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## Hot answers tagged binomial-tree

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 $$u = \exp(\sigma \sqrt{\... 5 In binomial tree models, there is no such a thing as a path. The binomial tree represents information about the distribution of the zero-curve at a given time and preserve enough information between different times to let you compute conditional expectations. Generally, you can not price path-dependant instruments in a model based on trees—because there is ... 5 Not all binomial trees take u=e^{\sigma\sqrt{\Delta t}}. Thinking of the binomial tree as a discrete approximation (on a grid) to a continuous process, it makes sense that a variety of choices for where to place grid points will work. For a listing of a few different choices of u, see the Tian Tree settings and others. From this Sitmo page you can see,... 5 The model here is the binomial option pricing model, so the second term in the brackets represents the expected future value of the option (under riskneutral probabilities). The aim of the option holder is always to maximize the value of his option. He can at any point sell the option at the fair market price E(V_{n+1}) or exercise it to get G_n. So if ... 5 You don't mention if the puts in question are exotic or vanilla, but assuming they are vanilla, you should read this paper by Chen and Joshi. In it, they find optimal performance by using smoothed, truncated Tian-parameter binomial lattices with Richardson extrapolation -- where the idea is to run one extra low-cost (long \Delta T) tree in order to ... 5 you don't need ud=1. In fact, there are now about 30 binomial trees which converge to Black--Scholes in the large step limit. Most of them do not have ud=1. All you need is$$ d < e^{r \Delta t} < u $$The tree recombines provided u and d don't change from step to step. See my book More Mathematical Finance for a comprehensive review and ... 5 There is a good quick well-known approximation for at-the-money options:$$\textrm{Call,Put} = 0.4 S \sigma \sqrt{T}.$$See further discussion at What are some useful approximations to the Black-Scholes formula?. 5 you have to be careful to distinguish between trinomial trees in a theoretical sense which do not give unique prices, and trinomial trees chosen as an approximation to the risk-neutral measure of the BS model. In the second case, they are an effective numerical method as are binomial trees. Trinomial trees are more useful when you want to ensure nodes lie ... 5 From the gentleman and scholar Emanuel Derman. Emanuel states "the last two pages answer the question asked". https://www.dropbox.com/s/cg299qsbquuqdru/TwitterNotesOnBDT.2017.pdf?dl=0&m= Please thank him directly on Twitter. 5 Assuming continuously compounded returns for a multi-period model with N being the number of periods: \begin{cases} &\log u \quad \text{with probability q}\\ &\log d \quad \text{with probability 1-q} \end{cases} given the stock price at maturity$$\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$$where ... 5 Note that the tree is recombining. You have u=1.2 and d=0.8 with ud=0.96. Your tree for the asset price reads as At time zero: 100 At time one: 80 or 120 At time two: 64 or 96 or 144 The transition probabilities are q_u=0.55 and q_d=0.45. For your put option with strike price K=104, you thus obtain by backward induction At time two: 40 or 8 ... 5 There is a deeper relationship between the two risk-neutral measures. Take any event in the binomial model with a finite number of steps and calculate the risk-neutral probability of it. Take the same event in the Black Scholes model and calculate the risk-neutral probability of it. For most events, the two probabilities are different. Now let the number of ... 4 For a martingale dX=a(X,t)\,dt+b(X,t) dW(t) where a and b are not constant, your tree will not recombine in general . This is the main issue. See for instance: Florescu, I. and F. G. Viens (2008, March). Stochastic volatility: Option pricing using a multinomial recombining tree. Applied Mathematical Finance 15 (2), 151-181. It deals with the case ... 4 The condition$$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{...

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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 $X_1 = \Delta S_1 + (1+r) \lambda$...

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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.

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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 ...

3

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(...

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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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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

All you need is to use the discretization to implement the MC approach. The following links should get you started: http://www.lcy.net/files/BDT_Seminar_Paper.pdf http://www-2.rotman.utoronto.ca/~hull/TechnicalNotes/TechnicalNote23.pdf http://www.iorcf.unisg.ch/Forschung/~/media/Internet/Content/Dateien/InstituteUndCenters/IORCF/Abschlussarbeiten/Frey%...

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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

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 ...

3

"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 ...

3

Your formula for $p$ is $$p = \frac{e^{{(\alpha - \delta})h} - d}{u - d},$$ where $\alpha$ is not expected return on stock but continuous risk free rate, i.e. 1%. If you use $\alpha$ as 1%, you will get $p=0.009125828$ which is within $[0,1]$ EDIT: With the information given in the question, it must satisfy following equality: S_0e^{\alpha - \delta}=...

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Quick answer The payoff you mention is that of a call spread, i.e. long a call $C_1$ struck at $K_1$ and short a call $C_2$ struck at $K_2$, with $K_2>K_1$. The price of the instrument is therefore: $V = C_1 - C_2$. [First way] If you are stuck because this payout seems 'unsual' to you, an easy way to reach your goal (assuming you know how to use ...

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You are at the beginning of a period and the stock price, worth $S$, can evolve in either of the 2 states: $S_u = u S$ or $S_d = d S$. The part you don't understand is related to forming so-called replicating portfolios. More specifically, using only the stock and a (risk-less) cash account, the question is 'How can one build a portfolio allowing to ...

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