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The following question is from my professor, I have no clue about it at all.

Let $(W (t))_{t\geq0}$ denote a standard Brownian motion. Consider a model of asset price given as

$$\mathrm{d} S(t)=\mu S(t) \mathrm{d} t+\sigma(t) S(t) \mathrm{d} W(t), \quad S(0)=s_{0}$$

where $\mu > 0$ is a constant and $\sigma(t)>0$ is a deterministic (non-random) function of time. Ito's isometry tell us that $\mathbb{E}\left[\left(\int_{0}^{t} f(s, W(s)) \mathrm{d} W(s)\right)^{2}\right]=\int_{0}^{t} \mathbb{E}\left[f^{2}(s,W(s))\right] \mathrm{d} s$ for a square integrable function f.

a). Using Itô’s isometry, compute the mean and variance of $\int_{0}^{t} \sigma(s) \mathrm{d} W(s)$.

b). Compute the mean of S(T).

For part (a), can I use the following process:

Let $Z(t)=\sigma(s)dW(s)$, $Z(t)$ is a martingale so $$\mathbb{E}(\int_{0}^{t} \sigma(s) \mathrm{d} W(s))=0$$

Moreover, based on the Ito's isometry, $$Var(Z(t))=\mathbb{E}\left(\int_{0}^{t} \sigma(s) \mathrm{d} W(s)\right)^2=\int_{0}^{t} \mathbb{E}\left[\sigma(s)^2\right] \mathrm{d} s=?$$

I am stuck here.

Evangelopoulos Foivos
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Ito_lover
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    You might be overthinking it. We are given that $\sigma(t)$ is a deterministic function of time. Therefore the expectation of $\sigma$ is merely itself and so the variance is $\int_0^t \sigma(s)^2 ds$. – Nap D. Lover Aug 20 '21 at 14:19
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    Do you mean $Z(t) = \int_0^t σ(u) dW_u $? If yes, then what you wrote so far is correct: $(Z(t))_{t \geq 0}$ is indeed a martingale since $σ(\cdot)$ is deterministic. To finish part a), observe that $ \mathbb E [σ(s)^2]= σ(s)^2$ since $σ(\cdot)$ is deterministic. Thus the integral becomes $ \int_0^t σ^2(s) ds. $

    Do you have any more information for b)? Perhaps that $S(t)$ is an Ito process so that you can use Ito's formula?

     – Evangelopoulos Foivos Aug 20 '21 at 14:23

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