W_t)=0d t+2W_td W_t+frac{1}{2}2dW_t^2$$

I stated before that the quadratic variation of $W$ gives that $dW_t^2=dt$. We have
$$d f(t,W_t)= 2W_td W_t+dt$$

By integration the equation , we have
$$f(T,W_T)-f(0,W_0)=2int_{0}^{T}{W_tdW_t}+T$$

or

$$ W_T^2=2int_{0}^{T}{W_tdW_t}+T$$

Finally, by switching the term in $T$ and dividing by $2$, we have what you want .

share|cite|improve this answer

answered Feb 27 '17 at 21:08

CanardiniCanardini

2,5001519

$endgroup$

add a comment | 

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3

$begingroup$

Roughly, let $f :(t,x) rightarrow f(t,x)$ from $mathbb{R^2}$ to $mathbb{R}$, a function smooth enough, classic differentiation gives $$Delta f(t,x)=frac{partial f}{partial t}Delta t+frac{partial f}{partial x}Delta x+frac{1}{2}frac{partial^2 f}{partial x^2}Delta x^2+frac{1}{2}frac{partial^2 f}{partial t^2}Delta t^2+mathcal{O(Delta x^2)}+mathcal{O(Delta t^2)}$$

Using infinitesimal $Delta x$ and $Delta t$, second order terms vanish, and we write
$$d f(t,x)=frac{partial f}{partial t}d t+frac{partial f}{partial x}d x$$

However, if $x$ refers now to a Brownian motion , moving to infinitesimal $Delta x$ and $Delta t$, $Delta t^2$ still vanishes, but $Delta x^2$ becomes $Delta t$ because of the non-nil quadratic variation of the Brownian motion. That is why you may hear that a Brownian motion is in terms of $sqrt{t}$.

Henceforth, the formula differs from the classic one as $Delta x^2$ is not negligeable anymore , and we have

$$d f(t,x)=frac{partial f}{partial t}d t+frac{partial f}{partial x}d x+frac{1}{2}frac{partial^2 f}{partial x^2}dx^2$$

This is the Ito-formula, and $x$ can be any Ito-process, a Brownian motion being one of them.

Now regarding your exercise :

Let $W$ a Brownian motion , and $f(t,x)=x^2$. You would agree that $W$ is a Brownian motion, hence an Ito process. Furthermore $f$ is twice differentiable.
You can apply the Ito's lemma, and use that $frac{partial f}{partial t}=0$, $frac{partial f}{partial x}=2x$ and $frac{partial^2 f}{partial x^2}=2$.
Here $x$ refers to the Brownian motion $W$

Thus,

$$d f(t,W_t)=0d t+2W_td W_t+frac{1}{2}2dW_t^2$$

I stated before that the quadratic variation of $W$ gives that $dW_t^2=dt$. We have
$$d f(t,W_t)= 2W_td W_t+dt$$

By integration the equation , we have
$$f(T,W_T)-f(0,W_0)=2int_{0}^{T}{W_tdW_t}+T$$

or

$$ W_T^2=2int_{0}^{T}{W_tdW_t}+T$$

Finally, by switching the term in $T$ and dividing by $2$, we have what you want .

share|cite|improve this answer

answered Feb 27 '17 at 21:08

CanardiniCanardini

2,5001519

$endgroup$

add a comment | 

3

$begingroup$

Roughly, let $f :(t,x) rightarrow f(t,x)$ from $mathbb{R^2}$ to $mathbb{R}$, a function smooth enough, classic differentiation gives $$Delta f(t,x)=frac{partial f}{partial t}Delta t+frac{partial f}{partial x}Delta x+frac{1}{2}frac{partial^2 f}{partial x^2}Delta x^2+frac{1}{2}frac{partial^2 f}{partial t^2}Delta t^2+mathcal{O(Delta x^2)}+mathcal{O(Delta t^2)}$$

Using infinitesimal $Delta x$ and $Delta t$, second order terms vanish, and we write
$$d f(t,x)=frac{partial f}{partial t}d t+frac{partial f}{partial x}d x$$

However, if $x$ refers now to a Brownian motion , moving to infinitesimal $Delta x$ and $Delta t$, $Delta t^2$ still vanishes, but $Delta x^2$ becomes $Delta t$ because of the non-nil quadratic variation of the Brownian motion. That is why you may hear that a Brownian motion is in terms of $sqrt{t}$.

Henceforth, the formula differs from the classic one as $Delta x^2$ is not negligeable anymore , and we have

$$d f(t,x)=frac{partial f}{partial t}d t+frac{partial f}{partial x}d x+frac{1}{2}frac{partial^2 f}{partial x^2}dx^2$$

This is the Ito-formula, and $x$ can be any Ito-process, a Brownian motion being one of them.

Now regarding your exercise :

Let $W$ a Brownian motion , and $f(t,x)=x^2$. You would agree that $W$ is a Brownian motion, hence an Ito process. Furthermore $f$ is twice differentiable.
You can apply the Ito's lemma, and use that $frac{partial f}{partial t}=0$, $frac{partial f}{partial x}=2x$ and $frac{partial^2 f}{partial x^2}=2$.
Here $x$ refers to the Brownian motion $W$

Thus,

$$d f(t,W_t)=0d t+2W_td W_t+frac{1}{2}2dW_t^2$$

I stated before that the quadratic variation of $W$ gives that $dW_t^2=dt$. We have
$$d f(t,W_t)= 2W_td W_t+dt$$

By integration the equation , we have
$$f(T,W_T)-f(0,W_0)=2int_{0}^{T}{W_tdW_t}+T$$

or

$$ W_T^2=2int_{0}^{T}{W_tdW_t}+T$$

Finally, by switching the term in $T$ and dividing by $2$, we have what you want .

share|cite|improve this answer

answered Feb 27 '17 at 21:08

CanardiniCanardini

2,5001519

$endgroup$

add a comment | 

$begingroup$

Roughly, let $f :(t,x) rightarrow f(t,x)$ from $mathbb{R^2}$ to $mathbb{R}$, a function smooth enough, classic differentiation gives $$Delta f(t,x)=frac{partial f}{partial t}Delta t+frac{partial f}{partial x}Delta x+frac{1}{2}frac{partial^2 f}{partial x^2}Delta x^2+frac{1}{2}frac{partial^2 f}{partial t^2}Delta t^2+mathcal{O(Delta x^2)}+mathcal{O(Delta t^2)}$$

Using infinitesimal $Delta x$ and $Delta t$, second order terms vanish, and we write
$$d f(t,x)=frac{partial f}{partial t}d t+frac{partial f}{partial x}d x$$

However, if $x$ refers now to a Brownian motion , moving to infinitesimal $Delta x$ and $Delta t$, $Delta t^2$ still vanishes, but $Delta x^2$ becomes $Delta t$ because of the non-nil quadratic variation of the Brownian motion. That is why you may hear that a Brownian motion is in terms of $sqrt{t}$.

Henceforth, the formula differs from the classic one as $Delta x^2$ is not negligeable anymore , and we have

$$d f(t,x)=frac{partial f}{partial t}d t+frac{partial f}{partial x}d x+frac{1}{2}frac{partial^2 f}{partial x^2}dx^2$$

This is the Ito-formula, and $x$ can be any Ito-process, a Brownian motion being one of them.

Now regarding your exercise :

Let $W$ a Brownian motion , and $f(t,x)=x^2$. You would agree that $W$ is a Brownian motion, hence an Ito process. Furthermore $f$ is twice differentiable.
You can apply the Ito's lemma, and use that $frac{partial f}{partial t}=0$, $frac{partial f}{partial x}=2x$ and $frac{partial^2 f}{partial x^2}=2$.
Here $x$ refers to the Brownian motion $W$

Thus,

$$d f(t,W_t)=0d t+2W_td W_t+frac{1}{2}2dW_t^2$$

I stated before that the quadratic variation of $W$ gives that $dW_t^2=dt$. We have
$$d f(t,W_t)= 2W_td W_t+dt$$

By integration the equation , we have
$$f(T,W_T)-f(0,W_0)=2int_{0}^{T}{W_tdW_t}+T$$

or

$$ W_T^2=2int_{0}^{T}{W_tdW_t}+T$$

Finally, by switching the term in $T$ and dividing by $2$, we have what you want .

share|cite|improve this answer

answered Feb 27 '17 at 21:08

CanardiniCanardini

2,5001519

$endgroup$

share|cite|improve this answer

answered Feb 27 '17 at 21:08

CanardiniCanardini

2,5001519

answered Feb 27 '17 at 21:08

CanardiniCanardini

2,5001519

answered Feb 27 '17 at 21:08

CanardiniCanardini

2,5001519