Tag Info

New answers tagged

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

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 ...


-1

I aim to give a careful mathematical treatment to this answer, whilst following the fantastic book "Basic Stochastic Processes" by Brzezniak and Zastawniak. The reason I am putting this answer on is twofold: first, to compliment @ William S. Wong's answer by adding greater mathematical intricacy for other users of the website, and secondly to confirm that ...


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 ...


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 $\hat P_t$ ...


0

The stock weight $\pi_t\geq0$ is nonnegative as $C_T=\max(S_T-K,0)$ is increasing in $S_T$ (always long). $\pi_t\leq 1$ cannot be greater than one, because one can receive at most 1 stock from the call at maturity, so you dont pay more than 1 stock price for it. Otherwise, one could buy the (cheaper) call and sell the replicating portfolio for riskfree ...


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, ...


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 ...


0

If you want a really intuitive answer, I thought of two things to explain the key idea: From a Masters student point of view: Assume $\xi \sim N(0, \sigma^2)$, prove that: $\mathbb{E}[f(\xi)\xi] = \sigma^2 \mathbb{E}[f'(\xi)] = \mathbb{E}[\xi^2]\mathbb{E}[f'(\xi)]$ A more intuitive financial explanation may go like this: Consider you have a binary ...



Top 50 recent answers are included