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I give you a brief outline about some key properties of Lévy processes. Lévy processes have three components: linear drift Brownian diffusion jumps. ($\to$ Lévy–Itô decomposition) You can model big jumps which occur as rare events (finite activity models) or consider a process which has infinitely many jumps during any finite time interval. Such ...


5

You are an investment bank. You trade a multitude of vanilla and exotic options. You want to make sure the option prices you quote as a client are arbitrage-free with respect to liquid option prices quoted in the market $-$ and also consistent between the different trading desks within your bank. Basically you want to avoid other market participants taking ...


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There is a deeper relationship between the two risk-neutral measures. Take any event in the binomial model with a finite number of steps and calculate the risk-neutral probability of it. Take the same event in the Black Scholes model and calculate the risk-neutral probability of it. For most events, the two probabilities are different. Now let the number of ...


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The answer in Implied Vol vs. Calibrated Vol as suggested by noob2 is more complete. But it may be slightly misleading in your last example. I've been a vanilla option market maker for ten years, so I'll chime in on what I would mean by that. If a market maker says he's calibrating his vols to the market it means exactly what you're saying: getting prices ...


2

Not sure if the problem is down to boundary conditions - could be a lot of other things, but the boundary conditions in the simplest form for call options are: For very large S: $V\left(t,S\right) \approx S$ For very small S: $V\left(t,S\right) \approx 0$ For put option, the equivalent boundary conditions are: For very large S: $V\left(t,S\right) \...


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It is poorly worded. Replace "options markets" with "stock markets" and it becomes clear that they're just noting typical spot vol correlation. When stock markets trade up volatility tends to fall and when stock markets trade down volatility tends to rise. This has been the case, on average, historically. For more on why, see here: Why does implied ...


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Cant talk specifically to stock pricing models but in foreign exchange the list in order of use goes: Geometric Brownian motion with time dependent vol and drift Local Volatility, either SABR or some other parametric or cubic-spline+Dupire Heston's stochastic volatility model Stochastic-Local hybrid volatility models, usually some from of parametric local ...


2

In the theoretical BSM case, where you are hedging continuously, there is no such risk. And in Geometric Brownian Motion there are no jumps. However once you rehedge at discrete time intervals (no matter how small) Gamma Risk shows up. It can be defined as the (first order estimate) of the P&L if the stock price moves by a finite amount $\Delta S$ in ...


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Bear in mind I am a business guy, not a quant -jump risk is the inaccuracy of the Delta caused by a large discontinuous move in the underlying. From what I recall of calculus 20+ years ago, Delta is the slope of the tangent line on the underlying (UL) price vs. option price curve. The tangent line's slope - Delta, is only completely valid at that one point. ...


1

Note that \begin{align*} Call_{\rm BTC}=\frac{1}{S}Call_{\rm USD}. \end{align*} The premium adjusted delta $Delta_{PA}$ is defined as the change of $Call_{\rm BTC}$ with respect to the change of the spot in BTC, that is, \begin{align*} Delta_{PA} &= \lim_{\Delta S\rightarrow 0}\frac{\Delta Call_{\rm BTC}}{\frac{\Delta S}{S}}\\ &=\lim_{\Delta S\...


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It's all about transposing prices into some space that changes more slowly, such that data you can garner from prices provided by someone else at some other point in time can be used to estimate value at some other point in time. Its effectively an interpolation and extrapolation tool. Say you have option prices at strikes of 10, 20, 30, 40, etc. And you ...


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$q$ is not delta hedge. $q$ is determined from the fact that $S_i$ is a martingale i.e. for $S_0$ $S_0=E(S_1)=quS_0+(1-q)dS_0$ (if no rates) This equation gives the same $q$ , dependent only on $u$ and $d$ , if calculated for $S_0$ , $S_1$ etc , thus $q$ is the same for all steps.


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yes. Assuming that $r$ and $r_f$ are the same, you should use continuous compounding: $$ F(t,T)=S_te^{r_f(T-t)} $$ and if the underlying is a dividend paying instrument, then $$ F(t,T)=S_te^{(y-r_f)(T-t)} $$ with $y$ the annualized continuous dividend yield.


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