# Tag Info

15

Gamma is the second partial derivative of the change in the price of the option wrt to the change in the underlying. Said another way, it is the change in delta. If you write down the Black-Scholes pricing formula, you's see the gamma term: $$...\frac{1}{2}\frac{\partial^2C}{\partial S^2}(\Delta S)^2...$$ Notice that the $\Delta S$ (change in stock price) ...

6

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 ...

6

Under the Black-Scholes model, \begin{align*} Gamma &= \frac{N'(d_1)}{S \sigma \sqrt{T-t}}\\ Vega &= SN'(d_1) \sqrt{T-t}. \end{align*} Then, it is easy to see that \begin{align*} Vega = S^2 \sigma (T-t) Gamma. \end{align*}

5

Since the volatility is not changing, we can assume that the only change is the underlying asset price $S$. Then \begin{align*} C(S+\Delta) &\approx C(S) + Delta \times\Delta +\frac{1}{2} Gamma \times \Delta^2 \\ &=11.50 + 0.58 \times 0.5 + \frac{1}{2}\times 2 \times (0.5)^2\\ &=12.04. \end{align*}

4

if you have a portfolio of calls and puts with the same maturity then your portfolio is gamma neutral if and only if it is vega neutral. The reasons is that the BS gamma divided by the BS vega is a function of $S$ and $T$ that does not vary with $K.$ So if you construct a linear combination that has zero gamma then the vega is zero too, and vice versa.

3

An important theme in N. Taleb's book(s) is that being short gamma is a dangerous situation, in the sense that you are subject to occasional very sharp losses, losses that could put you out of business or at least be very painful. You do make frequent small gains, but these are of less concern. (Recall that the P&L of an option when the underlying ...

3

Options on interest rates futures in the listed markets are always traded 1-yield (100-yield) just like the futures which are traded 1-yield. So negative rates aren't an issue and its always black volatility. In the OTC market, both normal and black volatility are quoted, but the common practice is to use black volatility is what is way more frequently used....

3

At a high level, just look at the delta. If it's so close to zero that it won't shift the price of the option by a penny, then you could say, "the option no long responds to the price of the underlying" Any price it has more or less a function of theta and vega only. In practice however, depending on the model you use, delta has some volatility input. To ...

3

I think what you are missing is simply the Vega-Gamma relation in the Black-Scholes model. Namely: $$Vega = \frac{\partial v}{\partial \sigma} = \sigma(T-t)S^2 \frac{\partial^2 v}{\partial S^2} = \sigma \tau S^2 \Gamma$$ Plugging this into your coverage error, you get its expression in terms of the Vega which is the most natural measurement of your ...

2

For ATMish options, as vol goes higher, the option looks even more ATM. That is, at higher vol, the difference between a 99% strike option and a 100% strike option is less pronounced than if vol were low; hence your deltas won't change as fast as spot moves and thus, less gamma. For far OTM options, its the opposite. They would have very little delta at ...

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

The problem with your formula is the equation sign $=$. The second order finite difference is only an approximation to the true gamma: $$f^{\prime \prime}(x) \approx \frac{f(x+h)-2f(x)+f(x-h)}{h^2}.$$ $h$ can not be a result. Ideally, it should be small (whatever that means), so your original choice of $1\text{bp}$ seems appropriate for this ...

2

He's saying that if you know the volatility, and you hedge continuously, you can lock in the exact Black-Scholes price. Any deviation from that delta hedging scheme must result in noise. ie the replicated price must have a distribution with some width around the theoretical value. This noise does not create systematic profit or loss, because it's just ...

1

Commonly used procedures: A) hedge when a 1 sd move has happened B). Hedge when your delta position exceeds some risk limit. C) hedge once a day D) hedge based on your desired delta position All are used. I personally prefer B.

1

I don't understand why you think the numbers dont match up. In my opinion it all works out. Perhaps best if you first convert all numbers to percentages and for 1 underlying instead of 100 multiplier. From OVML you have multiplier = 1 troy ounce S = 1075 K = 1075 r = 0.0033 T = 2/12 sig = 0.12 Convert all into percentages: S = 100 K = 100 r = 0....

1

I don't think your hypothesis is correct. If you have a very short dated ATM option, then your option will have close to infinite gamma but close to 0 vega. So this short dated ATM option is vega neutral but definitely not gamma neutral.

1

The "right" thing to do is to treat the options as derivative contracts. Let's say for simplicity that you are using Monte Carlo to compute VaR. Then you would simulate the equity prices on each iteration, and then apply an option-pricing formula to get the corresponding option prices on that iteration. This lets you obtain an accurate simulated portfolio ...

1

under the black scholes assumptions it is nearly impossible to move 50% in 1 day. A 1day move will be of size $\sigma \sqrt{1/252}$ (assuming 252 trading days). So you are talking about something that is about a 40 stdev move (assuming $\sigma$ = 0.2). that's why your gamma is 0.

1

It's a combination of too few sample paths and/or too small an increment. Your estimation error on the price is magnified by the $dS^2$. Try using a larger sample or a larger increment. Alternatively, you can use a multiplier instead of a fixed increment; in my experience, it usually yields better results.

1

Consider an instrument value $f(S_0^1, \ldots, S_0^n)$ that depends on $n$ spot levels. Let $$\overrightarrow{S}_0=[S_0^1, \ldots, S_0^n]^T$$ be an $n$-dimensional vector representing the spot levels. We can approximate the cross gamma \begin{align*} \frac{\partial^2 f\big(\overrightarrow{S}_0\big)}{\partial S_0^i \partial S_0^j} \end{align*} by a finite ...

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 $\frac{\partial^2 V}{\partial S^2}=\frac{\partial^2 C}{\partial S^2}=\Gamma_c$ ...

1

First, I think you made a mistake in your computations above. Where you wrote $(30-20)$, I think you really meant $(30-(-20))$ i.e. $30+20$, yielding a gamma P&L of $1000$ instead of $200$. Your total P&L over $[90,170]$ would then be $110$ instead of $-690$. It doesn't matter for my answer either way, just thought I'd point it out for confused ...

1

I've started thinking about this, too. My gedanken conclusion turned out to be too simple once I found what I was after: http://www.investment-and-finance.net/derivatives/o/option-beta.html, which I've confirmed in Black & Scholes (1973) p10 (eq 15). In short: $$\beta_{\text{option}} = \frac{S\cdot\Delta}{O}{\beta_S}$$ where $S$ is the underlying ...

1

I'd say that the shock size depends on the situation/asset. If your model produces somewhat noisy PVs, it is advisable to use a slightly larger $h$ to avoid numerical issues. You may also want to base your decision on empirical hedging performance. This may or may not help, but most bond index providers (Citi/Barclays) use a shock size of 25bp when reporting ...

1

I, personally, like to see gamma as change in dollar delta per percent (most systems have it as "GammaP"). This way, it's much easier to think about you delta position as the market is moving around. The number above is the BS gamma which is an unscaled 2nd derivative of delta. You need to rescale it to get gammap (delta change per percent). In general, it ...

1

Yes, if the two rates belong to two different currencies having different yield curves. Or in fact any two indices (bases) having different yield curves. e.g OIS vs LIBOR or LIBOR vs UST etc.

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