I'm currently reading Iain Clark's book Foreign Exchange Option Pricing and I got stuck at one sentence in the beginning of Section 3.3 that I feel is important to understand. He writes:

FX volatility smiles are characterized by providing volatilities, not as a function of strike, but as a function of delta. The choice of delta as the parameter describing the volatility smile is sensible, as otherwise a strike that might correspond to a considerably out-of-the-money option for small $T$ would be very close to at-the-money for large $T$.

where by $T$ he refers to the time left until expiry of the option. My question is: how do you know (or argue) that just because there is an option with expiry in 1 week that is out-of-the-money a similar option (with bigger $T$) will be very close to at-the-money?

Thanks in advance!


3 Answers 3


To a FX trader, "considerably out-of-the-money" means "low delta" and "close to at-the-money" means "close to 50% delta." That is, moneyness is measured in terms of delta.

A useful way to understand this is that delta measures the probability of finishing in the money[1]. A 10% delta option has 10% chance of finishing in the money; a 50% delta option has 50% chance of finishing in the money. But over larger maturities, larger spot moves are likely so a 10% delta option at long maturity has strike much farther from the forward than a 10% delta option at short maturity.

[1]Mathematically, the simple delta is a discount factor times the foreign-numeraire risk-neutral probability of finishing in the money. Psychologically, just think of it as probability.

  • $\begingroup$ This is not 100% correct, but is much more useful than "correct" expositions that are 99.3% to 99.5% correct at 10x the length and 100x the pomposity, $\endgroup$
    – demully
    Commented Jul 17, 2021 at 23:41

Delta is not the probability of finishing in the money as suggested in another answer, N(d2) is. The foot note mentions this. See Understanding N(d1) and N(d2): Risk-Adjusted Probabilities in the Black-Scholes Model by Lars Tyge Nielsen for a detailed explanation. If time and vol is low, $d1 \approx d2$ and delta will be closer to the risk-adjusted probability of the event that the option will finish in the money, which is $P(S_T > X) = N(d2)$.

Actual moneyness in terms of spot is unaffected by time. The same strike $K$ will always have the same moneyness against spot. E.g. is $S=1$ and $K= 1.05$, moneyness will be $1.05/1-1 = 5 \% \ OTMS$, where OTMS stand for out-the-money-spot. However, moneyness in terms of forward is affected by time, but does usually not make it more ATM either. Using covered interest rate parity and continuous compounding $$fwd(s,ccy2,ccy1,t) = s*exp^{(ccy2-ccy1)*t}$$ one can see how the value of the forward depends on the relative size of the two interest rates (and time). For a 10y forward, the moneyness in terms of forward would be $\approx 14\% \ ITMF$ with $F_{10y} \approx 1.22$ and $K=1.05$ as above. So this is actually further from ATM than spot moneyness (and it flips the side from OTM to ITM). enter image description here

I think Iain Clark is eluding to something similar to the effect of implied vol on delta. IVOL and time are very similar in that sense.

Here, only one of the two "forces" mentioned in the link really matters: The term $$\frac{log(\frac{S}{K})}{\sigma\sqrt t}$$ in d1 converges to 0 as $t \rightarrow \infty$ (the larger vol, the quicker it converges).

When t is very small, you can see that "time and vol adjusted (log) moneyness" is playing a significant role. If t increases, the difference between S and K eventually becomes negligible.

enter image description here

Just as delta is an increasing function in vol, it also grows in time. The probability of ending up in the money can counterintuitively turn zero for large vol and time. This is because the higher σ, the more the global maximum of the probability density function (the mode) shifts towards the lower bound of the lognormal distribution. enter image description here

Comparing a 1y 10D call vs a 10y 10D call, with same $\sigma = 10\%$, $r_{CCY1}=-1\%$ and $r_{CCY2}=1\%$ gives use as @q.t.f wrote a strike that is much farther from the forward for the long maturity option.

function GKMSpot(S, K,t,ccy1,ccy2,σ)
    d1 = ( log(S/K) +  ( ccy2 -ccy1 + 0.5*σ^2)*t ) / (σ*sqrt(t))
    d2 = d1 - σ*sqrt(t)
    c  = S*exp(-ccy1*t)*N(d1)-K*exp(-ccy2*t)*N(d2)
    delta = exp(-ccy1*t)*N(d1)
  return c, round(delta*100,digits=2)

enter image description here


It depends on how close your strike is to the forward (at expiry). Lets say you have an option which is expiring in a week, the forward will be close to the spot. Hence an out of the money strike for such as option will be closer to the at the money for an option expiring in 6 months (where the forward is pretty far from the spot).


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