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34

Many of them are on my website at emanuelderman.com. Others I probably have anyway. Feel free to email me


15

For large values of the spot S, this payout goes to infinity like the square of S. However, the hedging instruments available are vanilla options, which go like S to the first power. Mathematically, the payout can be replicated from a continuous portfolio of vanilla options, and this is what a bank would try to do. However, the weights of the vanilla options ...


12

Well, it's a topic which actually should have its own book dedicated. Unfortunately, existing literature is rare or not practical enough. Let me at least try to provide some key ideas and challenges you should consider when hedging this kind of structure. First, let's start with the question: How not to hedge an autocallable? What is not going to work is ...


11

I had read some of them; actually, it does not exist an on-line library that collected them (or, better, it existed here, but it seems the website does not work anymore). I reported here below some of them that you did not find: More Than You Ever Wanted To Know* About Volatility Swaps Model Risk The Volatility Smile And Its implied Tree Enhanced Numerical ...


11

I suspect this is because, conditional on being in-the-money, the payoff of your option is convex in stock price $-$ whereas for a vanilla call, the payoff is linear. As a consequence, the delta $\Delta$ and gamma $\Gamma$ hedge ratios are larger, in particular gamma becomes much more sizeable. Let us assume that rates are null to lighten notation. Then your ...


10

the problem is that the pay-off has discontinuous first derivative. Try a contract with pay-off that is twice differentiable and it will probably work. The problem is that all the value comes from the tiny number of paths within $\Delta S$ of the strike, and these paths have huge value. This is a well-known problem. As the bump size goes to zero, the ...


10

There's no easy answer to your question, as noob2 pointed out. You can look online for info from Universa. That fund does exactly what you are asking: https://www.universa.net/riskmitigation.html Of course, post a crash, such as the one we just experienced, the cost of hedges is larger than it is prior to such events. Understand that you aren't going ...


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 ...


9

With difficulty and high costs and secretively. Successful ones are the ones that are able to do it more cheaply. This is also the reason for their secretiveness: prices would go up. The costly but straightforward approach would be to buy equity index puts. However, I don't think anyone here can or will explain how you can tail hedge at scale significantly ...


9

Your question comes at this correctly, in my opinion. There is indeed a buyer and a seller behind every option; but the hedging behaviour of the two need not be equivalent... I used to work in an investment bank, and we used to call this (politely) "pin risk", or (less politely) "the gamma hammer". The idea (not perfect, but close enough) ...


8

The point is the following: Delta, $\Delta$, is defined as $\frac{\partial C}{\partial S}$, where $C$ is the value of the call option, and $S$ is the price of the underlying asset. So, given that the value of a call option for a non-dividend-paying underlying stock in terms of the Black–Scholes parameters is $$C = N(d_{1})S - N(d_{2})Ke^{-rT},$$ $$\Delta ...


8

There are more ways to approach this but the method I propose should work reasonably well in practice, especially if you increase the number of assets you hold. Calculate the beta of the stocks you're holding with respect to an index Buy $N_f$ (sell when $N_f$ is negative) future contracts on that index $N_f$ can be calculated as $$N_f = \frac{\beta_T - \...


8

Your simulation is basically fine, though you need to discount in USD. For hedging purpose, you need to use the instruments available in USD. Let $S=\{S_t, \, t\ge 0\}$ be the stock price process in EUR, $X=\{X_t, \, t\ge 0\}$ be the exchange rate process from one unit EUR to units USD, $r_f$ and $r_d$ be interest rates in EUR and USD. Moreover, let $B_t^f=...


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

You're right that the "real" greeks of a digital option are unwieldy, e.g. delta is zero everywhere except at the barrier where it is an impulse. So sell-side trading desks model/price digital options as tightly struck call/put spreads that will sit and play nicely with the rest of the book. Here's a simple example: let's say a bank sells a digital call on ...


7

You are long a vanilla option, so long gamma (positive gamma). If the stock price decreases, so does the delta of your option. Since you short-sold the stock to hedge, you now have short-sold too much since delta has decreased. As a consequence, you must buy back some stock.


7

You would be over hedged in your call position if it was delta neutral before the stock cratered. Since you are long delta on the call, you would have shorted stock to make the original position delta neutral. When the stock fell, your long call delta would have fallen, and you would buy to cover some of your short stock hedge. However, being long the ...


7

It depends a little bit what you're trying to do. If you can statically replicate the payoff of a position at $t=0$, then putting on that hedge will insulate you from all risk coming from the contract. Payoff doesn't need to be linear - for example, you can perfectly replicate a call option using a put option and a futures contract If you want to use only ...


6

Due to the lack of a carry arbitrage, VIX futures are actually the direct hedge for VIX Index options


6

Let $V(t, r_t, S_t)$ be the convertible bond price at time $t$, where \begin{align*} dS_t &= S_t(r_t dt + \sigma dW_t^1)\\ dr_t &=\kappa(\theta-r_t)dt+\Sigma dW_t^2, \end{align*} and where $\{W_t^1, \, t\ge 0\}$ and $\{W_t^2, \, t\ge 0\}$ are two standard Brownian motions with $d\langle W^1, W^2\rangle_t = \rho dt$. Then, \begin{align*} &\ dV(t, ...


6

This is a slightly extended version of my comment that summarizes the main result of the reference that I provided. This problem is discussed in detail in Chapter 12 of Wilmott (2006), which is based on the paper Ahmad and Wilmott (2005). See also the related Question 9 in Carr (2005). In your case, you are selling and delta hedging the option using the ...


6

A general hedging strategy Let assume that $S_1(t)$ and $S_2(t)$ are the price processes of your 2 stocks and that they follow a Geometric Brownian Motion (GBM): $$\forall \, i \in \{1,2\}, dS_i(t) =\mu_iS_i(t)dt + \sigma_iS_i(t)dW_i(t)$$ We assume both stocks have an instant correlation of $\rho$: $$dW_1(t)dW_2(t)=\rho dt$$ Let also $V(t)$ be the value ...


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

I would do as follows: A) First do PCA on an arbitrage-free monthly curve (assuming the most granular contract you will use is individual months). To ensure no arbitrages, you will need to drop out certain contracts, I would drop the most illiquid ones. To give you an example, if you are in Dec, you might see Jan, Feb and Mar quoted, but also Q1. In this ...


6

Presumably the option can be exercised for intrinsic at any point. Note the interviewer asked for a static hedge using the stock, not a dynamic hedge. Hence you must find a buy and hold portfolio that will always give you at least the value of the option (if you’re short it which I suppose is the question) until it is exercised. Note that the maximum ...


5

$\require{cancel}$ $$\text{PnL} = -[P(t+\delta t,S+\delta S)-P(t,S)] + rP(t,S)\delta t + \Delta(\delta S - rS \delta t + q S\delta t)$$ Assuming a pure diffusion, at the order 1 as $\delta t \to 0$ $$P(t+\delta,S+\delta S) = P(t,S) + \frac{\partial P}{\partial t}\delta t + \frac{\partial P}{\partial S}\delta S + \frac{1}{2}\frac{\partial^2P}{\partial S^2}(\...


5

The empirical relationship between the futures price $F$ and the spot price $S$ is $$ F = S e^{b\tau} $$ where $\tau$ is the time to expiry, and $b$ is the empirical basis, i.e. the number that makes the equation hold, given by $$ b = \frac{1}{\tau}\log(F/S) $$ It can be compared to the theoretical basis, $$ b_{\rm theor} = r - q $$ where $r$ is the ...


5

I nearly agree with @phlsmk's answer, but with some small differences. First off, the delta of a digital is not "zero everywhere except at the barrier where it is an impulse". This is what it is at $t=T$. before this, it is smoothed out, exactly like a regular option is. The problem is on what the delta may become. This is not the only place where it ...


5

The quoting convention must be explained somewhere in your book. For Eurodollar futures, this convention is 100 - yield, 92 means the yield is 8% per annum, so for one quarter you need to divide this discount by 4 to get the price (100% - (8% × (3month/12month)) = 100% - 2% = 98%


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 ...


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