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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

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

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 ...


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

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

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 ...


3

For a call option, the payoff is given by $(S_T-K)^+$. Note that the function $x^+$ is convex, then, by Jensen's inequality, the price $c$ satisfies \begin{align*} c &= e^{-rT}E\big((S_T-K)^+\big) \\ & \geq e^{-rT}\big(E(S_T-K)\big)^+\\ &=\big(S_0 - K \, e^{-rT}\big)^+. \end{align*} For the upper bound, note that \begin{align*} c &= ...


3

The No-Arbitrage bounds for a European put are: $$ (Ke^{-rT}-S)^+ \leq P \leq K e^{-rT}$$ This is because the maximum payoff at maturity is $K$ (discounted) and the minimum value is the discounted intrinsic value (since $E(e^{-rT}S_T)=S_t$ by the martingale condition and the payoff being always semi-positive).


3

Fact 1: if you are not good at pricing options, of course you can create a lot of arbitrage opportunities for the rest of the market. It does not matter whether the reason is in dividends or anything else. Fact 2: if you are good in pricing options, you price the dividend effect in advance. Consider the situation of the European calls, and suppose that both ...


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

First Question:This derivation is a special case of a PDE for general stochastic volatility models,described in books by Lewis (2000), Musiela and Rutkowski(2011) and others. The argument is similar to the hedging argument that uses a single derivative to derive the Black-Scholes PDE. In the Black-Scholes model, a portfolio is formed with the underlying ...


3

I know one article (download) that explaining how to calculate local vol surface from IV surface and also chapter 18 of this book is very good In this context. However you know that Dupire’s (1994) formula for local volatility is \begin{align} \sigma_L(k,T)=\sqrt\frac{\frac{\partial C}{\partial T}}{\frac{1}{2}K^2\frac{\partial^2 C}{\partial K^2}} \end{align} ...


3

When $\sigma=0$ , the boundary condition is little more complicated: \begin{align} P_t+(r-\delta)SP_S +\alpha P_r +\beta^2\frac{1}{2} P_{rr}-rP=0 \end{align} When $\sigma\rightarrow\infty$ , we have \begin{align} P(S,\infty,r,t)=0 \end{align} When $r=0$ , then \begin{align} P_t+aP_\sigma+\frac{1}{2}b^2P_{\sigma\sigma}+\sigma S b \rho_{12}P_{S\sigma}=0 ...


3

Answering my own question as it could be useful for others. Actually package fOptions is vectorized. The only constraint (and that make sense) is that you can't compute at the same time 2 different greeks, or mix up calls and puts. So assuming that you want to compute the delta of a set of puts, the code will be the following: ...


2

you can do the bounds without using a model or martingales. At maturity $$ 0 \leq C \leq S_T $$ with positive probability of strict inequalities. So before maturity, $$ 0 < C < S_t. $$ Since if these are violated, you can make an arbitrage. eg if $C \geq S_t$ hold $S_t - C$ to get a profit with positive probability and no chance of loss. Similarly, ...


2

One does not estimate the local volatility at a given $T$ and $K$. Instead, Dupire's formula actually gives $\sigma(T,K)$ for all $T$ and $K$. $$ \sigma^2(t_0,S_0;T,K)= \frac{\frac{\partial C}{\partial T} + (r - q)K \frac{\partial C}{\partial K} + qC}{\frac{1}{2} K^2 \frac{\partial^2C}{\partial K^2}} $$ where $C(t_0,S_0;T,K)$ are the call prices for ...


2

$V_t $ is the price of a tradeable. Because we can delta hedge it, $V_t =v(t,S_t) $ where $v$ is a solution of the PDE on some domain whose boundary corresponds to the exercise of the option. For a European option with payoff $g(S_T)$ at time $T$, the price function $v$ is the solution $$ \frac{\partial v}{\partial t} + ...


2

Generally no, because 'dividends' are already 'priced into' the options. Which means, if an ATM call cost 0.50, and stock price drops by 1.00(amount of dividend), the ATM becomes OTM, but it may still cost 0.50, because the initial price of 0.50 already factored in the dividend.


2

I would argue that there is some path-dependency involved. The BS model is considered the big breakthrough and it presented the world with some kind of tractable toy model. After that people saw that you had to adjust the model to account for all kinds of stylized facts (e.g. non-constant volatility for different strikes, over time and so on). Yet finite ...


1

Given two representations: $$ C = E_f[\varphi(X)] = \int \varphi(x) f(x)dx = \int \varphi(x) \frac{f(x)}{g(x)}g(x)dx = E_g[\varphi(X)\frac{f(X)}{g(X)}] $$ The difference of the variances of the MC estimators associated with the two expression is $$ Var[\widehat{C}^f_N] - Var[\widehat{C}^g_N] = \frac{1}{N}\int \varphi(x)^2 \left(1 - ...


1

A convex function is when the line between two points on the graph always lies above the graph. And this does hold for the put, its also sometimes called a sublinear function. Also see http://en.wikipedia.org/wiki/Convex_function So the author is correct in saying that $(K-s)^+$ is convex.


1

This is the Black Scholes Call Price: \begin{align} C(S, t) &= N(d_1)S - N(d_2) Ke^{-r(T - t)} \\ d_1 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S}{K}\right) + \left(r + \frac{\sigma^2}{2}\right)(T - t)\right] \\ d_2 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S}{K}\right) + \left(r - \frac{\sigma^2}{2}\right)(T - ...


1

It's a combination of too few sample paths and/or too small an increment. Your estimation error on the price is magnified by the $dS^2$. Try using a larger sample or a larger increment. Alternatively, you can use a multiplier instead of a fixed increment; in my experience, it usually yields better results.



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