- EdisonCruise
**Posts:**105**Joined:**

The above solution uses a simple method to calculate It-Is.Although the result looks quite correct to me, the process seems to have some problem. The last equation omit lots of terms. After t1, there should be more terms t2, t3... until t. If we write out the long series (rather than just 3 terms above) to give a strict proof, I have some difficulty to find the long series can converge to the final result Ws(Wt-Ws).

Last edited by EdisonCruise on January 11th, 2015, 11:00 pm, edited 1 time in total.

None of this looks right to me. The Ito integral [$]\int_0^t W_s dW_s = \frac{1}{2}W_t^2 - \frac{1}{2} t[$].When you first take calculus, Riemann integrals are defined in two ways: (I) as the limit of a sum, and (II) as an inverse to differentiation.Generally, you do a few tedious exercises to confirm method I for a few simple integrals, but then spend the rest of your life with method II.The analog of method II here is simply to check a putative answer with Ito's differentiation rule. The analog of method I might be interesting tosee if someone can post it for this example.

- EdisonCruise
**Posts:**105**Joined:**

Thank you Alan. I completely agree with you

Using an existing thread for following question:

I have an error term in a calculation that I'd like to simplify if possible, and it looks as follows:

[$] \text{error term} = E \left\{ \left[ \left(\int \sigma_u dW_u \right)^2 - \int \sigma_u^2 du \right] f (\bar\sigma) \right\} [$]

with [$] \bar\sigma^2 = \int \sigma_u^2 du [$]

If [$]f[$] doesn't depend on [$] \bar\sigma [$], or if [$] \bar\sigma [$] is deterministic (which it isn't) then the error term would be zero. But I don't think in general it is zero, and any help on how to proceed to approximate the error or calculate it would be appreciated.

Thanks

I have an error term in a calculation that I'd like to simplify if possible, and it looks as follows:

[$] \text{error term} = E \left\{ \left[ \left(\int \sigma_u dW_u \right)^2 - \int \sigma_u^2 du \right] f (\bar\sigma) \right\} [$]

with [$] \bar\sigma^2 = \int \sigma_u^2 du [$]

If [$]f[$] doesn't depend on [$] \bar\sigma [$], or if [$] \bar\sigma [$] is deterministic (which it isn't) then the error term would be zero. But I don't think in general it is zero, and any help on how to proceed to approximate the error or calculate it would be appreciated.

Thanks

I think you are correct in your feeling. I would be try to check my expectations putting [$]\sigma_t[$] a random function taking first two value and then when it is a discrete value function.

Thanks List, but I think taking discrete values for the vol will give zero for the integral.

Maybe easier to consider a specific example: is the following zero:

[$] E \left\{ \left( \left(\int \sigma_u dW_u \right)^2 - \int \sigma_u^2 du \right)\bar\sigma\right\} = 0? [$]

Maybe easier to consider a specific example: is the following zero:

[$] E \left\{ \left( \left(\int \sigma_u dW_u \right)^2 - \int \sigma_u^2 du \right)\bar\sigma\right\} = 0? [$]

It might make sense to consider an illustrative example

$$ E [ ( ( w_{2} - w_{1} ) w_{1} + ( w_{3} - w_{2} ) w_{2} )^{2} - ( w_{1}^{2} + w_{2}^{2} ) ] ( w_{1}^{2} + w_{2}^{2} )$$

to verify your expectation. He random sigma replaced by Wiener process. Here [$]\Delta t = 1[$].

$$ E [ ( ( w_{2} - w_{1} ) w_{1} + ( w_{3} - w_{2} ) w_{2} )^{2} - ( w_{1}^{2} + w_{2}^{2} ) ] ( w_{1}^{2} + w_{2}^{2} )$$

to verify your expectation. He random sigma replaced by Wiener process. Here [$]\Delta t = 1[$].

I dont understand, what is w1, w2, w3?It might make sense to consider an illustrative example

$$ E [ ( ( w_{2} - w_{1} ) w_{1} + ( w_{3} - w_{2} ) w_{2} )^{2} - ( w_{1}^{2} + w_{2}^{2} ) ] ( w_{1}^{2} + w_{2}^{2} )$$

to verify your expectation. He random sigma replaced by Wiener process. Here [$]\Delta t = 1[$].

Sorry,I dont understand, what is w1, w2, w3?It might make sense to consider an illustrative example

$$ E [ ( ( w_{2} - w_{1} ) w_{1} + ( w_{3} - w_{2} ) w_{2} )^{2} - ( w_{1}^{2} + w_{2}^{2} ) ] ( w_{1}^{2} + w_{2}^{2} )$$

to verify your expectation. He random sigma replaced by Wiener process. Here [$]\Delta t = 1[$].

[$]w_k = w ( t_k ) [$] , k = 1, 2, 3 .

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