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you have to be careful to distinguish between trinomial trees in a theoretical sense which do not give unique prices, and trinomial trees chosen as an approximation to the risk-neutral measure of the BS model. In the second case, they are an effective numerical method as are binomial trees. Trinomial trees are more useful when you want to ensure nodes lie ...


6

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

There is a good quick well-known approximation for at-the-money options: $$\textrm{Call,Put} = 0.4 S \sigma \sqrt{T}.$$ See further discussion at What are some useful approximations to the Black-Scholes formula?.


5

In binomial tree models, there is no such a thing as a path. The binomial tree represents information about the distribution of the zero-curve at a given time and preserve enough information between different times to let you compute conditional expectations. Generally, you can not price path-dependant instruments in a model based on trees—because there is ...


5

Not all binomial trees take $u=e^{\sigma\sqrt{\Delta t}}$. Thinking of the binomial tree as a discrete approximation (on a grid) to a continuous process, it makes sense that a variety of choices for where to place grid points will work. For a listing of a few different choices of $u$, see the Tian Tree settings and others. From this Sitmo page you can see,...


5

You don't mention if the puts in question are exotic or vanilla, but assuming they are vanilla, you should read this paper by Chen and Joshi. In it, they find optimal performance by using smoothed, truncated Tian-parameter binomial lattices with Richardson extrapolation -- where the idea is to run one extra low-cost (long $\Delta T$) tree in order to ...


5

The model here is the binomial option pricing model, so the second term in the brackets represents the expected future value of the option (under riskneutral probabilities). The aim of the option holder is always to maximize the value of his option. He can at any point sell the option at the fair market price $E(V_{n+1})$ or exercise it to get $G_n$. So if ...


5

you don't need $ud=1.$ In fact, there are now about 30 binomial trees which converge to Black--Scholes in the large step limit. Most of them do not have $ud=1.$ All you need is $$ d < e^{r \Delta t} < u $$ The tree recombines provided $u$ and $d$ don't change from step to step. See my book More Mathematical Finance for a comprehensive review and ...


5

From the gentleman and scholar Emanuel Derman. Emanuel states "the last two pages answer the question asked". https://www.dropbox.com/s/cg299qsbquuqdru/TwitterNotesOnBDT.2017.pdf?dl=0&m= Please thank him directly on Twitter.


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

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 ...


5

There is a deeper relationship between the two risk-neutral measures. Take any event in the binomial model with a finite number of steps and calculate the risk-neutral probability of it. Take the same event in the Black Scholes model and calculate the risk-neutral probability of it. For most events, the two probabilities are different. Now let the number of ...


5

I answer from a general discrete time/discrete state model point of view. This includes the binomial tree model as a special case. In finite dimensions, you can interpret asset payoffs and returns as vectors and retreat to linear algebra. Suppose you have $N$ states of nature and $J$ assets. Your payoff matrix is \begin{align*} A=\begin{pmatrix} X_1(\omega_1)...


4

The condition $$ud=1\text{, or equivalently }u=1/d$$ is necessary to ensure convergence of the Binomial tree's mean $\mu$ and standard deviation $\sigma$ to nonfinite values when $n$ (number of steps) goes to infinity. Cox-Rubinstein-Ross showed in their famous paper, that to achieve this, we must have: $$u=e^{\sigma\sqrt{t/n}}\text{, }d=e^{-\sigma\sqrt{...


4

For a martingale $dX=a(X,t)\,dt+b(X,t) dW(t)$ where $a$ and $b$ are not constant, your tree will not recombine in general [edit]. This is the main issue. See for instance: Florescu, I. and F. G. Viens (2008, March). Stochastic volatility: Option pricing using a multinomial recombining tree. Applied Mathematical Finance 15 (2), 151-181. It deals with the case ...


4

It's a pitty that you don't show in your question how you get to your value for $c_0$ but the idea is that you build a portfolio $X_0 = \Delta S_0 - \lambda$ and you infer the values for $\Delta$ and $\lambda$ so that $X_1 = c_1$ both in the up and down scenario. Then, because of the law of one price, $X_0 = c_0$. So for us $X_1 = \Delta S_1 + (1+r) \lambda$...


4

"But just for fun, let's say Pr(S1=Su)=1% and Pr(S1=Sd)=99%, in which case, on average, the call at time 1 would be worth 0.01*10 = 0.1$. How would anyone be willing to pay 9.28$ for that ? I'm pretty sure I'm missing something very basic, I hope someone can explain what it is." How would anyone pay 100 for the stock given these probabilities? You don't ...


4

there are many different trees. The first one, the CRR tree, used $$ u = e^{\sigma\sqrt{h}} $$ and $d = 1/u.$ However, you can take any real-world drift and still get the same prices in the limit so you can put $$ u = e^{\mu h +\sigma\sqrt{h}}, \text{ and } d = e^{\mu h -\sigma\sqrt{h}} $$ for any fixed $\mu.$ $\mu = 0$ is a poor choice for convergence. ...


4

You have forgotten the combinatorial factors for binomial probabilities on your terms. You need $$ {n\choose k} p^n(1-p)^{n-k},$$ not just $$ p^n(1-p)^{n-k}.$$ The second term should have a factor of $6$ and the third should have a factor of $15,$ etc.


4

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 ...


4

It appears that the motivation for $\mu = (\log K - \log S_0)/T$ may be that K is in the middle of the tree at $T$. I could see how this may improve accuracy since K is where the ‘action’ is. @noob2 I think that in the case of various choices of $\mu$, the up/down probabilities in the tree may be adjusted to give the correct risk neutral expectation for ...


4

This may not be answering your question - but it is worth noting that valuing barrier options on a binomial / trinomial tree is at best problematic. It is difficult to enforce the boundary conditions because nodes will not typically sit on the barrier itself, necessitating some kind of probability-weighted interpolation - which is unlikely to be numerically ...


3

The argument that the American and European call are worth the same is model independent. So it holds for the binomial model. So there is no need to check to see if the early exercise occurs because it won't. Of course, if you have written general purpose code, it is much easier to test for early exercise and always have the test fail than to try and deal ...


3

if you let $\delta t$ be small enough, this won't happen. So the solution is to take more steps. The CRR tree is very out dated in any case.


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In R you can use fOptions package to draw Binomial Tree graphs. Here is a simple code snippet #Install the package and load it install.packages('fOptions') library(fOptions) #Calculate the value of the option and plot optionVals<-BinomialTreeOption(TypeFlag="ce",S=100,X=100,Time=3,r=0.05,b=0,sigma=0.2,n=3,title="example binomial tree") BinomialTreePlot(...


3

You can calibrate the model by discretizing in time, and using a forward induction method as originally proposed by Jamishidian in 1991: F.Jamshidian, Forward Induction and Construction of Yield Curve Diffusion Models, J.Fixed Income 6, 62-74 (1991). Although he formulated this induction in the language of the binomial tree, the method is more general, and ...


3

Actually it is quite simple to demonstrate Ito's correction term in a binomial tree. Details can be found in my new paper (p. 8-10): von Jouanne-Diedrich, Holger: Ito, Stratonovich and Friends (April 21, 2017) Abstract This exposition should provide you with the bigger picture of stochastic calculus, especially stochastic integrals. It heuristically ...


3

To rule out arbitrage in the one-period model, we must assume $$ 0 < d < 1+r < u, $$ where $u$ is the up-factor, $d$ is the down-factor and $r$ is the risk-free interest rate. This chain of inequalities is the no-arbitrage condition. To see what happens if it doesn't hold, consider the case in which $$ 0 < 1+r < d < u. $$ Let $S$ denote ...


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