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I have the following exercise: A financial security pays off a dollar amount of $S_T^2$. Using Ito`s Lemma, what is the price today $V_t$ of this security? (S follows a Geometric Brownian Motion $dS = \mu Sdt + \sigma S dB$ )

I know that $V_t = e^{-r(T-t)}E^Q[V_T]$. So I tried to calculate $E^Q[V_T]$ with Ito`s Lemma:

  1. $dV/dt = 0$
  2. $dV/dS = 2S$
  3. $d^2V/dS^2 = 2$

So Itos Lemma yields $dV = (2\mu + \sigma^2)S^2 dt + 2\sigma S^2dB $. Now, I have to solve this: $\int_{t}^{T} S^2 = (2\mu + \sigma^2) \int_{t}^{T} S^2 ds + 2\sigma \int_{t}^{T} S^2dB $, but I don`t know how to get the result $V_t = S^2e^{(2r + \sigma^2)(T-t)}$.

Can anyone help me with this exercise?

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    $\begingroup$ Forget everything about brownian motion. you need expectation of square of a random variable whose distribution you know already. this is just statistics. $\endgroup$
    – Arshdeep
    Jul 19 at 16:48

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A few comments:

The question says "using Ito's lemma". We can also do it without using Ito's Lemma by simply calculating $E^Q(S_T^2)$ -- this should give you something that agrees with your form of $V_t$ that you were not sure how to get to.

To use Ito's Lemma in your calculation, I'm not entirely sure what approach the exercise is looking for, but if you recall solving the SDE of GBM, you can do something similar here to get an expression of the form $$\log(S^2_T) - \log(S^2_t) = (2r-\sigma^2)\int_t^Tdu + 2\sigma \int_t^Td\tilde{W_u} $$

which is a few short steps from the solution you're seeking.

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  • $\begingroup$ Exactly, I need some help to solve the SDE of GBM. I am still not sure how to get to your expression $\log(S^2_T) - \log(S^2_t) = (2r-\sigma^2)\int_t^Tdu + 2\sigma \int_t^Td\tilde{W_u}$. Can you please explain what steps you have done? $\endgroup$
    – Otto
    Jul 21 at 15:28
  • $\begingroup$ For the case of non-squared GBM, see here: en.wikipedia.org/wiki/Geometric_Brownian_motion#Solving_the_SDE For the case of squared GBM, this is very similar except of course we apply Ito's formula to $X_t = f(S_t)$ where $f(x) = \log(x^2) = 2\log(x)$, whereas the wikipedia post given just used $\log(x)$ $\endgroup$
    – Rylan
    Jul 21 at 15:45

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