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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{\...


5

Note that the tree is recombining. You have $u=1.2$ and $d=0.8$ with $ud=0.96$. Your tree for the asset price reads as At time zero: 100 At time one: 80 or 120 At time two: 64 or 96 or 144 The transition probabilities are $q_u=0.55$ and $q_d=0.45$. For your put option with strike price $K=104$, you thus obtain by backward induction At time two: 40 or 8 ...


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

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 $r_{USD}$.


2

A different way of thinking is risk-neutral pricing. Recall that a market is free of arbitrage if and only if there exists (at least) one risk-neutral probability measure. This is the (first) Fundamental Theorem of Asset Pricing. The risk-neutral probability is frequently set to be $$q=\frac{e^{r\Delta t}-d}{u-d}.$$ This expression defines a valid ...


2

In theory, we do not suppose there are transaction costs (or costs for short selling or even buying a security). In practice, effectively, you will have to pay the people that lend you the security you want to short (this activity is called security lending). What we notice, all the cases, is that if there exists a risk free rate such that $ 1 + r > u$ ...


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

Noob2 has given the answer, a recombining tree has $N+1$ nodes after $N$ time points. In our case, you have $8$ distinct possible values after $7$ time steps. I will next illustrate the difference between a recombining and non-recombining tree. For a binomial tree, if $u=\frac{1}{d}$, then an up-move followed by an down-move gives the same result as a down-...


1

Firstly, recall that American-style option may be exercised at every time point. If you model the stock as a tree, you need to check at every node, whether the investor would like to use his/her right of early exercise. So, how do you price derivatives in general? You build the tree for the stock price and then a second tree via so-called backward ...


1

Say the stock price is x, you short sell one unit of stock, and the proceeds earn r. In the u state, the stock investment will be worth $-ux$, and your bank account will be worth $x\left(1+r\right)$. So total profit and loss will be: $PL=-x u + x\left(1+r\right)= x\left(1+r-u\right)$ And in the d state, your PL will be: $PL=-x d+ x\left(1+r\right)= x\...


1

It essentially boils down to: same random variable, different probability measures. So when you set u and d, you fix the values that the random variable can take. Probability Measure does not change that- it only re-weights the probability in a way. The probability $p_1$ and $p_2$ are the probabilities of the two states under the P(physical) measure, and ...


1

As Alex said, the $X_i$ correspond to i.i.d. log-returns. The benefit is that $$\ln(S_T)=\ln(S_0)+\sum_{i=1}^N X_i.$$ And for sums of i.i.d. random variables, we know the limiting distribution due to the central limit theorem. Then, you can easily show that the stock price in the binomal model converges to a geometric Brownian motion (and hence, you end up ...


1

Answer was provided by Chris Taylor: the formula for the risk-neutral probability was off by a minus sign, it should be $$ p = \frac{e^{r \Delta t} S - S_m}{S_p - S_m} $$


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I think both are right in different ways! Risk neutral measure is the measure associated with the bank account, and the standard definition of bank account is indeed an adapted process. And technical conditions are indeed there for a reason. But then you can have, say the T-measure, which uses zero coupon, whose return is stochastic in each step, except in ...


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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).


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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.


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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.


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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)...


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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.


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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 ...


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