# Straddle Approximation - Directly from Integral

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?

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