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6

$$d Y \left(t\right) := d \left[\int_0^t{a \left(s\right)\mathrm{d}W_s}\right] = a \left(t\right) dW_t$$ Note that since $Y$ is a driftless process, it is a local martingale, and because $a$ is bounded, a true martingale. Its quadratic variation is given by $$\langle Y \left(\cdot\right)\rangle_t = \int_0^t{a^2 \left(s\right)\mathrm{d}s}$$ by definition ...

4

With your SDE for $F$, I get: $$dXdF = -a^2XFdt$$ $$FdX = rFdt + aXFdW$$ $$XdF = a^2XF dt -aXF dW$$ So, adding up: $$d(XF) = rF dt,$$ giving $$X_t = F^{-1}_t X_0 + rF^{-1}_t \int_0^t F_u du$$

4

Alternatively, we can use Ito isometry ($X$'s integrability and adaptability are assured by $a$'s boundness and adaptability, respectively): $$E[X_t|{\cal F}_s] = E[X_s\big|{\cal F}_s] + E\left[\left(\int_s^t a_udW_u\right)^2 - \int_s^t a_u^2du \big|{\cal F}_s \right]$$ $$= X_s + E\left[\left(\int_s^t a_udW_u\right)^2\big|{\cal F}_s\right] - E \left[ \... 3 Providing only a sketch here, using Ito integral definition (and commuting limit, summations and expectation), the result boils down to studying the expectation term:$$ E\left[ f_{t_{i-1}} (W_{t_i}-W_{t_{i-1}}) \cdot g_{t_{j-1}} (W_{t_j} - W_{t_{j-1}}) \right]. $$If the intervals don't overlap, i\not= j, and say t_i \leq t_{j-1}, then f_{t_{i-1}} g_{... 3 (As said in the comments, you need to put down some of your thoughts regarding the question too, like specifying the tools/theorems you would use or actual attempts to apply them, even if you can only cover early steps, not just the question itself.) Hints: We are given$$X_t^\theta:=\exp \left(\theta W_t-\frac{1}{2} \theta^{2} t\right)=\sum_{n=0}^{\infty} \...

2

Lets assume the price of the underlying equals the strike at some point prior to expiry. Then the probability of the price being still greater or equal the strike at expiry is 0.5. So the probability of the European option paying out is exactly half of the probability for the American option.

2

Hints: Show first that $$E[((W^1_t + W^2_t)-(W^1_s + W^2_s))^2] = (2+2\rho)(t-s)$$ Then conclude that $$[(2+2\rho)^{-1/2} (W^1 + W^2)]_t =t$$ On the other hand, show (using bilinearity of quadratic covariation) that $$[(2+2\rho)^{-1/2} (W^1 + W^2)]_t = [(2+2\rho)^{-1/2} (W^1 + W^2), (2+2\rho)^{-1/2} (W^1 + W^2) ]_t$$ $$= (1+\rho)^{-1} (t+ [W^1, W^2]_t)... 2 Note that SDE (4) does have a "closed-form" representation. Let X be$$X := S^p, $$so (4) is a geometric Brownian motion SDE$$dX = (p\alpha + 2^{-1}p(p-1) \sigma^2) X dt + p \sigma X dW, $$which, again due to Ito Lemma, is equivalent to$$ d \ln X = (p\alpha + 2^{-1}p(p-1) \sigma^2 - 2^{-1}p^2 \sigma^2) dt + p \sigma dW $$or$$ d \ln X = ...

1

If $s>0$, and the integral runs from $u=s$, then the integral only makes sense if we condition on what we know as of time $s$: we can write $W(s)=k$, where $k$ is some constant known at time $s$, i.e. the value of the Brownian motion $W_u$ known at time $u=s$ (can be zero, but doesn't need to be). Then, we have: \mathbb{E}\left[\int_{u=s}^{u=t}W_udu|\...

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