Mills Ratio Bounds for Gaussian Tails

Why Bound a Gaussian Tail?

For a standard normal random variable, the probability of exceeding a large positive threshold can be much smaller than its density at that threshold. The density is not itself a tail probability: it measures probability per unit length, while a tail probability accumulates density over an entire interval. A useful way to compare them is the Mills ratio, the upper-tail probability divided by the density. Bounds on this ratio give immediate, explicit bounds on Gaussian tail probabilities without evaluating an integral exactly.

These bounds are especially informative for thresholds in the far right tail. They also require care near zero: the density is not larger than the upper-tail probability for every positive threshold. The results below specify exactly what can be concluded from the bounds.

Definition: Let $Z\sim N(0,1)$, with density $\phi(x)=(2\pi)^{-1/2}\exp(-x^2/2)$ and distribution function $\Phi(x)=P(Z\le x)$. Its survival function is $\bar{\Phi}(x)=P(Z>x)=1-\Phi(x)$. For $x>0$, the Mills ratio is $R(x)=\bar{\Phi}(x)/\phi(x)$.

Because the standard normal distribution is continuous, $P(Z>x)=P(Z\ge x)$. The survival function has the integral representation $\bar{\Phi}(x)=\int_x^\infty \phi(t)\,dt$. We will compare this accumulated density with the density $\phi(x)$ at the threshold.

Deriving the Mills Bounds

For positive $x$, the standard Mills bounds are

\frac{x}{x^2+1}\,\phi(x) < \bar{\Phi}(x) < \frac{\phi(x)}{x}.

Both inequalities are strict. The upper bound follows by replacing the integrand with a larger one that can be integrated exactly. For the lower bound, integration by parts expresses the tail as an explicit leading term minus a remainder; that remainder can itself be bounded using the tail.

Theorem: For every $x>0$, $$ \frac{x}{x^2+1}<R(x)<\frac{1}{x}. $$ Equivalently, multiplying by the positive quantity $\phi(x)$ gives the Mills bounds for $\bar{\Phi}(x)$.

Proof of the upper bound. Since $t/x>1$ for $t>x$, and $\phi(t)>0$,

\bar{\Phi}(x) = \int_x^\infty \phi(t)\,dt < \frac{1}{x}\int_x^\infty t\phi(t)\,dt = \frac{\phi(x)}{x}.

The final equality follows from $\phi'(t)=-t\phi(t)$ and $\phi(t)\to 0$ as $t\to\infty$.

Proof of the lower bound. Since $(\phi(t)/t)'=-\phi(t)-\phi(t)/t^2$ for $t>0$, integration from $x$ to $b$, followed by $b\to\infty$, gives

\bar{\Phi}(x) = \frac{\phi(x)}{x} - \int_x^\infty \frac{\phi(t)}{t^2}\,dt.

This passage to the limit is valid: $\phi(b)/b\to0$, and the nonnegative remainder integral is bounded above by $\bar{\Phi}(x)/x^2$, since $t\ge x$ throughout the integration interval. More precisely, $1/t^2<1/x^2$ for $t>x$, so

\int_x^\infty \frac{\phi(t)}{t^2}\,dt < \frac{\bar{\Phi}(x)}{x^2}.

Substituting this strict bound into the integration-by-parts identity yields

\bar{\Phi}(x) > \frac{\phi(x)}{x} - \frac{\bar{\Phi}(x)}{x^2}.

Rearranging, with $1+1/x^2>0$, gives $\bar{\Phi}(x)>x\phi(x)/(x^2+1)$, as required. $\square$

Using the Bounds to Approximate Tail Probabilities

The two sides of the inequality give a guaranteed interval for the tail probability. A midpoint of that interval is one possible approximation; its absolute error cannot exceed half the interval's width. The midpoint is not asserted to be the exact probability or to be optimal for every purpose.

Worked Example: Bracketing a Standard Normal Tail at Three

Find Mills bounds for $P(Z>3)$, and use the midpoint of the bounds as an approximation.

First calculate the density:

\phi(3) = \frac{1}{\sqrt{2\pi}}e^{-9/2} \approx 0.0044318484.

The lower bound is $3\phi(3)/(3^2+1)=0.3\phi(3)$, and the upper bound is $\phi(3)/3$. Thus

$$ 0.0013295545 < P(Z>3) < 0.0014772828. $$

Their midpoint is $(0.0013295545+0.0014772828)/2\approx0.0014034187$. The interval width is approximately $0.0001477283$, so the midpoint's absolute error is at most $0.0000738642$. For comparison, direct numerical evaluation gives $P(Z>3)\approx0.0013498980$, which lies inside the derived interval.

The bounds become relatively close for large $x$. In fact, their ratio, lower bound divided by upper bound, is $x^2/(x^2+1)$, which tends to one. This observation gives a precise first-order comparison between the tail and the density.

Theorem: As $x\to\infty$, $R(x)\sim 1/x$, meaning that $xR(x)\to1$.

Proof. Multiply the Mills ratio bounds by $x>0$:

\frac{x^2}{x^2+1}<xR(x)<1.

The lower bound tends to one as $x\to\infty$, and the upper bound equals one. The squeeze theorem therefore gives $xR(x)\to1$. Equivalently, for large positive $x$, the tail is approximately $\phi(x)/x$ in the sense that $\bar{\Phi}(x)/(\phi(x)/x)\to1$. $\square$

Worked Example: A Tighter Relative Bracket at Four

Use the bounds to estimate $P(Z>4)$, and examine how closely they bracket one another.

The density is $\phi(4)=(2\pi)^{-1/2}e^{-8}\approx0.0001338302$. Therefore

\frac{4}{17}\phi(4) < P(Z>4) < \frac{1}{4}\phi(4),

or numerically,

$$ 0.0000314895 < P(Z>4) < 0.0000334576. $$

The ratio of the lower endpoint to the upper endpoint is $ (4/17)/(1/4)=16/17\approx0.9412$. Thus the lower bound is about 94.1% of the upper bound, a relatively narrow bracket. Numerical evaluation gives $P(Z>4)\approx0.0000316712$, inside the interval. The bounds also show that the tail is less than $\phi(4)/4$, so the density at four is more than four times the tail probability.

Comparison with the Density and a General Normal Variable

The upper bound implies a simple density-versus-tail comparison when $x\ge1$. For such $x$, $1/x\le1$, so

\bar{\Phi}(x)<\frac{\phi(x)}{x}\le\phi(x).

Hence the density is strictly larger than the upper-tail probability for $x\ge1$. This conclusion does not extend to every positive $x$. At zero, $\phi(0)=1/\sqrt{2\pi}\approx0.3989$, whereas $\bar{\Phi}(0)=1/2$. By continuity, the tail remains larger than the density for some positive values near zero.

Worked Example: Applying a Mills Bound to a Normal Location-Scale Model

Suppose $X\sim N(10,4)$, so its mean is $10$ and its variance is $4$. Bound $P(X>14)$.

The standard deviation is $\sigma=2$. Standardizing gives $P(X>14)=P((X-10)/2>2)=\bar{\Phi}(2)$. Since $\phi(2)\approx0.05399097$, the Mills bounds at $x=2$ give

\frac{2}{5}\phi(2) < P(X>14) < \frac{1}{2}\phi(2),

and hence

$$ 0.02159639 < P(X>14) < 0.02699548. $$

Direct evaluation gives approximately $0.02275013$, which is within the bracket. The variance parameter matters in the standardization: using $4$ as the standard deviation instead of $2$ would produce the wrong standardized threshold and therefore the wrong tail probability.

Common Mistake: Claiming Density Dominance Too Soon

A tempting but incorrect argument is: “For positive $x$, the upper bound gives $\bar{\Phi}(x)<\phi(x)/x$, and therefore $\bar{\Phi}(x)<\phi(x)$.” The final implication does not follow unless $1/x\le1$, which requires $x\ge1$. If $0<x<1$, the upper bound $\phi(x)/x$ is larger than $\phi(x)$, so it does not settle the comparison.

For instance, at $x=0.1$, $\phi(0.1)\approx0.39695$, while $\bar{\Phi}(0.1)\approx0.46017$. Here the tail is larger than the density. The correct conclusion from the upper Mills bound is that the density exceeds the tail for $x\ge1$; for general positive $x$, the Mills bounds compare the two quantities but do not by themselves establish density dominance.

Key takeaway: For every $x>0$, the standard normal tail lies strictly between $x\phi(x)/(x^2+1)$ and $\phi(x)/x$. These bounds yield the far-tail approximation $\bar{\Phi}(x)\sim\phi(x)/x$; density dominance follows directly for $x\ge1$, not for all positive $x$.

Check Your Understanding

Use the inequalities and distinctions developed above to answer these questions.

State the strict Mills bounds for $\bar{\Phi}(x)$, including the required condition on $x$.
In the lower-bound proof, why is $\int_x^\infty \phi(t)/t^2\,dt<\bar{\Phi}(x)/x^2$ valid?
What limit does the squeeze theorem give for $xR(x)$ as $x\to\infty$, and what does it say about the tail?
For what range of $x$ does the upper Mills bound directly prove that $\phi(x)>\bar{\Phi}(x)$?
For $X\sim N(5,9)$, what standard normal threshold corresponds to $P(X>11)$?