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## Hot answers tagged binomial

7

As anticlimactic as this may be, I'm going to answer my own question here.. I found this article that shows the connections between the two models.. http://epublications.bond.edu.au/cgi/viewcontent.cgi?article=1126&context=ejsie (mirror) that shows that prices do converge as N Periods increases. Also they provide all the Excel formulas to recreate ...

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 $$u = \exp(\sigma \sqrt{\... 5 The model here is the binomial option pricing model, so the second term in the brackets represents the expected future value of the option (under riskneutral probabilities). The aim of the option holder is always to maximize the value of his option. He can at any point sell the option at the fair market price E(V_{n+1}) or exercise it to get G_n. So if ... 4 The Black-Scholes price of this option is approximately 14.8. When I run a Monte Carlo simulation with 10000 paths and "exact" time stepping, I get results very close to this value. You are simulating the terminal asset price with the first-order Euler approximation over multiple time steps:$$S(t+\Delta t)= S(t) + rS(t)\Delta t + \sigma S(t)\sqrt{\...

4

The 1.04% are used in the calculation because it is 95% expected shortfall so you want to calculate the expectation on the 5% worst loss. In your problem there is 3 possible outcomes: loss of 200, 100 or 0. As the probability of loss of 200 or 100 is 0.04+3.92 = 3.96% < 5%, you need to take account of the loss of 0$for 1.04% part to reach the 5%. 4 Other than the binomial expansion, what is the relationship between$B$and$w$? As I understand correctly, there is no immediate relationship between the lag operator$B$and the weights$w$. However, going from the fractional differencing operator,$(1-B)^d$to$\hat{X}_tcan be done as follows: \begin{align} (1-B)^d X_t &= \sum_{k=0}^{\infty} \begin{... 3 As your code works for the short maturity case, I assume that it is correct. The volatility of80 \%$is simply huge. Thus the area covered by the paths is huge too. As you can read e.g. here the sampling error is proportional to the variance of the process, which is huge in your case. As a brute force solution you can just enlarge the number of samples. ... 3 From your answer to my comment, here is what I would do. Over the horizon$[0,\Delta t]$, the BS model tells you that the expected log-return is $$\Bbb{E}\left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] = \left(\mu-\frac{1}{2}\sigma^2\right)\Delta t$$ with a variance $$\Bbb{V}\left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] = \sigma^2 \... 3 This sounds like the first chapter of Björks book am I right? It treats a single-stage model. Simply put, if 1+r \leq d you buy the stock and have V_1\geq 0 with positive probability of making a profit. If 1+r \geq u you want to sell the stock short and buy the bond from the proceeds. The result is the same. Edit: To show that the condition is ... 3 Since all your options have the same strike, you do not have any "explicit" skew or smile exposure in your portfolio. If I had to guess, almost all of your P&L can be explained by primary exposures, with some Theta losses offset by your Gamma scalping and Vega gains. An example of a book with an explicit smile exposure would be a vega-neutral fly - you ... 3 It is quite common to see non-smooth convergence in tree models and this is not specific to digital options. The problem usually that the tree is constructed independent of the contract to be priced. Thus, the location of the strike relative to the two surrounding nodes might vary widely between two successive step sizes. For European plain vanilla options, ... 3 The pricing of options is married with the concept of a hedging strategy that replicates the effect of the option. If you can only long or short a stock that will not replicate the greeks, it only creates delta. It is the commitment to the strategy that achieves it. For example if the price goes up and you are committed to buying more to increase your delta ... 3 I think you have the correct understanding. The arbitrage is only possible if the risk-neutral probability distribution of the stock is perfectly known, as it is in your simple binomial model. In the real world you can never know the precise distribution, so you cannot create a true arbitrage between an option and its underlying stock in this manner. ... 3 Im going to hazard a guess that your problem is u**(N-i). Large exponents are notoriously poor performers, I would first look to restructure that aspect of the code and then isolate other poorly performant sections afterwards. For example you might observe that: S_T[i] = S0 * u**(N-i) * d**(i) is equivalent to: S_T[i] = S0 * u**N * (d/u)**i then u**N ... 3 Say there are just two periods: Payoff at n, and premium/price at n-2. We know the current stock price, say S_{n-2}, and we know in the next epoch, it will either be: S_{n-1}^u=uS_{n-2} (up state), or S_{n-1}^d=dS_{n-2} (down state). We need to make the decision at epoch n-2 as to how many units of the stock to buy or sell to hedge the option, and ... 2 FYI, the binomial distribution converges to normal as n goes to infinity, which is a nice way of thinking about the relationship between BS and the binomial tree models. see here 2 first you have to find the p, u and d that match standard deviations in part (I). The problem is a little underdefined in that even if you match mean and standard deviation precisely, there are multiple solutions. Impose an extra condition to get all three. Once this is done compute q via$$ q = \frac{e^{r \delta t} - d}{u-d}. $$The state-price ... 2 You agree that the proposition is proven if the equations have a unique solution. You agree that there is a unique solution if u>d. Then we just have to show that u>d. But the definition of u and d is that we have a binomial model where there are two possible outcomes for the stock, a higher outcome su and a lower outcome sd. Hence u>=d by assumption , ... 2 There is one condition under which the risk neutral probability of an event can be zero: if the real world probability is zero. If not then any contract that pays off in that event must go down in price if the contract is modified as to not pay off or pay off less in that event. Otherwise, one can buy one and sell the other... it's arbitrage in the "free ... 2 The call is worth more unless the risk free rate is zero. Let p be the probability of S_0 going up, r be risk free rate, T is the one time step. Then no arbitrage means$$ S_T = S_0 \exp(r T) = p S_0 u + (1-p) S_0 d$$. I am assuming u and d, which you did not say, are the up and down factors. Then obviously$$ S_0 \exp(r T) >= (S_0 \exp(r T) +... 2 I have checked the answer to my side, even using the alternative, the original formula$u=e^{\sigma \sqrt{\delta t}} $, I still get$1.05943$I suspect it is not really a logic error, but more of a truncation error in calculating the$\sqrt{1/12}$2 Going through his code, seems like he has used this formula to come up with the figure 1.0604:$u=\frac{1}{2}\left(e^{-r \delta t}+e^{(r+\sigma^2) \delta t}\right)+\sqrt{\frac{1}{4}\left(e^{-r \delta t}+e^{(r+\sigma^2) \delta t}\right)^2-1}$and then$d=\frac{1}{u}$Here is the relevant part of Wilmott’s code: 1 hmm, Trigeorgis seems to be saying that the value of being able to switch at one of the times 1,2 and 3 is the same as the sum of being able to switch at each of the times. This seems wrong to me since if you switch at time 1, the value of the ability to switch at time 2 becomes worthless. Valuing an exchange option on a binomial tree is pretty easy -- ... 1 Look at slide 3 of Mod 4: Fixed Income Derivatives: Bond Forwards, that's the relevant equation. $$G_0 = \frac{E_0^\mathbb{Q}[Z^j_t/B_t]}{E_0^\mathbb{Q}[1/B_t]}$$ Where$G_0$denotes the price of the forward at$t=0$,$E_0^\mathbb{Q}$is the risk-neutral price at$t=0$,$Z^j_t$denotes the ex-coupon price of the bond at time$t$and state$j$, and$B_t$... 1 Let's illustrate with a one step tree. Take a call option. Without even making a specific assumption about the payout of the option, except that it will be greater in case of an uptick than a downtick:$f_u>f_d$. The price at time 0 for the option will be$f=(1+r)^{-1}f_u$by the risk-neutral valuation formula, since you assume$1+r=u$. Sell the option ... 1 Black Scholes can be seen as the continuous limit of a binomial model when the number of steps go to infinity. (It can be seen as a result of Donsker's theorem) Thus it is normal that your call price in the one-period model is different than the one in the BS model. If you have$n$steps in your binomial to describe the period$[0,T]\$ and if your ...

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I think the proof has already been provided at the end of the proof in Shreve's Theorem 4.4.5. Specifically, note that, since \begin{align*} \frac{1}{(1+r)^{n \wedge \tau^*}}V_{n \wedge \tau^*}. \end{align*} is a martingale, \begin{align*} \tilde{\mathbb{E}}\left(\frac{1}{(1+r)^{N \wedge \tau^*}}V_{N \wedge \tau^*}\right) &= V_0 = \max_{\tau \in S_0} \...

1

It is as simple as just taking the max(). The problem is that you took the wrong one. You must consider the max between the intrinsic value of the option on the one hand and its discounted continuation value (which is an expectation in the risk-neutral world) on the other. In your final loop, you should therefore replace the line W(j+1,1) = max(K-W(j+2,1)...

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Increase the number of paths in your simulation for the getting the terminal prices, and at some point your monte carlo option price will finally converge to Black scholes option price as you are using a very longer maturity call option i.e. 10 year call option.

1

you need a positive dividend rate or a negative interest rate. Without these, it is a model-free result that early exercise is never optimal for a call option.

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