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6

the problem is that the pay-off has discontinuous first derivative. Try a contract with pay-off that is twice differentiable and it will probably work. The problem is that all the value comes from the tiny number of paths within $\Delta S$ of the strike, and these paths have huge value. This is a well-known problem. As the bump size goes to zero, the ...


5

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 ...


5

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 ...


5

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 ...


5

By definition the fair value of an option is given by an expectation value of the payoff, $\mathbf{E}\left[\textrm{payoff}(\textit{paths})\right]$. The probability distribution of the paths is the risk neutral measure. This is just an integral expression of the form you wrote. This applies to all option prices. Many options are, of course, special in the ...


5

importance sampling is well known to be tricky. See the extensive discussion in Glasserman's book. I presume that you are simply meanshifting and multiply by the ratio of normal densities. For this sort of problem, I'd use a more stratified algorithm instead and force every path to end in the money. To do this I'd compute the uniform that goes to the ...


4

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

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

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

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 ...


4

You Know that $dB_t=r_tB(t)dt$ . Ito's formula give us \begin{align} dZ(t)=\frac{1}{B(t)}d\,\Pi(t)-\frac{\Pi(t)}{B\,^2(t)}dB(t)+0 \end{align} As your teacher mentioned, $d\Pi(t)=r(t)\Pi(t)dt+\sigma(\Pi(t),t)dW(t)$,Thus we have \begin{align} & dZ(t)=\frac{1}{B(t)}[r(t)\Pi(t)dt+\sigma(\Pi(t),t)dW(t)]-\frac{\Pi(t)}{B\,^2(t)}r(t)B(t)dt\\ & ...


4

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


4

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 ...


4

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 ...


4

I know two papers explaining how to calibrate this kind of models, and one of them explain the impact of the quality of the fit on a pricing model: Aït-Sahalia, Y. (2002, January). Maximum likelihood estimation of discretely sampled diffusions: A closed-form approximation approach. Econometrica 70 (1), 223-262. Azencott, R., Y. Gadhyan, and R. Glowinski ...


4

The Heston model is represented by the bivariate system of stochastic differential equations (SDE) \begin{align} & d{{S}_{t}}=rS_tdt+{\sqrt\upsilon_t} d{{W}_{1}}(t) \\ & d{{\upsilon}_{t}}=\kappa(\theta-\upsilon_t) dt+\sigma{\sqrt\upsilon_t}d{{W}_{2}}(t) \\ \end{align} The most popular way to estimate the parameters of the Heston model is with loss ...


4

First we write dynamic of ${{x}_{t}}=\ln ({{S}_{t}})$ \begin{align} & d{{x}_{t}}=({{r}_{t}}-\delta -\frac{1}{2}\sigma _{t}^{2})t+{{\sigma }_{t}}d{{W}_{1}}(t) \\ & d{{\sigma }_{t}}=a({{\sigma }_{t}},t)dt+b({{\sigma }_{t}},t)d{{W}_{2}}(t) \\ & d{{r}_{t}}=\alpha ({{r}_{t}},t)dt+\beta ({{r}_{t}},t)d{{W}_{3}}(t) \\ \end{align} Let \begin{align} ...


4

we should first define some notation before discussing pricing. Let $t_0$ be initial time and $ t_1, . . . , t_M$ be pre-specified exercise dates with $t_0 < t_1 < · · · < t_M = T$ , the final maturity, and $Δt = t_m−t_{m−1}$. Without a loss of generality it is assumed exercise dates are equidistant. To price a Bermudan option, its value is split ...


3

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 ...


3

This is a bit of an old question, but I thought I'd contribute to add more weight to to what some people have been saying. A CSO (calendar spread option) is NOT a calendar spread of options. If you read it carefully, you can see the Hull quote Max Li posted is talking about a calendar spread, not a CSO. A CSO needs to be priced the same way as a spread ...


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

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


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 ...


3

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

this is probably the most asked question in quantitative finance... There are many answers. One nice example to consider is what if the calls were struck at zero. The call then pays the stock price at time $T$ and so it's value today must the stock price today since we can replicate by holding one unit of stock. This will be true regardless of the drift of ...



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