# Tag Info

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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

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 ...

4

You have to look at the terms and conditions on your individual bond. The way the specifications usually work is that a call will result in accrued interest being paid, effectively making up for the lost coupon. Sometimes there's even an extra penalty. A put will result in a loss of coupon in almost all cases, and so is almost always done just after a ...

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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 ...

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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 ...

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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 ...

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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{, ... 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 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) ... 4 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 ... 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") ... 3 A condition for correct calibration of the short rate model is that it exactly reproduce the present values of fixed (option-free) cashflows - that is, that it give the same answer as ordinary discounting at the spot rate. If it doesn't, you've done something wrong - sort of like using a model that violates put-call parity. (Actually, it's exactly like ... 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 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 ... 3 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?. 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. 3 You know that the Ho-Lee model is represented by the stochastic differential equations \begin{align} dr_t=\lambda_t\,dt+\sigma\,dW_t \end{align} In order to Implementation our binomial tree, we use the Euler discretization. \begin{align} r_t=r_{t-\Delta t}+\lambda_{t-\Delta t}\,\Delta t+\sigma\,\sqrt {\Delta t} \,Z \end{align} where Z is a standard normal ... 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 Your formula for p isp = \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 - ...

2

I would create separate estimates for volume and choice of debt instrument. There are tools to estimate these simultaneously but I do not see a compelling advantage here. I assume the volume is conditional on the choice of debt issuance so you might start by predicting choice of debt issuance and use this as an input to the volume model. The volume model ...

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Let's start with question (2). If you are not obtaining $S=1.5295e+009$ after backwardation, then you have a bug in your binomial tree code. You may wish to find and eliminate that before proceeding. One simple check is to make all the terminal nodes have value 1.0. You should obtain that the initial node has value $e^{-rT}$. This assumes, of course, ...

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Under the typical Black-Scholes model, you "cannot" do it, because the assumption is that each of the securities in the portfolio has a lognormal terminal distribution, and the sum of lognormally distributed variables it not itself lognormally distributed. In theory one needs an N-dimensional tree (or grid) to treat an N-element portfolio. I write "cannot" ...

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I see your porblem - Hull unfortunately does not explain the reasoning behind the approach. The hint the books gives is correct. Using Taylor series $e^x$ can be written as $e^x = 1 + x + \frac{x^2}{2}+...$. Hull also incoporates a dividend rate $q$ but we can disregard it here. $p$ is given by $p=\frac{e^{r\Delta }-d}{u-d}$. We also have $u=\frac{1}{d}$. ...

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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 ...

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It's more likely that people are familiar with the below representation - John Hull - of a 2-step binomial tree that you described. However, by analogy, your representation (although not conventional) is fine too, as long as S,u,d are materialized by the index increments [i,j] at each node. i++ for each upward / downward trend (1 increment) j++ for each ...

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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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"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 ...

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I think they have expanded the third term on the LHS and simplified.

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As I mentioned above, I am not sure what the variable $r$ is. If we ignore that, or assume the questioner wanted to say its the risk free interest rate, then it has no effect on the number of paths. Then it is clear that after 50 steps going from \$1024 to \$2500 requires a net of 4 up movements with the given $x=y^{-1}=1.25$. Thus the number of steps ...

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I read the question as follows: You have one stock $S_0$ and after one period it either goes up to $S^+$ where the option takes the value $f^+$ or it goes down to $S^-$ where the option takes the value $f^-$. The bond grows from $B_0$ to $B_1 = B_0 \exp(r)$. Then you need to solve $$a S^+ + b B_1 = f^+ \\ a S^- + b B_1 = f^-$$ for $a,b$ which are $2$ ...

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