Tail Behavior of the Inverse-Gaussian Distribution

Why examine both ends of the support?

The inverse-Gaussian distribution has positive support, but its density behaves very differently near the two ends of that support. Close to zero, the leading exponential term depends on the reciprocal of the response; far to the right, it depends on the response itself. Finding both forms makes the endpoint behavior precise and shows why a single description of “the tail” is not enough.

Write $Y\sim IG(\mu,\lambda)$, with mean parameter $\mu>0$ and shape parameter $\lambda>0$. The density is defined for $y>0$. In this tutorial, $g(y)\sim h(y)$ means $g(y)/h(y)\to 1$ at the stated endpoint. This ratio-based definition distinguishes a leading asymptotic form from an approximation that is merely close at one particular value of $y$.

A factorization that reveals the endpoint behavior

Definition: For $\mu>0$, $\lambda>0$, and $y>0$, the inverse-Gaussian density is $$ f(y;\mu,\lambda) = \left(\frac{\lambda}{2\pi y^3}\right)^{1/2} \exp\left\{-\frac{\lambda(y-\mu)^2}{2\mu^2y}\right\}. $$ Define the positive constants $a=\lambda/(2\mu^2)$, $b=\lambda/2$, and $A=\sqrt{\lambda/(2\pi)}\,e^{\lambda/\mu}$. Then an equivalent form is $f(y;\mu,\lambda)=A y^{-3/2}\exp(-ay-b/y)$.

To verify the factorization, expand the quadratic expression:

\frac{(y-\mu)^2}{\mu^2y} = \frac{y}{\mu^2}-\frac{2}{\mu}+\frac{1}{y}.

Substitution into the density gives the exponent $-\lambda y/(2\mu^2)+\lambda/\mu-\lambda/(2y)$. The middle term is constant in $y$, so it is absorbed into $A$. The resulting form separates the two endpoint contributions: $-b/y$ is decisive near zero, while $-ay$ is decisive as $y$ tends to infinity. The factor $y^{-3/2}$ is present at both ends.

Worked Example: Exposing Both Rates

Let $Y\sim IG(2,3)$. Find the constants in the factored density and identify the leading exponential term at each endpoint.

Here $\mu=2$ and $\lambda=3$, so $a=3/(2\cdot 2^2)=3/8$, $b=3/2$, and $A=\sqrt{3/(2\pi)}e^{3/2}$. Thus

f(y;2,3) = \sqrt{\frac{3}{2\pi}}e^{3/2} y^{-3/2}\exp\left(-\frac{3y}{8}-\frac{3}{2y}\right).

As $y\downarrow0$, the term $-3/(2y)$ determines the exponential decay; as $y\to\infty$, it is $-3y/8$. The polynomial factor in both cases is $y^{-3/2}$. Neither endpoint rate can be obtained by treating the other exponential term as if it were the dominant one.

Density asymptotics at zero and infinity

Theorem: For $Y\sim IG(\mu,\lambda)$, with $\mu,\lambda>0$, and $A,a,b$ as defined above, $$ f(y;\mu,\lambda)\sim A y^{-3/2}e^{-b/y}\quad(y\downarrow0), \qquad f(y;\mu,\lambda)\sim A y^{-3/2}e^{-ay}\quad(y\to\infty). $$ Equivalently, the near-zero exponential rate is $b=\lambda/2$ in $1/y$, and the far-right exponential rate is $a=\lambda/(2\mu^2)$ in $y$.

Proof. From the exact factorization, the ratio of the density to the proposed near-zero leading form is

\frac{f(y;\mu,\lambda)}{A y^{-3/2}e^{-b/y}}=e^{-ay}\longrightarrow 1 \qquad(y\downarrow0),

because $a$ is a fixed positive constant. At the other endpoint,

\frac{f(y;\mu,\lambda)}{A y^{-3/2}e^{-ay}}=e^{-b/y}\longrightarrow 1 \qquad(y\to\infty),

because $b$ is fixed and positive. Both limits establish the stated asymptotic equivalences. $\square$

Taking logarithms gives another way to read the result:

\log f(y;\mu,\lambda) = \log A-\frac{3}{2}\log y-ay-\frac{b}{y}.

Near zero, the term $-b/y$ dominates the logarithmic and linear terms; at infinity, $-ay$ dominates the logarithmic and reciprocal terms. In particular, $\log f(y)/(-1/y)\to b$ as $y\downarrow0$, and $\log f(y)/(-y)\to a$ as $y\to\infty$. These limits state the exponential rates without suggesting that the two endpoints have the same functional form.

Worked Example: Checking the Approximation at Finite Values

For $Y\sim IG(1,2)$, compute the density and compare each endpoint approximation with the exact density at a value near the relevant endpoint.

The constants are $a=1$, $b=1$, and $A=e^2/\sqrt{\pi}$. Therefore

f(y;1,2)=\frac{e^2}{\sqrt{\pi}}y^{-3/2}e^{-y-1/y}.

The near-zero approximation is $(e^2/\sqrt{\pi})y^{-3/2}e^{-1/y}$. The exact density divided by this approximation is $e^{-y}$. At $y=0.1$, the ratio is $e^{-0.1}\approx0.9048$, so the exact density is about $90.48\%$ of that leading form there.

The far-right approximation is $(e^2/\sqrt{\pi})y^{-3/2}e^{-y}$. The exact density divided by it is $e^{-1/y}$. At $y=10$, that ratio is also $e^{-0.1}\approx0.9048$. These finite-value calculations illustrate the limiting ratios; the theorem’s claims concern the limits, not exact equality at $0.1$ or $10$.

From density tails to probability tails

A density value and a tail probability are different quantities. Integrating the density changes the polynomial power in the leading term. For the inverse-Gaussian density, the corresponding probability-tail equivalents can also be obtained explicitly.

Result: With $A,a,b>0$ as above, let $F(y)=P(Y\le y)$. Then $$ F(y)\sim \frac{A}{b}y^{1/2}e^{-b/y}\quad(y\downarrow0), \qquad P(Y>y)\sim\frac{A}{a}y^{-3/2}e^{-ay}\quad(y\to\infty). $$

Derivation. For the lower tail, substitute $u=b/t$ in the integral of the factored density from $0$ to $y$. With $x=b/y$, this gives

F(y) = \frac{A}{\sqrt b} \int_x^\infty u^{-1/2}e^{-u}e^{-ab/u}\,du.

For $u\ge x$, the extra factor satisfies $e^{-ab/x}\le e^{-ab/u}\le1$. As $x\to\infty$, both bounds tend to $1$, so this factor does not change the integral’s leading equivalent. Also, $\int_x^\infty u^{-1/2}e^{-u}\,du\sim x^{-1/2}e^{-x}$. For example, this last equivalence follows by l’Hôpital’s rule: the derivatives of the integral and of $x^{-1/2}e^{-x}$ have ratio $1/(1+1/(2x))\to1$. Consequently, $F(y)\sim (A/\sqrt b)(b/y)^{-1/2}e^{-b/y} =(A/b)y^{1/2}e^{-b/y}$.

For the upper tail, define $H(y)=(1/a)y^{-3/2}e^{-ay-b/y}$. Both $P(Y>y)$ and $H(y)$ tend to zero. Their derivatives are

\frac{d}{dy}P(Y>y)=-A y^{-3/2}e^{-ay-b/y}, \qquad H'(y)=H(y)\left(-\frac{3}{2y}-a+\frac{b}{y^2}\right).

As $y\to\infty$, $H'(y)\sim-y^{-3/2}e^{-ay-b/y}$, so the ratio of the two derivatives tends to $A$. L’Hôpital’s rule therefore gives $P(Y>y)/H(y)\to A$, which is the claimed upper-tail equivalent. Thus the density has power $y^{-3/2}$ at both endpoints, while integration gives power $y^{1/2}$ for the lower probability tail and $y^{-3/2}$ for the upper one. $\square$

Worked Example: Equal Right-Tail Rates, Different Near-Zero Rates

Compare $IG(1,2)$ and $IG(2,8)$. Determine whether their exponential rates match at each endpoint.

For $IG(1,2)$, $a=2/(2\cdot1^2)=1$ and $b=2/2=1$. For $IG(2,8)$, $a=8/(2\cdot2^2)=1$ and $b=8/2=4$. Both densities therefore have the same far-right exponential rate $e^{-y}$. Their leading density forms are

f(y;1,2)\sim \frac{e^2}{\sqrt{\pi}}y^{-3/2}e^{-y}, \qquad f(y;2,8)\sim \frac{2e^4}{\sqrt{\pi}}y^{-3/2}e^{-y} \qquad(y\to\infty).

The rates and polynomial powers match, but the constants do not. Near zero, the rates differ: the first density has the factor $e^{-1/y}$, while the second has $e^{-4/y}$. The parameters $a$ and $b$ describe distinct endpoint features, so matching one rate does not imply matching the other.

Common mistake: treating both endpoints as ordinary exponential tails

A tempting but incorrect shortcut is to call the inverse-Gaussian density “exponentially decaying” and use the same form at both ends. That overlooks the reciprocal term: as $y\downarrow0$, the controlling factor is $e^{-b/y}$, not $e^{-by}$. Nor does the shared factor $y^{-3/2}$ mean it controls the overall decay. At either endpoint, the relevant exponential term dominates the logarithm of the polynomial factor.

A second pitfall is to transfer the density’s power directly to a probability tail. The lower-tail density has a $y^{-3/2}$ factor, but its integral has a $y^{1/2}$ factor. The upper-tail probability, in contrast, retains the $y^{-3/2}$ power. When discussing endpoint behavior, specify whether the object is the density or a probability.

Key takeaway: The inverse-Gaussian density factors exactly as $A y^{-3/2}e^{-ay-b/y}$. Its far-right rate is $a=\lambda/(2\mu^2)$; its near-zero rate is $b=\lambda/2$. Keep these distinct, and distinguish density asymptotics from integrated-tail asymptotics.

Check Your Understanding

Use the factored density to answer the following questions.

For $IG(3,6)$, calculate $A$, $a$, and $b$, and state the leading exponential term at each endpoint.
Why does the ratio of the exact density to $A y^{-3/2}e^{-b/y}$ tend to $1$ as $y\downarrow0$?
For $IG(1,4)$, what is the far-right exponential rate, and what is the near-zero exponential rate?
How does the power of $y$ in the lower-tail probability equivalent differ from the power in the density equivalent near zero?
Two inverse-Gaussian models have the same $a$ but different $b$. Which endpoint rate matches, and which endpoint rate differs?