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A marginal statistical likelihood uses a statistic whose marginal sampling distribution depends on the parameter of interest but does not involve the nuisance parameter. It integrates out other observed coordinates. A profile likelihood instead keeps the full data and, for every fixed interest parameter, fits the nuisance parameter by maximizing the full likelihood. Profiling supplies a likelihood function but usually not a normalized sampling probability density function. Here the marginalization is of data coordinates; Bayesian model evidence averages parameters against a prior and is a separate construction.

Use shape-rate notation for the gamma distribution. The independent sums have $X\sim \mathrm{Gamma}(m,\psi \lambda )$ and $Y\sim \mathrm{Gamma}(n,\lambda )$, so their joint probability density for $x,y>0$ isFor $x=tu$, $y=u$ the absolute Jacobian determinant is u. ThereforeIntegrate u using the gamma integral ${\int }_0^{\infty }{u}^{m+n-1}{e}^{-bu}du=\Gamma (m+n)b^{-(m+n)}$. The lambda factors cancel, leavingThus T gives a nuisance-free marginal likelihood. Discarding only terms independent of psi, its logarithm isThe factor t in the logarithm is required, as in the original PDF.

The full log-likelihood is $m\log \psi +(m+n)\log \lambda -\lambda (\psi x+y)+c(x,y)$. Its lambda derivative vanishes uniquely atThe second derivative in lambda is negative. Substituting this maximum gives ${\ell }_p(\psi )=m\log \psi -(m+n)\log (\psi x+y)+c^{\prime }(x,y)$. Since $\psi x+y=y(1+\psi t)$, it agrees with ${\ell }_m(\psi ;t)$ up to a psi-independent constant. This proves the Gamma-ratio marginal and profile likelihood identity. Their common maximizer is $\hat{\psi }=m/(nt)=my/(nx)$, if an interest-parameter estimate is also wanted.