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4

In the integral $$\int_0^t S_u dW^{*}_u \, ,$$ $dW^{*}_u \equiv W^{*}_{u+du} - W^{*}_u$ is independent from the integrand $S_u$. So, $\mathbb{E}\left[ \int_0^t S_u dW^{*}_u\middle\vert \mathcal{F}_0\right] = \int_0^t \mathbb{E}\left[S_u \middle\vert \mathcal{F}_0\right]\mathbb{E}\left[dW^{*}_u\middle\vert \mathcal{F}_0\right] = 0$, since ...


3

A portfolio $V_t(\alpha_t,\beta_t)$ (for stock $S_t$ and zerobond $B_t$) is self-financing iff: $$V_t=\alpha_tS_t+\beta_t B_t$$ It further implies $$dV_t=\alpha_tdS_t+\beta_tdB_t$$ To replicate a derivative $C(S_t,t)$ by a self-financing portfolio of stock and bond, set: $$dV_t=dC_t$$ The dynamics of $dC$ can be specified using Ito's Lemma on ...


2

$ \sigma S $ is in units of dollars per square root of a unit of time. $ \sigma $ is usually quoted as an annual or daily percentage. $ dX ^2 $ is in units of time, as $ E[(dX)^2] = dt $. Here is an online tutorial which you may find helpful. EDIT by kotozna: $\sigma$ has dimensions 1/(square root of time) and $dX$ has dimensions square root of time. ...


1

Let's distinguish: measure $P$ gives probability over the paths, hence you can't really say that it has certain mean and variance: that would apply to a measure $P_t$ which restricts $P$ to the time instance $t$. Now, if you know that $P_t$ has density $f$ (with respect to Lebesgue measure) and $\frac{\mathrm d\hat P_t}{\mathrm d P_t} = g$ then $\hat P_t$ ...


1

Part 1: Show that there exists a trading strategy which replicates a European Call. Proof: I am actually going to prove a stronger statement: that there exists an admissible trading strategy which replicates any payoff in this market. By the First Fundamental Theorem of Asset Pricing, there is no arbitrage if there exists a change of measure such that, ...


1

If we are going to have the form \begin{align*} dr = A dt + BdW_t, \end{align*} Then both A and B are functions of $t$ and $r_t$, otherwise, $r_t$ is normal. However, note that \begin{align*} r_t = \exp\Bigg(\frac{1}{\sigma(t)}\bigg(\int_0^t \theta(s)\sigma(s) ds +\sigma(0)\ln r_0 + \int_0^t\sigma^2(s) dW_s\bigg)\Bigg). \end{align*} That is, $r_t$ is ...



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