# Tag Info

## Hot answers tagged binomial-tree

5

Assuming continuously compounded returns for a multi-period model with $N$ being the number of periods: \begin{cases} &\log u \quad \text{with probability q}\\ &\log d \quad \text{with probability 1-q} \end{cases} given the stock price at maturity $$\log\left(\frac{S_T}{S_0}\right)=i\log u+(N−i)\log d=i\log\left(\frac{u}{d}\right)+N\log d$$ where $... 5 one of the most fundamental results states that the binomial model converges towards the Black Scholes model if the step size$\Delta t$converges to zero. The Black Scholes model is an option pricing model where the underlying is given by $$S_T = S_0 \cdot \exp \Bigl(\sigma W_T - \frac 12 \sigma^2 T \Bigr).$$ By choosing$$u = \exp(\sigma \sqrt{\... 3 What you say is perfectly true and there is no contradiction. Arbitrage means risk free profit , so your ‘statistical arbitrage’ is not arbitrage at all. It just says that if you take risk, your expected returns can be higher than the risk free rate. How much higher depends in the risk aversion of market participants. 3 If I understand your question correctly, another way to word it is: if an event that has probability 0 under the physical measure$\mathbb{P}$, how can it have a positive probability under the risk-neutral measure$\mathbb{Q}$? The answer is simply: it cannot! According to the theory of risk-neutral pricing through no arbitrage arguments, we require that$\...

2

Actually recombining binomial trees are only a particular case of an explicit FDM scheme. But they have obvious limitations, the foremost being that they cannot accomodate local volatilities. Also 1/2 explicit 1/2 implicit FDM schemes (Crank-Nicolson) have faster convergence with respect to the size of the time step. And FDM schemes can accomodate all sorts ...

2

You compare the result of an analytical solution (european call) with the numerical solution for the american option. It seems as if you use to few steps to calculate your American option price. Just try to increase the number of steps and see what happens. Or just compare the european price based on the same binomial tree with the american one and you ...

2

from the look of it your discounting is incorrect because as you increase M you should discount with 1/(1+r0*t) (assuming r0=0.0214 is the annual interest rate where as you seem to discount by 1/(1+r0*T)) also seems there is a typo in your "r" variable, probably you want it to be r=r0=0.0214 instead of r=0.0214*0.25. now a working rewrite of your code for ...

1

Yes it is equivalent of the dividend rate. The b in the function is cost of carry, so here it would be: $b=r_{USD}-r_{AUD}$ And r in the function is $r_{USD}$.

1

Exactly what @Alex C said. It's the time homogeneous diffusion proprety. You can't state such an argument in models where volatility is no longer time homogeneous ( that's being time independant and depending only on the underlyings).

1

Risk neutral valuation tells you to discount the expected payoff in the risk free world. Therefore, it should be $\sum_{i=0}^{N-1}R^{i+1}(1-p)^i p$, where $N$ is the number of periods.

1

A tree and a lattice are the same thing and are most often used interchangeably. Both a tree and a lattice can either be recombining (an up move followed by a down move is equal in value to a down move followed by an up move) or not recombining. A non-recombining lattice is sometimes considered a bush.

1

I think stuff does not cancel out because there was a mistake in the initial formula, which should instead read: $exp(r \delta t) V(S,t)=pV(uS,t+\delta t)+(1-p)V(dS,t+\delta t)$ Developing the function V to second order Taylor terms: $exp(r \delta t) V(S,t)=p \big[V(S,t)+\frac{\partial V}{\partial S}(uS-S)+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}(uS-S)... 1 These tree are uncorellable. Solving the equations will lead to$c(i,j) = 0\$. It kind of makes sense since the probabilities of migration are independant. I wonder what solutions exists to price products where underlyings are correlated and have such an high variance that Monte-Carlo approach is not considerable ?

1

Cetaris parabus, American options will always have a higher price. The option to exercise at any point is worth > 0. I cant speak much to the MATLAB functions themselves or their implementation, though. It looks like one is Black-Scholes and the other is Cox-Ross-Rubinstein so they differ fundamentally in some way.

1

1) First of all many future option contracts are European, so for those there's no modeling problem. Just use BS. Now certain contracts like quarterly ES option are american for historical reasons 2) Futures do not entitle the long holder to the dividends of the underlying. That's the difference with other type of derived instruments (like SPY or QQQ) and ...

Only top voted, non community-wiki answers of a minimum length are eligible