# Tag Info

7

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 Hypothesis....

6

We assume that the process $\{r(t), \, t \ge 0\}$ satisfies an SDE of the form \begin{align*} dr(t) = \big( \theta(t) - a(t) r(t) \big)dt + \sigma(t) dW_t, \quad t > 0, \end{align*} where $\{W_t, \, t \ge 0 \}$ is a standard Brwonian motion. Note that \begin{align*} d\left(e^{\int_0^t a(u) du}r(t) \right) &=a(t) e^{\int_0^t a(u) du}r(t) dt + e^{\int_0^...

5

Suppose that the given condition is true. You want to construct an arbitrage portfolio to take advantage of this. Now, $d$ is an interest rate, and the condition suggests that $d$ is too high. So you will want to receive $d$ in order to profit. If you could, you would borrow money at $r$ and lend it to the stock broker or exchange to collect the interest ...

5

This is not quite true, in either direction. If you have an arbitrage free implied vol surface, you might not have a well-defined local vol surface. An example comes from a discrete model. Consider a spot dynamics where the spot is a martingale that jumps up or down by integer amounts. The spot distribution is discrete, with zero density in between ...

4

The paper could be clearer indeed. It is a slightly confusing topic, but the important step here is to understand the consequence of the derivative $C$ in the portfolio being priced at the assumed vol $\sigma$. This implies (by Black-Scholes) that it will by definition be true that: $\theta_t + \frac{\partial{C}}{\partial{S}}rS_t+ \frac{1}{2}\frac{\partial^... 4 The whole point of no-arbitrage pricing in a complete market is that a general underlying model of the form $$d S_t = \mu(S_t,t)\, dt + \sigma(S_t,t) \, dW_t$$ can be replaced with the risk-neutral process. $$d S_t = (r - \sigma^2/2)\, dt + \sigma(S_t,t) \, dW_t$$ for the purpose of finding the theoretical fair option price. This, of course, follows ... 4 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 >...

4

What you need to do is to first make a variable change such as $u = \frac{x-\xi}{2\sqrt{t}}$. Then change the order of the limit and the integral.

3


1

Possibly a misunderstanding from my side. I understood that a formal proof of the not-existence of a "blameless" algorithm was asked. My approach was first to create a bitstream from price data for ex. if(price_n > price_n-1 ) then bit_n=1 else bit_n=0. Then it obvious that a formal proof is only possible if the bitstream is "Kolmogorov-compressible". Thus i ...

1

I would consider analyzing the wins of a sample of trading systems over random intervals of time (i.e. minutes, hours, days, months, years). It could be theorized that each trading system has a “win rate” between 0 and 1. It is expected that your sample of trading systems and their “win rates” would follow a logit function with the “win rate” coming close ...

1

Let lnA be N(0,1) and lnB be N(0,k) where we will let k tend to zero. Then B has all of its density at 1, so A+B>1 in the limit. Hence A+B is not lognormal.

1

You're correct - the delta in terms of futures contracts is discount factor * N(d1) for a European call-on-future. Consider a zero-strike call as an example, the delta would be 1.0 multiplied by the discount factor. Maybe the author meant to say the delta in terms of forward contracts is N(d1)?

1

If I have read the question correctly then I will assume that $a$, $b$, $c$, $d$, $T_i$, and $k_i$ are constants. If this is the case then the only term which we need to show is bounded is $$\big(a + b(T_i - t)\big)\exp\big(-c(T_i-t)\big).$$ If we assume that we are only considering the temporal domain $0 \leq t \leq T_i$ such that $T_i - t \geq 0$ then ...

1

You're half way there. When $R < D \ (< U)$, the return of the stock dominates the risk-free return in all states of the world. To benefit from that, just borrow cash and invest in the stock. At $t=0$ this requires no net investment: borrowing cash means your account is credited $S_0$, while subsequently buying the stock suggests it is debited $S_0$...

1

The practical problem would be that you yourself would own larger and larger parts of the market and influence the market prices with your buying and selling decisions in an unfavourable way. There will be fewer and fewer counterparties from which you can earn your money so your system would automatically stop working. So you don't have to have infinite ...

1

I just comment your second point, because in the definition i now of the LOP the state price vector (martingale measure) is involved. assuming the LOP holds then: state price vector is unique <=> the market is complete. to proof "=>" look at the trinomial model, show that the model is not complete by trying to find a hedge for a call, afterwards ...

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