FX Delta-Neutral ATM Volatility and Smile Quotes

# ATM volatility for FX options


# 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

## Answer by will (score 2)

https://quant.stackexchange.com/a/51633

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 $Z(K-F)$.

## Answer by AKdemy (score 1)

https://quant.stackexchange.com/a/63546

Table 3 shows the math needed to get Delta neutral strikes. P.a. stands for premium adjusted which is also explained in the paper. It is a very good paper to read.

Equity VOL surfaces are fundamentally different. Generally, derived from listed option prices (with all complexities involved in that).

FX is relatively simple. ATM Delta neutral Straddles (DNS), Risk Reversals (RR) and Butterfly (BF) quotes completely describe the entire surface. ATM the level, RR the skew and BF the kurtosis. There are some complications like premium included / excluded Delta, Delta style (spot vs forward), spread models used etc but equity is generally more complicated with de-Americanizing option prices, finding implied forwards and dividends and the like.

Delta neutral is neat in itself. In terms of the strike, it's simply the K such that

```
call delta = - put delta
```

Take a simple example with assumes premium excluded:

```
Days to expiry = 30
Vol = 8%
Fwd = 1.25
```

$$Fwd*exp^{0.5*(Vol/100)^2*(days/365)} = 1.2503$$

Daycount may vary obviously. Could be ACT/365 or BD/252 (commonly used in BRL) for example.

On a side remark, it's interesting that you have a closed-form solution for Delta as well as Strike with premium adjusted delta. However, going from Delta to strike itself is not possible and would require a root solver in this case.

To get actual vols for calls and puts, one needs to transform these and solve for call / put vol.

```
RR = Vol of an OTM Call Option (C) - Vol of an OTM Put Option (P)
BF = ( C + P ) / 2 - Vol of ATM DNS
```

it's simple to show that

```
C = ATM + BF + RR/2
P = ATM + BF - RR/2
```

where you get say a 25 Delta call if you use 25D RR and BF respectively.

The above uses smile BF but brokers frequently quote market BF (short broker fly). These are a bit more involved and have equal notional strangle strikes whereas the notional on the ATM straddle is set such that the package is initially vega neutral. FX Volatility smile construction by Dimitri Reiswich and Uwe Wystup explains this in more detail (if this link does not work anymore, it can be searched easily).

## Answer by nbbo2 (score 0)

https://quant.stackexchange.com/a/51635

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.

## Answer by user42108 (score 0)

https://quant.stackexchange.com/a/59864

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"