# ATM volatility for FX options

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

• This is wrong even for equities... – CABLE Mar 18 '20 at 2:55

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 $$Z(K-F)$$.

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

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

https://en.wikipedia.org/wiki/Greeks_(finance)#Relationship_between_call_and_put_delta

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.

• hmmmmm. this looks suspiciously like another answer... – will Mar 14 '20 at 17:40
• We may be talking about different Deltas, $\frac{d}{dS}$ versus $\frac{d}{dF}$ – noob2 Mar 14 '20 at 17:43

From an old (~2005) UBS paper, "What does the term "At-The-Money" Really Mean?": "What is the quoting convention used for ATM? In the inter-bank market, options traders quote Delta Neutral volatilities and this is what they are referring to when using the term "at-the-money" or ATM. It must be specified before trading what premium currency is being quoted for, but a default is usually assumed and is fairly well agreed among market participants...Many customers prefer the ATMF strike, as it is very transparent what strike is solved for"