Showing the Difference Between an Ito Integral & Riemann Integrals.

In summary, the problem at hand involves solving a question using Ito Integrals and Riemann integrals without directly using Ito's Lemma. The Ito integral and Riemann integral have fundamental differences, with the former taking into account the randomness of the process being integrated. To solve the problem, one can use the definition of the Ito integral and its properties, such as linearity and the equivalence to the Riemann integral for deterministic functions. The extra term in the Ito integral arises from the randomness of the Brownian motion increments, and by manipulating the terms, one can prove the desired result.
  • #1
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Hi Everyone,
A problem I have here. I am trying to solve a problem involving Ito Integrals and Riemann interals.

Homework Statement


Prove

[tex]\int^{T}_{0} tdW(t) = TW(T) -\int^{T}_{0} W(t)dt [/tex]

Homework Equations


I want to solve this question WITHOUT using Ito's Lemma directly.

The Attempt at a Solution


OK I know that in general, the Ito integral and the Riemann integral are going to be slightly different. The Ito Integral will have an extra term.

So in my question, I know if I was to generally just integrate the Riemann integral, I would get:
[tex]\int^{T}_{0} tdW(t) = TW(T) [/tex]
But I know the Ito integral adds the extra:
[tex] -\int^{T}_{0} W(t)dt [/tex]
But how does this arise? Is it possible to show how using the mean square error ? I keep reading the mean square error when reading about this but don't really see the connection. Maybe someone could kindly help out please?

Thanks
 
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  • #2

I would like to offer some guidance on how to approach this problem. Firstly, it is important to understand the fundamental differences between the Ito integral and the Riemann integral. The Ito integral is a stochastic integral, meaning that it is defined as the limit of a sequence of Riemann sums. This means that it takes into account the randomness of the process being integrated.

To solve this problem without using Ito's Lemma directly, you can use the definition of the Ito integral and the properties of stochastic integrals. Firstly, recall that the Ito integral is defined as a limit of Riemann sums:

\int^{T}_{0} f(t)dW(t) = \lim_{n\to\infty} \sum_{i=1}^{n} f(t_{i})\left(W(t_{i}) - W(t_{i-1})\right)

where t_{i} = \frac{iT}{n}. From this definition, we can see that the extra term in the Ito integral arises from the fact that the increments of the Brownian motion, W(t_{i}) - W(t_{i-1}), are random variables. Therefore, when taking the limit, we need to account for the randomness by including the term -\int^{T}_{0} f(t)dt.

To prove the desired result, you can start by expanding the Ito integral using the definition above and then manipulating the terms to show that it equals TW(T) -\int^{T}_{0} W(t)dt. You can also use the properties of stochastic integrals, such as linearity and the fact that the Ito integral of a deterministic function is equal to the Riemann integral.

I hope this helps in solving the problem. Good luck!
 

FAQ: Showing the Difference Between an Ito Integral & Riemann Integrals.

What is an Ito integral?

An Ito integral is a type of stochastic integral used in stochastic calculus to calculate the value of a stochastic process at a particular point in time. It is named after the Japanese mathematician Kiyoshi Ito.

How is an Ito integral different from a Riemann integral?

An Ito integral is designed to integrate stochastic processes, which are random in nature, while a Riemann integral is used to integrate deterministic functions. The main difference is that an Ito integral takes into account the random fluctuations of a stochastic process, while a Riemann integral does not.

What are some practical applications of Ito integrals?

Ito integrals are commonly used in financial mathematics to model and analyze the behavior of financial markets, such as stock prices and interest rates. They are also used in physics, biology, and other fields to model and study systems with random fluctuations.

Can Ito integrals be approximated by Riemann sums?

Yes, Ito integrals can be approximated by Riemann sums, but the approximation may not be accurate, especially for processes with large fluctuations. This is because Riemann sums do not take into account the random nature of the process, while Ito integrals do.

Are there any famous theorems related to Ito integrals?

Yes, the most famous theorem related to Ito integrals is the Ito isometry, which states that the expected value of the square of an Ito integral is equal to the integral of the square of the integrand over the time period. This theorem is a fundamental tool in stochastic calculus and has many applications in finance and other fields.

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