Inverse-Gaussian and Gamma Tail Comparison

Introduction

Inverse-Gaussian and gamma responses are both positive and right-skewed, so their densities can look similar over a substantial range of values. But their behavior near zero and far into the upper tail is governed by different terms in their density formulas. Those differences matter when assessing whether a family can assign appreciable density to very small or very large responses.

There is no parameter-free statement that one family always has the heavier upper tail. The comparison depends on the parameters, and even matching the means and variances need not match the upper-tail decay rates. We will compare densities first, keeping that distinction from tail probabilities explicit.

Parameterizations and Density Ratio

Definition: An inverse-Gaussian random variable with mean $\mu>0$ and shape parameter $\lambda>0$, written $Y\sim\operatorname{IG}(\mu,\lambda)$, has density $$ f_{\mathrm{IG}}(y)=\sqrt{\frac{\lambda}{2\pi y^3}}\exp\!\left[-\frac{\lambda(y-\mu)^2}{2\mu^2y}\right],\qquad y>0. $$ A gamma random variable with shape $\alpha>0$ and rate $\beta>0$, written $Y\sim\operatorname{Gamma}(\alpha,\beta)$, has density $$ f_{\Gamma}(y)=\frac{\beta^\alpha}{\Gamma(\alpha)}y^{\alpha-1}e^{-\beta y},\qquad y>0. $$ Here $\Gamma(\alpha)$ is the gamma function. Both densities are zero outside the positive real line.

For the inverse-Gaussian density, expand the quadratic in the exponent:

-\frac{\lambda(y-\mu)^2}{2\mu^2y} =-\frac{\lambda}{2\mu^2}y+\frac{\lambda}{\mu}-\frac{\lambda}{2y}.

Thus its upper-tail exponential decay rate is $c=\lambda/(2\mu^2)$, while its density near zero includes the factor $\exp[-\lambda/(2y)]$. By contrast, the gamma density has ordinary exponential decay $e^{-\beta y}$ at infinity and only a power of $y$ multiplying it near zero. Dividing the two densities gives the exact ratio

\frac{f_{\mathrm{IG}}(y)}{f_{\Gamma}(y)} = \frac{\sqrt{\lambda/(2\pi)}\,\Gamma(\alpha)e^{\lambda/\mu}}{\beta^\alpha} y^{-\alpha-1/2} \exp\!\left[-\frac{\lambda}{2y} -\left(\frac{\lambda}{2\mu^2}-\beta\right)y\right].

Two Tail Comparison Results

Theorem: For any $\mu,\lambda,\alpha,\beta>0$, the inverse-Gaussian density is asymptotically smaller than the gamma density as $y\downarrow0$: $$ \lim_{y\downarrow0}\frac{f_{\mathrm{IG}}(y)}{f_{\Gamma}(y)}=0. $$ At the upper end, set $c=\lambda/(2\mu^2)$. If $c<\beta$, the ratio tends to infinity; if $c\geq\beta$, the ratio tends to zero. Thus, for $c<\beta$, the inverse-Gaussian density eventually exceeds the gamma density, whereas for $c\geq\beta$, the gamma density eventually exceeds the inverse-Gaussian density.

Proof. In the exact ratio, all factors except $y^{-\alpha-1/2}\exp[-\lambda/(2y)]$ have finite limits as $y\downarrow0$, with the $e^{-(c-\beta)y}$ factor tending to one. Put $t=1/y$. The relevant variable factor becomes $t^{\alpha+1/2}e^{-\lambda t/2}$. Its logarithm is $(\alpha+1/2)\log t-\lambda t/2$, which tends to $-\infty$ because $\log t/t\to0$. The ratio therefore tends to zero.

For the upper end, take the logarithm of the ratio. It is a constant plus $-(\alpha+1/2)\log y-(c-\beta)y-\lambda/(2y)$. If $c<\beta$, the positive linear term $(\beta-c)y$ dominates the logarithmic term, so the log ratio tends to $+\infty$. If $c>\beta$, the negative linear term dominates and the log ratio tends to $-\infty$. If $c=\beta$, the linear term vanishes, but $-(\alpha+1/2)\log y\to-\infty$. These limits give the stated ratio limits and density ordering. $\square$

The equality case is informative: matching the exponential rates does not make the densities asymptotically equal. The inverse-Gaussian density has a $y^{-3/2}$ factor, whereas the gamma density has $y^{\alpha-1}$; their ratio still tends to zero for every $\alpha>0$. The exact ratio makes this polynomial distinction visible.

Worked Comparisons

Worked Example: Matching Mean and Variance Does Not Match the Upper Tail

Take $Y_{\mathrm{IG}}\sim\operatorname{IG}(2,4)$ and $Y_{\Gamma}\sim\operatorname{Gamma}(2,1)$. For these parameterizations, the inverse-Gaussian mean and variance are $\mu$ and $\mu^3/\lambda$, while the gamma mean and variance are $\alpha/\beta$ and $\alpha/\beta^2$. Both choices therefore have mean $2$ and variance $2$.

For the inverse-Gaussian density, the prefactor is $\sqrt{4/(2\pi)}=\sqrt{2/\pi}$. Expanding the exponent gives \[ -\frac{4(y-2)^2}{2(2^2)y} =-\frac{(y-2)^2}{2y} =-\frac{y}{2}+2-\frac{2}{y}. \] Hence

f_{\mathrm{IG}}(y)=\sqrt{\frac{2}{\pi}}\,y^{-3/2} \exp\!\left(2-\frac{2}{y}-\frac{y}{2}\right), \qquad f_{\Gamma}(y)=y e^{-y}.

The ratio is

\frac{f_{\mathrm{IG}}(y)}{f_{\Gamma}(y)} =\sqrt{\frac{2}{\pi}}\,e^2 y^{-5/2} \exp\!\left(-\frac{2}{y}+\frac{y}{2}\right).

At $y=1$, the inverse-Gaussian density is $\sqrt{2/\pi}e^{-1/2}\approx0.484$, while the gamma density is $e^{-1}\approx0.368$. Near zero the ratio tends to zero. At infinity, the inverse-Gaussian rate is $c=4/(2\cdot 2^2)=1/2$, less than the gamma rate $\beta=1$, so the ratio tends to infinity. The inverse-Gaussian density is therefore eventually larger in the upper tail, despite the matched means and variances.

Worked Example: Equal Upper-Tail Exponential Rates

Let $Y_{\mathrm{IG}}\sim\operatorname{IG}(1,2)$ and $Y_{\Gamma}\sim\operatorname{Gamma}(1,1)$. The inverse-Gaussian rate is $c=2/(2\cdot1^2)=1$, equal to the gamma rate $\beta=1$. Expanding the inverse-Gaussian exponent yields $-\frac{2(y-1)^2}{2y}=2-y-\frac{1}{y}$. The densities and their ratio are therefore

f_{\mathrm{IG}}(y)=\frac{1}{\sqrt{\pi}}y^{-3/2} \exp\!\left(2-y-\frac{1}{y}\right), \qquad f_{\Gamma}(y)=e^{-y},

\frac{f_{\mathrm{IG}}(y)}{f_{\Gamma}(y)} =\frac{e^2}{\sqrt{\pi}}y^{-3/2}\exp\!\left(-\frac{1}{y}\right).

As $y\to\infty$, the final exponential tends to one and the power $y^{-3/2}$ tends to zero. Thus the gamma density eventually exceeds the inverse-Gaussian density, even though their upper-tail exponential rates are equal. At the other endpoint, the factor $\exp(-1/y)$ drives the ratio to zero as well.

Worked Example: A Gamma Density That Dominates in the Upper Tail

Take $Y_{\mathrm{IG}}\sim\operatorname{IG}(1,8)$ and $Y_{\Gamma}\sim\operatorname{Gamma}(2,1)$. For the inverse-Gaussian density, the prefactor is $\sqrt{8/(2\pi)}=2/\sqrt{\pi}$, and the exponent expands as \[ -\frac{8(y-1)^2}{2y}=8-4y-\frac{4}{y}. \] The gamma density is $y e^{-y}$. Dividing gives

\frac{f_{\mathrm{IG}}(y)}{f_{\Gamma}(y)} =\frac{2e^8}{\sqrt{\pi}}y^{-5/2} \exp\!\left(-\frac{4}{y}-3y\right).

Here $c=8/2=4$, which is greater than $\beta=1$, so the ratio tends to zero in the upper tail. As a numerical check at $y=1$, the inverse-Gaussian density is $2/\sqrt{\pi}\approx1.128$, and the gamma density is $e^{-1}\approx0.368$. The ratio is $2e/\sqrt{\pi}\approx3.067$, agreeing with $1.128/0.368$ up to rounding. The inverse-Gaussian density is larger at this particular value, but the gamma density eventually dominates as $y$ increases.

Interpreting the Comparison Carefully

A density comparison is a pointwise comparison of density heights. It is not, by itself, a comparison of probabilities such as $P(Y>u)$. A density has units inverse to those of the response, and the probability over an interval is an integral of the density over that interval. The limits above establish which density is larger sufficiently far into the indicated region; they do not assert an ordering at every positive $y$.

The lower-tail result is especially strong in one respect: for any fixed positive parameters, the inverse-Gaussian density is suppressed near zero by $\exp[-\lambda/(2y)]$, while the gamma density approaches zero, a finite positive value, or infinity according to whether $\alpha>1$, $\alpha=1$, or $0<\alpha<1$. In all three cases, the inverse-Gaussian-to-gamma density ratio still tends to zero.

For the upper tail, compare rates before comparing polynomial powers. The inverse-Gaussian rate $c=\lambda/(2\mu^2)$ is determined by both its shape and mean, while the gamma rate is $\beta$. If $c<\beta$, inverse-Gaussian exponential decay is slower and wins over any fixed power difference. If $c>\beta$, gamma exponential decay is slower. At equality, the gamma density wins by the power comparison. Neither equal means nor equal variances alone determines which of these rate cases holds.

Common Mistake

Pitfall: “The two models have the same mean and variance, so they must have the same tail behavior.” Matching these moments does not equate the densities or their decay rates. In the first worked example, the matched moments give $c=1/2$ for the inverse-Gaussian density and $\beta=1$ for the gamma density, so the inverse-Gaussian density eventually dominates in the upper tail. The correction is to expand the density exponents, compare $c$ with $\beta$, and, if the rates are equal, compare the remaining powers of $y$.

Key takeaway: Near zero, the inverse-Gaussian density is smaller relative to any gamma density with fixed positive parameters. In the upper tail, compare $c=\lambda/(2\mu^2)$ with the gamma rate $\beta$: the inverse-Gaussian density dominates only when $c<\beta$; otherwise the gamma density dominates eventually.

Check Your Understanding

Use the exact density ratio and its limiting behavior to answer these questions.

For fixed positive parameters, what factor in the inverse-Gaussian density forces its density ratio with a gamma density to zero near the origin?
If $c>\beta$, which density is eventually larger as $y\to\infty$, and why?
When $c=\beta$, what remaining factor determines the upper-tail density ordering?
For $\operatorname{IG}(2,8)$, calculate $c$. Compare it with the rate of a $\operatorname{Gamma}(3,1)$ density and state the resulting upper-tail density ordering.
Why does an eventual ordering of density heights not, by itself, establish an ordering of probabilities $P(Y>u)$?