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After I went through the derivation to get the skew in Backus et al., I had two questions:

  1. In the proof, it mentioned the application of the arbitrage condition and then obtained equation (31): $$\mu_n = (r_{nt} - r^*_{nt})n - \sigma_n^2 /2 - \sigma_n^3\gamma_{1n}/3! -\sigma_n^4\gamma_{2n}/4!,$$ I don't know what's that condition and how we can obtain equation (31) from that condition.

  2. For the identity equation (32): $$S_te^{-r_{nt}^*n}\phi(d) = Ke^{-r_{nt}n}\phi(d-\sigma_n),$$ I can verify this by substituting the expression of $d$. But what's the explanation for this identity?

Thank you so much for your time and help!

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    $\begingroup$ 2. is a relatively well known identity that comes up again and again in deriving the Greeks quant.stackexchange.com/questions/42360/… esp. Theta and Delta IIRC. If you have been asked to derive these from the BS formula you will have used this. $\endgroup$ – noob2 Jul 27 '20 at 1:04
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    $\begingroup$ @noob2 Thanks! I know how to show this identity. I was wondering if there is any explanation or intuition behind this identity. $\endgroup$ – JuniorQuant Jul 27 '20 at 1:14
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For (32), under Black-Scholes model ($r^*$ foreign interest rate, in FX world, or continuous dividend, in equity world), we have Gamma

$$\frac{\partial^2 C}{\partial S^2} = \mathrm{e}^{-r^*\tau}\frac{\phi(d)}{S\sigma\sqrt{\tau}} $$

and Dual Gamma

$$\frac{\partial^2 C}{\partial K^2} = \mathrm{e}^{-r\tau}\frac{\phi(d -\sigma\sqrt{\tau})}{K\sigma\sqrt{\tau}} $$

So identity (32),

$$ S\mathrm{e}^{-r^*\tau}\phi(d) = K\mathrm{e}^{-r\tau}\phi(d-\sigma\sqrt{\tau}),$$

can be interpreted as stating the relationship between Gamma and Dual Gamma under Black-Scholes model:

$$S^2\frac{\partial^2 C}{\partial S^2} = K^2 \frac{\partial^2 C}{\partial K^2}$$

For (31), note that the arbitrage condition in equality (11), under lognormality assumption,

is $$\mu_n + \sigma_n^2 /2 = (r_{nt} -r^*_{nt} )n$$

and that correction $\sigma_n^2 /2$ is coming from cumulant generating function calculation

$$\ln \mathbf{E}\left[\mathrm{e}^{\sigma_n w} \right] = \sigma_n^2 /2 $$ (for $w$ standardized normal).

Once $w$ gets a Gram-Charlier expanded density, if we recompute the cgf, we get:

$$\ln \mathbf{E}\left[\mathrm{e}^{\sigma_n w} \right]= \sigma_n^2 /2 + \sigma_n^3\gamma_{1n}/3! + \sigma_n^4\gamma_{2n}/4!$$

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  • $\begingroup$ Thanks, ir7! Any comments for the first question? $\endgroup$ – JuniorQuant Jul 27 '20 at 4:30
  • $\begingroup$ I see: the first CGF uses a normal distribution, so there is no skew and kurtosis. But then the non-arbitrage requires us to have an extra correction. Thanks, ir7! $\endgroup$ – JuniorQuant Jul 27 '20 at 23:42
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I don't know "what this identity intuitively means" but I can tell you an anecdote about how I encountered it. (This is a true story, though I won't mention the school and the professor).

The professor wrote on the blackboard the BS equation (with no dividends):

$$C=S N(d_1) - K e^{-rT} N(d_2)$$

and asked: what is Delta, i.e. what is $\frac{dC}{dS}$?

Student in the first row saw that $S$ appears in the first term, where it is multiplied by $N(d_1)$ so he answered: "the derivative is $N(d_1)$"

Professor smiled, said very good and wrote on the board: $$\Delta\equiv\frac{dC}{dS}=N(d_1)$$

Later in the hallway or lounge waiting for the next class, the smartest student in the class (who was also a bit of a troublemaker) said: "Professor so and so is very superficial and you are foolish to go along with it. $S$ not only appears at the start of the first term but also appears "inside" $d_1$ and $d_2$ and this has to be taken into account in taking the derivative".

In fact you can see this by writing out the complete equation like so:

$$S_1 N(\frac{\ln(S_2/K) + (r + \sigma^2/2)\tau}{\sigma\sqrt{\tau}}) - K e^{-rT} N(\frac{\ln(S_3/K) + (r - \sigma^2/2)\tau}{\sigma\sqrt{\tau}})$$

I have written $S_1,S_2,S_3$ to identify the three places where $S$ appears but of course these three are all equal in value. It seemed clear that we had only taken into account the contribution of $S_1$ to the derivative and neglected the contributions of $S_2$and $S_3$.

If you compute the additional terms from $S_2$ and $S_3$ in the Delta it is a long calculation but you will find they are $$ S\phi(d_1) - K{e}^{-r\tau}\phi(d_2)$$

So does this mean that the professor was wrong? Had he dropped two important terms? It turns out that the above expression is equal to zero (this is identity 32 that we are talking about). So the equation the professor wrote is correct (and it is also given in many authoritative sources). The professor probably knew the whole story but he did not bother to explain it because he liked to go quickly (and in this he was perhaps a little superficial); he is the author of a textbook and I think he understood the issue. Maybe he was purposely playing a little trick on us to make us think.

In any case the meaning of Identity 32 is that "the contributions of $S_2$ and $S_3$ to Delta cancel each other out. It is OK to differentiate w.r.t. $S_1$ only". If someone has a further explanation why this is so I would like to hear it.

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    $\begingroup$ Not an explanation why it holds but another idea: a call option can be decomposed into exercise probabilities under different numéraires, $C=S_0\mathbb{Q}_S[\{S_T\geq K\}]-Ke^{-rT}\mathbb{Q}[\{S_T\geq K\}]$. The call's delta then equals $\mathbb{Q}_S[\{S_T\geq K\}]$ which is $N(d_1)$ in the Black-Scholes world. Denoting the corresponding probability density functions by $\varphi_{S_T}$, do we also have $S_0\varphi_{S_T}^{\mathbb{Q}_S}(K)-Ke^{-rT}\varphi_{S_T}^{\mathbb Q}(K)=0$ for a general model? $\endgroup$ – Kevin Jul 27 '20 at 21:09
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    $\begingroup$ Tried to answer your question below. I'm sticking with duals and homogeneity. :) $\endgroup$ – ir7 Jul 27 '20 at 22:13
  • $\begingroup$ @KeSchn Sorry, I am not very familiar with the numeraires. What does $\phi_{S_T}^{\mathbb{Q}_S}$ mean? $\endgroup$ – JuniorQuant Jul 27 '20 at 23:43
  • $\begingroup$ @JuniorQuant I simply meant the standard probability density function of the random variable $S_T$ under the probability measure $\mathbb{Q}_S$. The latter is an artificial probability measure under which you express prices of assets with respect to the stock price (as opposed to a locally risk-free bank account). $\endgroup$ – Kevin Jul 27 '20 at 23:46
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    $\begingroup$ @JuniorQuant First and second derivatives of call price wrt to strike are needed to make sense of the 'vol skew'. In particular, you can look into Dupire's 'dual' PDE for local volatility. There are other places too. You could make it an SE question. The word 'dual' may not be very popular, but strike remains a variable just like any other in a price function (BS or others), hence sensitivities wrt to it are unavoidable. $\endgroup$ – ir7 Jul 30 '20 at 15:05
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(An attempt to answer @noob2's question posed in answer above.)

Black-Scholes is homogeneous: $$ xC(S,K) = C(xS,xK) $$

for all $x>0$. This is true even if one triplicates $S$ (new variables but taking the same value as $S$), as the two copies are always divided by $K$.

Taking derivative wrt $x$ gives:

$$ C(S,K)=S(\partial_1C)(xS,xK) + K(\partial_2C)(xS,xK) $$

Setting $x=1$, we get:

$$ C(S,K)=S(\partial_1C)(S,K) + K(\partial_2C)(S,K) $$

that is, the sum of (dollar) spot Delta and dual Delta.

With triplicates, we have:

$$ C(S,K, S_3,S_4)=S(\partial_1C)(xS,xK,xS_3,xS_4) + K(\partial_2C)(xS,xK,xS_3,xS_4) + S_3(\partial_3C)(xS,xK,xS_3,xS_4) + S_4(\partial_4C)(xS,xK,xS_3,xS_4)$$

and for $x=1$

$$ C(S,K, S_3,S_4)=S(\partial_1C)(S,K,S_3,S_4) + K(\partial_2C)(S,K,S_3,S_4) + S_3(\partial_3C)(S,K,S_3,S_4) + S_4(\partial_4C)(S,K,S_3,S_4)$$

When triplicates get set to the same value $S$, the sum of contributions from $\partial_3$ and $\partial_4$ must disappear.

Note: One other route to explore is put-call symmetry for Black-Scholes, where $K$ switches roles with $S$.

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