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Consider reading Lorenzo Bergomi's excellent book -- or at least the first chapter available here for download --, it will help you clarify things. Some remarks as to your original question: It is well known that, under a pure diffusion assumption, the total P&L of a delta hedged European option (i.e. an option whose payoff only depends on the value of ...

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The value of European binary call, paying \$1 if$S_T > K$or nothing otherwise, is $$c_t=e^{-r(T-t)}N(d_2)$$ where,$d_2=\frac{ln(S_t/K)+(r-\sigma^2/2)(T-t)}{\sigma \sqrt{T-t}}$Delta of your binary call option is $$\Delta_t=\frac{\partial c_t}{\partial S_t}=\frac{e^{-r(T-t)}N'(d_2)}{\sigma S_t \sqrt{T-t}}$$ Derivation We need to compute $$\Delta_t=\... 13 Regarding your 1st question, jumps are indeed unhedgeable. From a theoretical point of view, you might want to look at Merton's "Option pricing when underlying stock returns are discontinuous", the original paper that adapted Black-Scholes framework to include jumps. If you look at page 7, just after equation (9): Unfortunately, in the presence of the ... 11 Options have an asymmetric payoff profile: The payoffs are zero for almost all cases and positive else (as we well know). If the option is OTM, most of its payoffs are zero. A rise in volatility will hence increase the likelihood for instead positive payoffs from a change in the underlying price (i.e. delta increases). If the option is already ITM, most(... 10 First when transaction costs are involved the trader has to make a tradeoff between return and risk. Continuous rebalancing/hedging could lead to infinite transaction costs but provides (in theory) a perfect hedge. Discrete hedging enables to minimize transaction cost but leads to hedging errors and more risk. To find a price one must introduce an ... 9 By delta hedging you are saying that you have a view on the path and the volatility of the option you are trading, but not on its direction; in your case, that being short delta. From a theoretical perspective, all options are priced fairly and not delta hedging simply increase the variance of your payouts. In your example, selling a call and delta ... 8 Calendar spreads have a number of disadvantages for trading Vega: Vega in different months are generally not additive, some traders use root-time-Vega but it does not remove the additional risk. You are trading time spread not just volatility, so be careful Calendar spreads are affected by dividends and rate changes - another source of risk. A gamma-neutral ... 8 Generally speaking, in the real world, you'd always want to use the correct implied vol. But you should think of your question in terms of: (1) Vega mark-to-market (m2m) PnL vs. theta/gamma profile (2) Change in risk and PnL due to higher order risks (vanna, volga) Vega mark-to-market PnL vs. theta/gamma profile In a simple, pure Black Scholes world ... 7 The differential equation has a trend due to the interest rate. When you discount you take this trend away:$$ \frac{d}{dt} (e^{-rt}Z_t) = -re^{-rt}Z_t + e^{-rt} \frac{d}{dt}Z_t = e^{-rt}\frac{1}{2}S_t^2\Gamma_t(\hat{\sigma}^2-\beta_t^2) $$Z doesn't appear on the rhs anymore and you can integrate$$ e^{-rT}Z_T - e^{-r0}Z_0 = \int_0^T e^{-rt}\frac{1}{2}... 7 Assume you buy a plain vanilla call option at the price$V$and the spot$S$. You immediately delta hedge buy selling$\partial V / \partial S$units of the underlying asset. The underlying asset now instantaneously jumps form$S$to$S' = S + \Delta S$. The new value of the call option is$V'. Your total p&l is \text{P&L} = V' - ... 7 This question has been asked many times and some clarifications appear needed. As pointed out in an answer to this question, the portfolio \begin{align*} \Delta_t^1 S_t + \Delta^2_t C, \end{align*} where\Delta_t^1 = -\frac{\partial C}{\partial S}$and$\Delta_t^2 =1$, is, generally, neither self-financing nor locally risk-free. To derive the Black-... 6 Due to the lack of a carry arbitrage, VIX futures are actually the direct hedge for VIX Index options 6 Assuming all else remains equal (implied vol has not changed and very little time decay has occurred), Gamma scalping can best be explained by Gamma (or realized volatility) enhancing the value of a delta hedged portfolio. For example: If you are long an at-the-money call option, you are long 0.5 Delta and long Gamma. If you hedge this position, you will ... 6 Yes, in the sense that it is assumed that the delta will be passed between participants at time of execution. Not necessarily. A non delta neutral trade may be used for speculation , or for hedging. 6 In the Black-Scholes-Merton model, with model option price$V$as a function of underlying price$S_t$, strike price$X$, continuously compounded risk-free rate$r$, continuously compounded dividend yield$y$, time-to-maturity (in year fractions)$\tau$and implied volatility$\sigma$, our$\Deltais defined as $$\Delta\equiv \frac{\partial V}{\partial S_t}... 5 If it wasn't clear from the previous answers, the answer they want is that the delta becomes infinite. That's because a tiny move in the stock will change the payout by 100 so your delta hedge must be enormous. 5 To answer your questions: Is the trading p&l meant to be the delta-hedging p&l? Yes, in his example it concerns delta hedged pnl. how come p&l is raising steadily even when stock price is rising? the trader should be losing money on the delta hedging because he is short gamma? He is short gamma but long theta. He is initially making money ... 5 If you could hedge continuously with zero transaction costs, the gamma would be irrelevant: you would perfectly replicate with delta hedging and be done. In practice, hedging is discrete and there is a certain amount of slippage giving a random outcome with mean zero. The larger the gamma, the bigger the variance of slippage. Trading more frequently ... 5 You should have a look at the following paper: Ahmad, Riaz and Paul Wilmott (2005) "Which free lunch would you like today, Sir? Delta hedging, volatility arbitrage and optimal portfolios," Wilmott Magazine, Nov. 2005, pp. 64—79 which tackles this exact issue. 5 As long as you live in a world where implied and realized vol are the same, there is no net profit (or loss) from gamma scalping. However, if they are different, then you make a gain or loss which is not path dependent. This is all still in a hypothetical world of course with continuous trading. In reality when rehedging less frequently, pnl becomes random ... 5 Gamma scalping (being long gamma and re-hedging your delta) is inherently profitable because you make 0.5 x Gamma x Move^2 across the move from your option. (You get shorter delta on downmoves, so you buy underlying to hedge, you get longer on upmoves, so you sell on upmoves, etc.) Because it's inherently profitable across any move, you must pay for the ... 5 We work in a Black-Scholes world. Consider the following delta-hedged portfolio:$$ \Pi_t=V_t-\frac{\partial V}{\partial S}S_t$$We assume the portfolio is self-financing^{\text{(a)}}, therefore:$$ \begin{align} \text{d}\Pi_t &= \text{d}V_t-\frac{\partial V}{\partial S}\text{d}S_t \\[3pt] \tag{1} & = \left(\frac{\partial V}{\partial t}\text{d}t+\... 5 It is true that when FX options are traded, the delta is often traded as well. That is a practice specific to the FX option market. It is called an "exchange of delta". You can undo it by selling the delta in the spot market immediately after the option trade. Or you can request no delta exchange at the time you make the trade. Some say this practice arises ... 5 I assume you want to price a derivative product that pays\int_0^T\ln S_tdt$at maturity time$T$, from time$t=0$. I'll ignore generalization to time$t$because it is trivial (split the integral in two, before and after$t$as you did). The first trick it to do an integration by part on$\ln S_t dt$:$d(t\ln S_t) = \ln S_t dt + \frac{t}{S_t}dS_t + \frac{t}...

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I think you need to go even one step further than vonjd went in his reply. If liquid trading of the underlying is not possible, not only the arbitrage argument underlying risk neutral pricing breaks down. In that case there is simply no reason why the prices of those two assets (the option and its underlying) should be related in any way at all. So in my ...

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An Investment Bank earns a profit by selling you an option at a slightly higher price than the theoretical price, or buying it back from you at a slightly lower price. They call this "earning a spread". Then they hedge the option, so as not to make any [further] gains or losses on it (other than the risk free rate). Another way they could earn a profit is ...

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As has been remarked in the comments already, the standard deviation of your hedging error should approach zero as your re-hedging frequency (the number of time steps) increases. Here is a sample plot of how it should behave like. It was generated using $T = 1$, $K = 100$, $S_0 = 100$, $r = 5\%$, $\sigma = 20\%$ and 1,000 sample paths. Just as a sanity ...

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As indicated by the name, delta one products have a delta of exactly 1 (at least theoretically) with respect to the underlying; moreover, AFAIK the delta has to be constant, i.e. a product with optionality that happens to have $\Delta = 1$ for some period won't be classified as delta one (otherwise it would be a nightmare for traders to manage their books!). ...

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We denote by $C(S_0, K)$ the price for a call option with payoff $(S_T-K)^+$ at the option maturity $T.$ Here $S_0=100$ is the spot stock price. Generally, \begin{align*} C(S_0, K) \ne (S_0-K)^+. \end{align*} Moreover, \begin{align*} C(S_0+\Delta, K)-C(S_0, K) \approx \frac{\partial C}{\partial S_0} \Delta, \end{align*} where $\frac{\partial C}{\partial ... 4 Main references As explained in my comments, the correct approach to derive the hedging portfolio would be the one described in Gordon's answers to the following questions: Derivation of BS PDE problem using Delta hedging Black Scholes differential The hedging portfolio$C_t-(\partial C/\partial S)S_t\$ is not self-financing We can check that the hedging ...

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