I am quite familiar with equity implied volatility and smiles.

However, I find it quite confusing and unclear when it comes to FX. I read many materials but could not get a grasp of the notion of ATM volatility in FX option markets.

I am refering to the ATM delta neutral one. I saw that is defined as the the strike such that straddles have zero delta.

My thinking: Given that a long straddle is being long a call and long a put with the same strike and maturity, the straddle always has a zero delta. Isn't it ? I draw this conclusion from the fact that calls and puts with the same strike and maturity have the same implied volatility. And since the delta of a call is of opposite sign as the delta of a put this will give zero delta for the combination made to construct the straddle.

Could anyone tell me where I am wrong and help me understand the definition for ATM delta neutral volatility?

Thank you

  • 1
    $\begingroup$ This is wrong even for equities... $\endgroup$ – CABLE Mar 18 at 2:55

Your logic is wrong.

Put call parity states that $C_k - P_k = Z \cdot (F - K )$ If we take the derivative of all of this (with respect to the fwd):

$$\frac{\mathrm{d}C_k}{\mathrm{d}F} - \frac{\mathrm{d}C_k}{\mathrm{d}F} = \frac{\mathrm{d}}{\mathrm{d}F} Z \cdot (F - K ) = Z$$

$$\Delta^C_k - \Delta^P_k = Z$$

So if you write the delta of a put(call) in terms of the call(put) at the corresponding strike: $$\Delta^P_k = \Delta^C_k - Z$$ $$\Delta^C_k = Z + \Delta^P_k$$

So the delta of a straddle can be written any of the following ways: $$\Delta^S_k = \Delta^C_k + \Delta^P_k$$ $$\Delta^S_k = \Delta^C_k + (\Delta^C_k - Z) = 2\Delta^C_k - Z$$ $$\Delta^S_k = (Z + \Delta^P_k) + \Delta^P_k = 2\Delta^P_k + Z$$

And when we say that straddle has a delta of zero, then you basically drop out that $$\Delta^{C/P}_k = \pm\frac{Z}{2}$$

i.e. 50 delta.

Your error was the thinking that the delta of a call is the negative of that of a put. Think about a call and put with strikes of a million on an underlying with a value of 100. The call is wildly out of the money and has no delta, while the put will pay out $Z(K-F)$.

| improve this answer | |

No, not every straddle has zero delta, the two deltas do not usually cancel out.

For a non-dividend paying stock the relationship is $\Delta^C−\Delta^P=1$


Keep in mind though that a foreign currency pays a dividend (the foreign interest rate q) in which case I believe the relationship is $\Delta^C−\Delta^P=e^{qT}$

Then the Delta of a Straddle is $$\Delta^S = 2 \Delta^C-e^{qT}$$

when $q$ is zero we get that the Delta of a Straddle ranges between -1 and +1.

| improve this answer | |
  • $\begingroup$ hmmmmm. this looks suspiciously like another answer... $\endgroup$ – will Mar 14 at 17:40
  • $\begingroup$ We may be talking about different Deltas, $\frac{d}{dS}$ versus $\frac{d}{dF}$ $\endgroup$ – noob2 Mar 14 at 17:43

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