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6

At the first glance, what you are asking for is a model admitting arbitrage, so there is a zero chance of losing money and positive chance of yielding profits. Well, many equilibrium models start with assuming arbitrage is not possible (otherwise it would be trivial wouldn't it). But, in my opinion, what you actually seek is the Efficient Markets ...


5

Swap Just to be clear, (3.4c) leads to (3.5a) when we assume lognormal $R(\tau)$. Lognormal $R(\tau)$ means we can write $$R(\tau) = R_0 e^{-\frac{1}{2}\sigma^2 \tau + \sigma \sqrt{\tau} Z}$$ with $Z$ normal, and I'm assuming a zero mean -- which I think is required. Then for (3.4c) we have for the expectation value: $$ E\left[(R(\tau) - R_0)^2 \right] = ...


4

Trinomial trees give incomplete markets so there is a range of possible risk neutral prices. So you have to find the possible probabilities that make the tree risk-neutral and see what prices you get. You have the correct expressions. Now just have to parametrize the set of solutions. It is one-dimensional and all the probabilities are positive so you need ...


4

All the topics you've mentioned are wonderful and shouldn't be eschewed by reading some finance-oriented review book. I recommend these instead. Linear algebra: Hoffman and Kunze and Halmos Set theory: Halmos Measure theory: Rudin and Tao


3

First, we have $P(t)+S(t)=C(t)+B(t,T)\cdot K$, Then, $\frac{\partial P(t)}{\partial S(t)} + \frac{\partial S(t)}{\partial S(t)} = \Delta^{\text{put}}_{t}+1$ and $\frac{\partial C(t)}{\partial S(t)} + \frac{\partial [B(t,T)\cdot K]}{\partial S(t)} = \Delta^{\text{call}}_{t}+0$. Finaly, $\Delta^{\text{call}}_{t}-\Delta^{\text{put}}_{t}=1$. This relationship ...


3

I would recommend the books from Steven Shreve. Here is a link to some one of his older online pdf's (1997 but nevertheless true) so you can check if that fits the bill. http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.137.6951&rep=rep1&type=pdf


2

For a martingale $\{M_t \mid t\geq 0\}$ and the stochastic integral \begin{align*} I_t = \int_0^tZ_s dM_s, \end{align*} we have that \begin{align*} E((I_t)^2) = E\bigg( \int_0^tZ_s^2 d\langle M\rangle_s\bigg), \end{align*} where $\langle M\rangle$ is the quadratic variation. That is, the ito's isometry holds for a martingale integrator only. However, in ...


2

The claim payoff you describe, $g(M)$, looks to me like a tight butterfly spread that pays off only in one state of the world. Can't you just replicate that by short two calls with strike $K_0$ and long two calls, with strikes one either side at $K_0\pm 1$? Then the price of your option would be $C(K_0+1)+C(K_0-1)-2\cdot C(K_0)$. This is effectively the ...


2

The above question was a typo due to the author -- the expression should be evaluated as \begin{equation} E(t|\mathcal{F}_{s}^{W}) = t \end{equation} due to the reasoning in the question. Sorry for the noise.


2

Sorry, but despite being used as a popular example in machine learning, no one has ever achieved a stock market prediction. It does not work for several reasons (check random walk by Fama and quite a bit of others, rational decision making fallacy, wrong assumptions ...), but the most compelling one is that if it would work, someone would be able to become ...


2

This will depend on the nature of your tree. For a re-combining binomial tree, the number of nodes, including the initial one, will be \begin{align*} \sum_{i=1}^n i = \frac{n(n+1)}{2}. \end{align*} For the paths, as at each time $j$, there are two possibilities from each node, the total path number is $2^n$.


1

For question a). From the assumptions, in particular, that $R=0$, \begin{align*} \pi_l + \pi_m + \pi_u &=1\\ \frac{1}{2}\pi_l + \pi_m + 2\pi_u&=1. \end{align*} Set $\pi_m=x$, and solve for $\pi_l$ and $\pi_u$, \begin{align*} \pi_l &= \frac{2}{3}(1-x)\\ \pi_m &= x\\ \pi_u &= \frac{1}{3}(1-x), \end{align*} where $0<x<1$. The option ...


1

Let $B_t$ be the value of the risk-free asset at time $t$. Then $B_0=1$ and $B_{t+1} = (1+R) B_t$. Moreover, let $\beta_t$ be units invested in the risk-free asset at time $t$. It is clear that $\beta_0 = w_0 - \Delta_0 S_0$. Since the strategy is self-financing, \begin{align*} \Delta_{t-1} S_{t-1} + \beta_{t-1} B_{t-1} = \Delta_t S_{t-1} + \beta_t ...


1

This doesn't really suffice as an existence proof, but you can start with a series of mathematical results collectively known as no free lunch theorems. The linked paper proves the average performance of any optimization algorithm over arbitrary problem domains is independent of the algorithm. That is, no single algorithm can ever be better than others on ...


1

This is where one needs the concept of no free lunch with vanishing risk (NFLVR), whose proof you can find in: Delbaen & Schachermayer (1994). Though, as a warning, I should mention it is pretty involved.


1

Why do you think such a theorem would exist? I will give you a counterexample: You have two assets A and B. Both are completely identical in every respect, except price: The price of A is USD 1 the price of B is USD 2. Your strategy is simple: You (short) sell B for a gain of 2 and buy A for 1. This strategy requires no capital and leaves you with an ...


1

I believe a few things need to be said here. First, returns are usually calculated (END_VALUE-BEGIN_VALUE)/BEGIN_VALE. There are other ways, but this is what is usually used, and much arguments can be had on what "value" actual is. Second, data frequency should be aligned so daily standard deviation should be aligned to daily expected returns. Third, the ...


1

Let \begin{align*} w_t = \frac{1}{\sqrt{\sigma_1^2+\sigma_2^2 -2\sigma_1\sigma_2 \rho}}(\sigma_1\tilde{w}_t^1-\sigma_2\tilde{w}_t^2). \end{align*} Then, using Levy's characterization, we can show that $\{w_t \mid t \geq 0\}$ is a standard Brownian motion.


1

A convex function is when the line between two points on the graph always lies above the graph. And this does hold for the put, its also sometimes called a sublinear function. Also see http://en.wikipedia.org/wiki/Convex_function So the author is correct in saying that $(K-s)^+$ is convex.


1

Could you please be more specific with your question and post the text here? This will be more helpful for other people visiting the site. Now as far as to where the 1/2 went, usually people put 1/2 in front of the second order term because this will simplify to 1 after the derivation: $$ \frac{\partial x^2}{\partial x} = 2x $$ vs $$ \frac{1}{2} \cdot ...


1

You might want to give us the exact statement of the author. Let the Wiener process $W_{s}$ be a r.v. from $\left(\mathcal{F}_{s},\Omega\right)\to\left(\mathcal{B}\left(\mathbb{R}\right),\mathbb{R}\right)$. The Borel-$\sigma$-algebra $\mathcal{B}\left(\mathbb{R}\right)$ contains all intervals of the form $\left[x,y\right]$ for $x\neq y\in\mathbb{R}$, ...



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