The ATMF straddle approximation formula, given by

$V_\text{Str}(S, T) \approx \sqrt{\frac{2}{\pi}} S_0 \sigma \sqrt{T}$

where $S_0$ is the current underlying spot price, $T$ is the time remaining until expiration, and $\sigma$ is the volatility of log returns. The derivation using the Black-Scholes formula and Taylor expanding the normal cdf is easily found, for instance here: https://brilliant.org/wiki/straddle-approximation-formula/.

However, we have that for $Z \sim N(0, 1)$, $\mathbb{E}[\sigma|Z|] = \sigma \sqrt{2/\pi}$ (the MAD). The ATMF straddle value is, under risk-neutral measure, $\mathbb{E}[|S - K|]$ (assuming $S$ follows GBM of course). My question is, how to argue this approximation directly from integration?

My attempt so far: since $K = S_0 e^{rT}$

$\int_0^\infty |S - S_0 e^{rT}| p_T(S) dS = S_0 \int_0^\infty |S/S_0 - e^{rT}| p_T(S) dS$

where $p_T$ is the risk neutral pdf of prices (lognormal as defined by GBM). Next, let $y = \log(S/S_0)$ so $dy = dS/S$. Then, after altering the limits of integration based on the substitution,

$\int_{-\infty}^\infty |e^y - e^{rT}| \varphi\left(\frac{y - (r - \sigma^2/2)T}{\sigma \sqrt{T}} \right) dy$

where $\varphi$ is the standard normal pdf. In other words, $Y = (r-\sigma^2/2)T + \sigma \sqrt{T} Z$. Now, we can Taylor expand $e^{rT} \approx 1 + rT$ and $e^y \approx 1 + y$ so, performing another variable change to convert to standard normal notation

$\int_{-\infty}^\infty |1 + (r-\sigma^2/2)T + \sigma\sqrt{T}Z - (1+rT)| \varphi(z) dz = \int_{-\infty}^\infty |\sigma\sqrt{T}Z - \sigma^2T/2| \varphi(z) dz$

which seems to suggest that the result holds when $\sigma^2T/2 \approx 0$, i.e. $\sigma$ and $T$ are not too large. This relates to the more common derivation, since we see that the linear approximation of the normal cdf gets worse at higher variances here: https://www.desmos.com/calculator/rvmmhlxxx9.

See any issues? Or a different way to accomplish it?

  • $\begingroup$ Not sure you did it right. Shouldn’t that be a $e^y$ rather than a $y$ in the integral ? $\endgroup$
    – dm63
    Commented Nov 5, 2023 at 11:12
  • $\begingroup$ That's it! Lmk if you have any thoughts after the edit $\endgroup$
    – aarongroff
    Commented Nov 5, 2023 at 18:54
  • $\begingroup$ You reached the right conclusion. $\endgroup$
    – dm63
    Commented Nov 5, 2023 at 22:25


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.