# Tag Info

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When you say 'overprice' I assume you mean model price > market price. In my experience this is true for all reasonable models. It's due to excessive supply of the Bermudan structure in the market.

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The most rigorous approach I have seen so far eliminating the risk premium is this one: Emanuel Derman: The Perception of Time, Risk and Return During Periods of Speculation (2002) Equation 2.23 on page 11 derives $\mu$ ~ $r$ but it only holds in the limit when you hypothesize countless uncorrelated stocks in a diversifiable market. Still an interesting ...

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well there are lots of things to get right... first you need to the non-callable version right, to get that right requires getting the smile right since a callable range accrual is really just a bunch of digitals with timing effects. these days discounting and forwarding are done with different curves so you'll need to get that right too. then you'll ...

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This question is extremely interesting and not that straightforward. See answer here. From a financial perspective this is very much like pricing an American call (stopping rule = intrinsic value from exercice (i.e. current cash earned) > continuation value (i.e. what you can expect to gain). Note that you can never win more than 13 nor lose (at worst you ...

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Let $\sigma(F,K)$ be the SABR implied vol. In the shifted model, the formula essentially becomes $\sigma(F+x,K+x)$ (you have to shift the strike as well). So to answer your question in the ATM vol calibration you take $K=F$ in order to have $F+x=K+x$. There is no need to "reconcile" anything as it is just a model. Once you have your model, you have to ...

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There is a logical fallacy in your argument. The price of a European call expiring 1 day before a dividend payment may well be greater than that of a call expiring after it. In other words, claiming that $$C_E (S_0,K,t_D-1\text {day}; D, t_D) < C_E (S_0,K,T; D, t_D)$$ is not necessarily true. Try the above inequality with a huge dividend (e.g. $D ... 1 I think you got it. Wrapping up: Usually denoted by$(\mathcal {F}_t)_{t \geq 0}$, a filtration is a series of adaptive subsets of the$\sigma$-algebra$\mathcal{F}$that keeps track of what really happened as time went by (i.e. fixed$\omega$). Over the probability space$(\Omega, \mathcal{F}, \mathbb{P})$, a random variable$X_t $is measurable iff ... 0 I don't agree with the contention that market prices are always used as the benchmark upon which to base model performance. I think this is model dependent. Market prices make sense (for example for modelling an underlying predictor), but for example, for a derivatives model, I would argue the values of those derivatives at expiry (or earlier for path ... 6 $$\begin{array}{rcl} (1) & \partial_KC_t(T,K) & \leq 0 \\ (2) & \partial^2_KKC_t(T,K) & > 0 \\ (3) & \partial_T C_t(T,K) & \geq 0 \\ \end{array}$$ If$(1)$doesnot hold, it exists$K_1<K_2$such that$C_t(T,K_1)<C_t(T,K_2)$. Then as barrycarter said in his comment, you sell$C_t(T,K_2)$and you buy$C_t(T,K_1)$, so your ... 1 Under the risk neutral measure, the expected present value of the butterfly payoff is: $$V_0 = e^{-rT} * \int_{S_T=K_1}^{K_3}P(T,S_T)f_{S_T}dS_T$$ And if we assume that$f_{S_T}$is constant from$K_1$to$K_3, then: $$V_0 = e^{-rT} * \dfrac{1}{\Delta K} \int_{S_T=K_1}^{K_3}P(T,S_T)dS_T = e^{-rT} *\dfrac{\delta^2}{\Delta K}$$ 4 Peter Jaeckel has written various papers on this. "by implication" and "Let's be rational" are the most recent ones. He also provides code on his website www.jaeckel.org. (Note: the question asked for literature.) 1 Look on Google for Asymptotic behavior of Implied Volatility Near Infinity you will find results like : $$I(K) \stackrel{K\to\infty}{=} \sqrt{\frac{2}{T}}\left(\sqrt{\ln \frac{K}{C(K)}}-\sqrt{\ln\frac{1}{C(K)}}\right) +\text{O}_{K\to \infty}\left(\frac{\ln\ln\frac{1}{C(K)}}{\sqrt{\ln\frac{1}{C(K)}}}\right)$$ 0 We assume that the inequality is given by \begin{align*} B > N C(K-1/N, T) - N C(K, T).\tag{1} \end{align*} The argument for the case with the inequality \begin{align*} B < N C(K, T) - N C(K+1/N, T) \end{align*} is similar.  For the binary option, \begin{align*} \pmb{1}_{\{S_T \ge K\}} = \begin{cases} 1, & \textrm{if } S_T \ge K,\\ 0, & ... 0 I think you are confused with what's exactly log-normally distributed. The distribution of option prices can't be normal or log-normal because the prices can't be negative. In general, we don't model option prices, we model the underlying stochastic processes (i.e: geometric brownian motion, mean-reverting etc). We then use the distribution of those ... 0 In general, an option payoff cannot be normal, as the payoff is generally positive, while a normal variable can be negative. For a standard call option, the distribution function can be computed from the distribution of the underlying stock. Specifically, consider the vanilla European option payoffX=(S_T-K)^+$. Then, for$x < 0, \begin{align*} P(X \le ... 0 Let's take a call option on a stock with exercise price K. What is the risk-neutral probability of a payoff x? Probability (x=0) = p(stock <=K). Also we have prob(payoff = x>0) = probability (stock=x+K). Hence the required density function f has two parts, (a) an accumulation point at zero representing the probability of being out of the money and ... 2 As I mentioned above, I am not sure what the variabler$is. If we ignore that, or assume the questioner wanted to say its the risk free interest rate, then it has no effect on the number of paths. Then it is clear that after 50 steps going from \$1024 to \$2500 requires a net of 4 up movements with the given$x=y^{-1}=1.25$. Thus the number of steps ... 1 Two hints : The number of paths never going up to$3125$when starting from$1024$and stepping up by a multiplicative factor of$5/4$and down by a multiplicative factor$4/5$is the same as the number of paths starting from$0$and and stepping up by an additive factor$+1$and stepping down by an additive factor of$-1$and never going up to$5$Let ... 7 This is an interesting and not so easy question. Here's my 2 cents: First, you should distinguish between mathematical models for the dynamics of an underlying asset (Black-Scholes, Merton, Heston etc.) and numerical methods designed to calculate financial instruments' prices under given modelling assumptions (lattices, Fourier inversion techniques etc.). ... 1 [Short Answer] You write$E [S_T]=S_0(1+r)^T $but you actually compute the RHS as$X (1+r)^T$in your numerical application. [Long Answer] The stock price is a martingale in an equivalent measure using the risk-free asset as numeraire i.e. $$E [S(T)] = (S_0 u) q + (S_0 d) (1-q) = S_0 (1 + r ) \Delta t$$ In that case, dividing each member by$S_0\$ and ...

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