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$$ \textbf{Preface} $$ I am assuming log normal asset but this is not clear from the question? Or rather I have misinterpreted the question! Well as I see it from a a purely mathematical exercise $$ d\left(\dfrac{S_t}{M_t}\right) =\frac{1}{M_t}dS_t - \frac{S_t}{M_t^2}dM_t +O(dt^2) $$ using Ito's lemma. Then we can sub in the original processes yields ...


0

nice point. One way of looking at it, I think, is that you have just two Brownian motions, so in a sense your space is just 2 dimensional. Thus, as long as $V$ and $V_1$ are not "linearly dependent", you're spanning the space, and you're done, and it doesn't really matter what $V_1$ you're choosing. Now, this is of course a very hand-waving argument, in ...


1

very deep, elaborate question. It seems to me, though, that you're somewhat missing the point: The fundamental theorem of asset pricing is that in a no-arbitrage environment, asset prices divided by some numeraire are martingales (with respect to a measure dependent on the numeraire, with the measure being unique if the market is complete with respect to ...


0

Note that $$\frac{dQ_{T_p}}{dQ}|_{T_0} = \frac{P(T_0, T_p)}{P(0, T_p)}\frac{A(0, T_0, T_n)}{A(T_0, T_0, T_n)}$$. Then $$E^{Q_{T_p}}\big(S(T_0, T_n)\big) = E^Q\bigg(S(T_0, T_n) \frac{P(T_0, T_p)}{P(0, T_p)}\frac{A(0, T_0, T_n)}{A(T_0, T_0, T_n)}\bigg) \\ = \frac{A(0, T_0, T_n)}{P(0, T_p)} E^Q\bigg(S(T_0, T_n) \frac{P(T_0, T_p)}{A(T_0, T_0, T_n)}\bigg).$$ That ...


3

Consider an (arithmetic) Ornstein-Uhlenbeck process as a model of the asset price $X_t$: $$dX_t = \kappa(\mu-S_t)dt + \sigma dW_t$$ where $\mu$ is the mean-reversion level, $\sigma$ is a volatility parameter, $W_t$ is Brownian motion, and $\kappa$ is the reversion speed. An Ornstein-Uhlenbeck process will revert to the mean infinitely often if $\kappa ...


3

For Q1, the function $a(t)$ is the instantaneous correlation. The form given by (2) is basically the Cholesky decomposition. Of course, you may directly show, uisng Levy's characterization, that $$ \widetilde{W}(t) = \int_0^t\bigg[\frac{1}{\sqrt{1-||a(t)||^2}} dZ(t) -\frac{a(t)^T}{\sqrt{1-||a(t)||^2}} dW^B(t) \bigg] $$ is a standard scalar Brownian motion ...


3

Q1: $$(1)\rightarrow(2)$$ (1): $a(t)$ is the instantaneous correlation of $\rho(Z_t,W_t)$ because: $$\rho(dZ_t,dW_t)=\dfrac{Cov(dZ_t,dW_t)}{\sigma_{dZ_t}\sigma_{dW_t}}=\dfrac{E(dZ_t\cdot dW_t)}{\sqrt{dt} \sqrt{dt}}=\dfrac{\langle dZ_t, dW_t\rangle}{t}=a(t)$$ $\Rightarrow$ (2) holds as following, in the 1-dim case: $dZ_t\sim N(0,dt),$ ...



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