Evaluating Curran’s Shifted-Normal Integral for Asian Options
Summary
The document derives a tail-probability expression used in Curran’s approximation for arithmetic Asian options on several futures. It corrects an initial distribution mix-up: the density in the integral is normal, with mean μ(G) and variance σ²(G), rather than lognormal. Standardizing the integration variable converts the density to the standard normal form.
The exponential factor can then be combined with the normal density by completing the square. This shifts the standardized variable by σⱼ,ₗ,G/σ(G), so the lower integration bound shifts by the corresponding amount. The remaining integral is the upper tail of the standard normal distribution, expressed as one minus its cumulative distribution at the shifted bound. The answer is an algebraic derivation rather than a numerical example or validation. Its result depends on the stated normal density and parameter definitions; the original question’s lognormal description would not justify this calculation as written.
Key ideas
- The density f(x) in the derivation is normal with mean μ(G) and variance σ²(G).\nStandardizing x gives a standard normal integration variable.\nCombining the exponential tilt with the density is equivalent to shifting the normal variable.\nThe integral evaluates to a standard normal upper-tail probability at the shifted lower bound.
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# Arithmetic Asian Call Option under Curran Model
# Arithmetic Asian Call Option under Curran Model
I am reading Curran's Method for Approximating Arithmetic Average Option for Several Futures. I have a difficulty on this integral on Page 17. How to evaluate this integral?
$\int_{lnK}^{\infty} e^{\frac{\sigma_{j,l,G}}{\sigma^2(G)}(x - \mu(G)) - \frac{1}{2}\frac{\sigma^2_{j,l,G}}{\sigma^2(G)} } f(x)dx = (1-\Phi(\frac{ln{K} - \mu(G) - \sigma_{j,l,G}}{\sigma(G)}))$
$f(x)$ is the PDF of the lognormal distribution, $\Phi$ is the CDF of normal distribution.
Update @ 24 Mar 2025: I read the paper again. The probability density function $f(x)$ is pdf of normal distribution instead of that of lognormal distribution.
## Answer by StupidMan (score 0, accepted)
https://quant.stackexchange.com/a/82203
$\int_{lnK}^{\infty} e^{\frac{\sigma_{j,l,G}}{\sigma^2(G)}(x - \mu(G)) - \frac{1}{2}\frac{\sigma^2_{j,l,G}}{\sigma^2(G)} } f(x)dx$, where $f(x)$ is the probability density function of normal distribution $N(\mu(G), \sigma^2(G))$
$=\int_{lnK}^{\infty} e^{\frac{\sigma_{j,l,G}}{\sigma^2(G)}(x - \mu(G)) - \frac{1}{2}\frac{\sigma^2_{j,l,G}}{\sigma^2(G)} } \frac{1}{\sigma(G)\sqrt{2\pi}}e^{-\frac{(x-\mu(G))^2}{2\sigma^2(G)}}dx$
Let $z = \frac{x - \mu(G)}{\sigma(G)}$ implies $dx = \sigma(G)dz$
$=\int_{\frac{lnK - \mu(G)}{\sigma(G)}}^{\infty} e^{\frac{\sigma_{j,l,G}}{\sigma(G)}z - \frac{1}{2}\frac{\sigma^2_{j,l,G}}{\sigma^2(G)} } \frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}dz$
$=\int_{\frac{lnK - \mu(G)}{\sigma(G)}}^{\infty} \frac{1}{\sqrt{2\pi}}e^{\frac{-z^2 + 2\frac{\sigma_{j,l,G}}{\sigma(G)}z - \frac{1}{2}\frac{\sigma^2_{j,l,G}}{\sigma^2(G)}}{2}}dz$
$=\int_{\frac{lnK - \mu(G)}{\sigma(G)}}^{\infty} \frac{1}{\sqrt{2\pi}}e^{\frac{-(z - \frac{\sigma_{j,l,G}}{\sigma(G)})^2}{2}}dz$
Let $y = z - \frac{\sigma_{j,l,G}}{\sigma(G)}$ implies $dy = dz$
$=\int_{\frac{lnK - \mu(G) - \sigma_{j,l,G}}{\sigma(G)}}^{\infty} \frac{1}{\sqrt{2\pi}}e^{\frac{-y^2}{2}}dy$
$ = (1-\Phi(\frac{ln{K} - \mu(G) - \sigma_{j,l,G}}{\sigma(G)}))$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.