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This is in fact a tricky matter. As you say one way is to calculate delta by an analytic formula, i.e. calculate the first derivative of the option pricing formula you are using with respect to the underlying's spot price. The second way is to do it numerically, i.e. change the spot price by a small value $dS$, calculate the value of the option and then ...

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I assume you mean that the hedge error should go to zero when the time-step size goes to zero. This is the case! I have a BS delta hedge simulator here: http://www.christian-fries.de/finmath/applets/HedgeSimulator.html and the source code here: http://www.finmath.net/java/ which shows that the delta hedge converges. However, in order to have that result, ...

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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 transaticton cost but leads to hedgint errors and more risk. To find a price one must introduce an ...

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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 $$\Delta_t=... 4 If you could hedge continuously with zero transaction costs, the gamma would be irrelevant: you would perfectly replicate with delta hedging and be done. In practice, hedging is discrete and there is a certain amount of slippage giving a random outcome with mean zero. The larger the gamma, the bigger the variance of slippage. Trading more frequently ... 4 Due to the lack of a carry arbitrage, VIX futures are actually the direct hedge for VIX Index options 3 The differential equation has a trend due to the interest rate. When you discount you take this trend away:$$ \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}... 3 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 ... 3 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 ... 3 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(... 3 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-... 3 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 ... 3 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 ... 3 Assuming you already have a way to obtain hedge ratios and the like, your best available choice is probably blotter (used to be just quantstrat). You will find that it isn't necessarily oriented toward options. Generally for options backtesting, pros end up making their own or buying commercial software. There are tons of commercial providers, but I don'... 2 I think you incorrectly calculate portfolio values. For me the easiest way to keep track of portfolio positions and avg price is to separately calculate the sum of volume traded on the long and short side and to also calculate an average price separately for buys and sells. You obviously must update the avg price and size of the particular side (long/short)... 2 It may be the case with certain exotics that greeks are derived analytically through approximations. In that case at certain boundaries you may get different results from such approximation over the numerical approach. Why do you not approach the numerical case similarly than most banks and hedge funds when they "shock" their options books: Simply shift your ... 2 Your portfolio composition is not clear. To simplify, we assume that it consists of units of a stock and options on this stock. What you can do is to sell 4000 units of options that will bring it to gamma neutral, and then to balance the delta, you can buy 2,400-450=1,950 units of the stock. 2 First we write dynamic of${{x}_{t}}=\ln ({{S}_{t}})\begin{align} & d{{x}_{t}}=({{r}_{t}}-\delta -\frac{1}{2}\sigma _{t}^{2})t+{{\sigma }_{t}}d{{W}_{1}}(t) \\ & d{{\sigma }_{t}}=a({{\sigma }_{t}},t)dt+b({{\sigma }_{t}},t)d{{W}_{2}}(t) \\ & d{{r}_{t}}=\alpha ({{r}_{t}},t)dt+\beta ({{r}_{t}},t)d{{W}_{3}}(t) \\ \end{align} Let \begin{align} &... 2 Assuming zero interest, the put option has the price \begin{align*} KN(-d_2)-S_0N(-d_1), \end{align*} and delta-N(-d_1)$. When$N(-d_1)$units of stocks are shorted and invested in bonds, the total value in bonds is$KN(-d_2)$, which is indeed greater than the option price. However, as you have shorted$N(-d_1)$units of stocks, your portfolio value is \... 2 "does the underlying usually see increased trading?" Not necessarily. Most market makers do not re-hedge much in the underlying. In many markets the delta is exchanged (off-exchange) alongside the options trade at initiation, making both parties delta neutral at the outset. Re-hedges in large vol books are generally accomplished through other options and ... 2 As I commented, I think this is simply a way to prove that both statements are equivalents, that is when the implication goes in both directions. There are no such things as a definition, it's all about the assumption that you make. Actually, a more general point could be the following:$u$and$d$are define such that after one period the asset gets$u$... 2 Delta of a digital (or binary) option is like the normal distribution probability function , approaching 0 at far OTM / ITM conditions and representing a very high peak at ATM. The peak at ATM approaches infinity as we approach the maturity. This is never 0.5 like a vanilla option since the payoff never simulates the payoff of the underlying. If you want ... 2 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.

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We assume that \begin{align*} dX_t &= X_t(rdt + \sigma_x dW^x_t)\\ dY_t &= Y_t\Big[rdt + \sigma_y\Big(\rho dW^x_t + \sqrt{1-\rho^2} dW^y_t \Big)\Big], \end{align*} where $\rho$ is the correlation, which we assume to be less than 1, and $W^x$ and $W^y$ are two independent standard Brownian motions. Then \begin{align*} d(X_t/Y_t) &= \frac{1}{Y_t}...

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suppose you sell a K = 105 call. When the stock reaches exacty 105 you buy 1 stock at 105. Now suppose the stock moves to 104.99, using your logic you sell 1 share at 104. You lost $0.01. Again, after a while stock reaches 105 you buy 1 stock. After some time it goes up, but eventually it goes down again below 105. Thus you sell 1 share below 105. Again ... 1 It just simply means you have to borrow money 1 Yes. If you use the BS Model for computing deltas and the same model for evolving the stock price then you should replicate the pay-off of any contract. 1 A perfectly hedged portfolio should not make any profits different from the risk free interest rate. However, you won't be able to hedge perfectly in the real world. Delta hedging for example requires continous trading and adjusting (this is one way to derive the black -scholes formula: thex hedge the stock perfectly and therefore obtain a risk -free rate ... 1 let$\frac{\partial C}{\partial S}=\delta_c$let$\frac{\partial^2 C}{\partial S^2}=\Gamma_c$let$\frac{\partial C_0}{\partial S}=\delta_0$let$\frac{\partial^2 C_0}{\partial S^2}=\Gamma_0$we want$\frac{\partial V}{\partial S}=\frac{\partial C}{\partial S}=\delta_c$and$\frac{\partial^2 V}{\partial S^2}=\frac{\partial^2 C}{\partial S^2}=\Gamma_c\$ ...

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It appears that you are plotting your analytical delta as a % of the delta of the underlying. This is why the delta converges to 100% As for the numerical delta, it could be that you are not adjusting for the DV01 of the underlying. This would explain why the numerical delta still increases as the option gets more in the money and why the distortion is ...

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