I put this proof together while studying T cell movement patterns, where I needed a clear comparison between the tail behaviours of the lognormal and gamma distributions. Although the result that the lognormal has heavier tails is often stated in textbooks, articles, and blog posts but the details are left as a heuristic or are implied from moment-based arguments. This specific proof, using a direct comparison of log-densities and asymptotic limits, does not appear to have been published in this form. I have written it out here both for clarity and as a reference for others looking for an explicit derivation.
Proof
Let \( f(x) \) be the gamma probability density function, and \( g(x) \) the lognormal probability density function.
To compare their tail behaviours, we analyse the limit:
\[
\log f(x) - \log g(x)
\]
as \( x \to \infty \).
The gamma density is asymptotically: \[ \log f(x) = (\alpha - 1)\log x - \frac{x}{\beta} + \text{const} \]
The lognormal density is: \[ \log g(x) = -\log x - \frac{(\log x - \mu)^2}{2\sigma^2} + \text{const} \]
Expanding the quadratic term: \[ \log g(x) = -\log x - \frac{(\log x)^2 - 2\mu \log x + \mu^2}{2\sigma^2} + \text{const} \]
Taking the difference: \[ \log f(x) - \log g(x) = (\alpha - 1 + 1)\log x - \frac{x}{\beta} + \frac{(\log x)^2 - 2\mu \log x + \mu^2}{2\sigma^2} + \text{const} \]
Simplifying: \[ \log f(x) - \log g(x) = -\frac{x}{\beta} + \frac{(\log x)^2}{2\sigma^2} + \text{lower-order terms} \]
As \( x \to \infty \), the linear term dominates: \[ -\frac{x}{\beta} + \frac{(\log x)^2}{2\sigma^2} \to -\infty \]
Therefore: \[ \log \left( \frac{f(x)}{g(x)} \right) \to -\infty \quad \Rightarrow \quad \frac{f(x)}{g(x)} \to 0 \]
Hence, the gamma density decays faster than the lognormal in the tail: \[ f(x) \in o(g(x)) \quad \text{as } x \to \infty \]
This confirms that the lognormal distribution has heavier tails than the gamma distribution.
Q.E.D.