# Tag Info

## Hot answers tagged option-pricing

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.). ...

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 ...

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.)

3

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 ...

2

As I mentioned above, I am not sure what the variable $r$ 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

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 ...

1

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 ... 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: $$V_0 = e^{-rT} * \dfrac{1}{\Delta K} \int_{S_T=K_1}^{K_3}P(T,S_T)dS_T = e^{-rT} *\dfrac{\delta^2}{\Delta K}$$ 1 Look on Google for Asymptotic behavior of Implied Volatility Near Infinity you will find results like : $$I(K) \stackrel{K\to\infty}{=} \sqrt{\frac{2}{T}}\left(\sqrt{\ln \frac{K}{C(K)}}-\sqrt{\ln\frac{1}{C(K)}}\right) +\text{O}_{K\to \infty}\left(\frac{\ln\ln\frac{1}{C(K)}}{\sqrt{\ln\frac{1}{C(K)}}}\right)$$ 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 ... 1 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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