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5

Q: What does the risk-neutral price represent if the option is not replicable? In an incomplete market, there is no unique martingale measure but instead a set $Q$ of equivalent martingale measures. Consequently, there is an interval of arbitrage-free prices: $ \Big( inf_{\mathbf{Q} \in Q} E_{\mathbf{Q}}[DX], sup_{\mathbf{Q} \in Q} E_{\mathbf{Q}}[DX] ...


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The following paper gives you really all of the missing steps in a very detailed form: Black-Scholes Option Pricing Formula by Michael Tomas and Ravi Shukla From the paper: "This presentation is purely for pedagogical purposes. In the course of doing work on option pricing, we found no complete solution for the Black-Scholes model. By complete, we mean ...


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In three bullet points: Efficiency: the obtained prices maximize assumed utilities of different agents. In their paper "The Valuation of Option Contracts and a Test of Market Efficiency", Cohen, Black and Scholes compare the theoretical value of options to their market price. The efficiency is in this sense: can agents obtain more or less in practice than ...


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Most of the time, when you have a simple SDE without a drift, it's a martingale because the Wiener process itself is a martingale. In your example, you have a constant with the Wiener process, therefore the whole process must also be a martingale because the expectation is clearly X(t). However, we can't conclude a driftless SDE is always a martingale. ...


4

I would use the following arguments: If the option were on the first throw of the dice, then we would price it using the expectation, which is $3.5$ (= $(1+2+\cdots+6)/6$. Now we have a 2 stage game: First throw : if the player throws more than $3.5$ points, i.e. $4,5,6$, then there is no sense in throwing again. If he throws $1-3$ then it makes sense to ...


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{, ...


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Very simply, Ross' framework assumes a great deal to extract the true pricing kernel. Time homogeneity, additively separable state dependent utility, (discrete time Markovian structure - though these have been relaxed.) In particular, there are two schools of criticism, one is that time homogeneity makes little sense in the real market. In fact, the Recovery ...


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it doesn;t imply $ \ln S_T=\ln S_0+rT+σW^Q_T$ it implies $ \ln S_T=\ln S_0+(r-0.5\sigma^2)T+σW^Q_T$ look up Ito's lemma. This is covered in just about any book on financial maths including my own Concepts etc


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Your characterisation is correct but incomplete. 1) The most important part of Black-Scholes is not the model but the more general framework of dynamic hedging: you can replicate your payoff by continuously trading the underlying and the amount (delta) you should hold is the derivative of the current premium with respect to the current spot. This is a much ...


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This drift comes from making the discounted stock a martingale in the risk-neutral measure $\mathbb Q$ You start with a stock in $\mathbb P$ having this form: $$ dS_t = \mu S_t dt + \sigma S_t dW_t $$ You also have a discount factor $e^{rt}$. The idea is to remove the drift of the discounted process in $\mathbb Q$ so you get (after applying Girsanov's ...


3

Assume the price follows a lognormal process. We can convert it into a problem of finding the probability of a standard Brownian motion particle starting from $0$ and hitting $x$ before time $t$, or its first passage time $\tau_x$ being less than $t$. This can be derived through the reflection principle. The paths crossing $x$ are exactly paired up by the ...


3

The error is, you are not storing the random numbers for the same path at the end: xbefore = x + c*tau + sigma*sqrt(tau)*randn() A = muA + sigmaA*randn(); xafter = xbefore + A; But then at end you set a different path here by creating a new random number: xT = log(S0)+(c+muA*lambda)*T+sqrt((sigma^2+(muA^2+sigmaA^2)*lambda)*T)*randn(); randn() ...


3

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 ...


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These options can be priced by adding an early exercise premium value to the intrinsic value: http://www.statistics.nus.edu.sg/~stalimtw/PDF/lb-float.pdf


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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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There are lots of papers online and here are a few I would suggest math.umn riskworx G. Dimitroff, J. de Kock Nowak, Sibetz I you have matlab there is an step step example to calibrate SABR model. Since it uses the financial toolbox of matlab for a few functions I dont think you can replicate it in any other language. There must be C++ code available ...


3

You can view the price of an option as the cost to dynamically replicate it. The more volatility, the more costs you will have trading the underlying to keep your delta equal to 0 (I'm assuming you sold the option, hence a negative gamma position). So, if at any spot, any date your local vol is above 0.194, rebalancing the portfolio will be constantly more ...


3

Think of moving volatility in the other direction. As volatility approaches zero, any call strike strictly smaller than the ATM strike, $K<K_{ATM}$, will have zero probability of ending in the money, and the corresponding option value will be zero. An infinitesimally small change in stock price will not move $K$ past $K_{ATM}$, so the option value ...


3

I found and answer to my own question. So, I post it here for people who maybe have the same problem. The answer, however, is quite intuitive. The last observation used for the estimation of the physical density is also the time point where the investors know the most about the physical density because at this point the most possible historical observations ...


3

Simply put, no. Vega depends on a variety of factors (including the level/price of the underlying asset). However, vomma/volga/vega convexity (whatever you want to call dVega/dIV) is always positive. So as IV increases, the vega of an option increases - I think this might have been what you were getting at. It's important to understand that IV is an input ...


3

Since American style options allow early exercise, put-call parity will not hold for American options (unless they are held to expiration). In practice, there is also a difference between calls and puts for European options as well. The full description is here: What causes the call and put volatility surface to differ?


3

I don't know the BS formula you are trying to use. The price is the expected value of the discounted payoff under the risk neutral probability measure (I.e. Under which S is a martingale) So the you need to compute the risk neutral probabilities for S to go up or down. The probabilities given in the problem have no impact. They are just there to trick the ...


3

The above equation is the price of a call option. It has nothing stochastic inside it. It only depends on the current price and the time. So no Ito is needed. You should just compute the derivatives of your solution v (like you do for any deterministic multivariable function), plug them into the PDE and verify that it's satisfied.


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?.


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You only have one asset in your portfolio which means that you can only statically hedge. By the definition of self financing, $V_0=\phi_0 S_0$, $V_1=V_0+\phi_1 (S_1-S_0)$, and $V_1= \phi_1 S_1$. Putting these last two together, $V_0=\phi_1 S_0 $. Hence $\phi_1=\phi_0$ and you have a static position. Intuitively, this is because you cannot trade in ...


2

LSM is very fiddly. The most important things in my view are 1) don't believe anyone who says that the choice of basis functions doesn't matter. 2) implement an upper bounder, eg Andersen--Broadie (2003) or Joshi-Tang (2014) so you can tell if your prices are good 3) do two passes, one to build the strategy, one to price, if they give very different ...


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Pricing a bond futures contract is already a very difficult task (because of the embedded delivery option), not to mention an American option on it. Bottom line, you'll need to build either a tree/lattice or run some monte carlo simulations. Here's a sketch of how you could go about doing it – 1) Using a term structure model, generate the distribution of ...


2

The problem with your formula is the equation sign $=$. The second order finite difference is only an approximation to the true gamma: $$ f^{\prime \prime}(x) \approx \frac{f(x+h)-2f(x)+f(x-h)}{h^2}. $$ $h$ can not be a result. Ideally, it should be small (whatever that means), so your original choice of $1\text{bp}$ seems appropriate for this ...


2

Historical returns are not to be used 'untreated' for the calculation of option prices. The expectation that you will be using in Monte Carlo will take the form $$ C(K,T) = E^Q\{D(T)\ \max[0, S_T-K, 0]\} $$ where $T$ is the maturity, $K$ is the strike price, $S$ is the stock price and $D$ is the discount factor. But the expectation is taken under the 'risk ...


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I think you are right. Now when I check papers I've used for my thesis I don't see almost any with empirical data section. Maybe this one will be helpful: Roswell E. Mathis, III, Gerald O., Bierwag Pricing Eurodollar Futures Options with Ho and Lee and Black, Derman, and Toy Models: An Empirical Comparison



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