Gaussian, Logistic, and Laplace Tail Comparison

Why Compare Survival Tails?

A distribution’s center and spread do not fully describe the chance of an unusually large observation. The survival function $P(X>x)$ measures the probability of exceeding a threshold $x$. Its rate of decay matters when assessing rare large values, but comparisons require care: distributions with different scale parameters can assign very different probabilities at the same threshold.

Here we compare the upper tails of normal, logistic, and Laplace distributions. First we derive their large-threshold decay rates. Then we use a common calibration—mean zero and variance one—to make direct finite-threshold comparisons. The conclusions concern the upper tail; by symmetry, the same calculations describe the lower tail after reflecting the threshold.

Survival Functions and Parameters

Write $t=x-\mu$ for a threshold measured above a common location $\mu$. For a normal distribution, the scale parameter $\sigma$ is its standard deviation. For the logistic distribution, $s>0$ is its scale parameter; for the Laplace distribution, $b>0$ is its scale parameter. Their upper survival functions for $t>0$ are as follows.

Definition: If $X$ has a continuous distribution, its survival function at $x$ is $P(X>x)$. For the three location-scale families considered here, with $t=x-\mu>0$, $$ \begin{aligned} X\sim N(\mu,\sigma^2):\quad P(X>x)&=\overline{\Phi}\!\left(\frac{t}{\sigma}\right),\\ X\text{ logistic with location }\mu\text{ and scale }s:\quad P(X>x)&=\frac{1}{1+\exp(t/s)},\\ X\text{ Laplace with location }\mu\text{ and scale }b:\quad P(X>x)&=\frac{1}{2}\exp(-t/b). \end{aligned} $$ Here $\overline{\Phi}(z)=1-\Phi(z)$ is the standard normal survival function. The logistic formula follows from its CDF $F(x)=1/(1+\exp(-(x-\mu)/s))$; the Laplace formula follows by integrating its density $(2b)^{-1}\exp(-|x-\mu|/b)$ over $x$ to infinity.

These formulas already show an important distinction. The logistic and Laplace tails decay exponentially in $t$, with rates $1/s$ and $1/b$. The Gaussian tail decays on a quadratic exponential scale, involving $\exp(-t^2/(2\sigma^2))$. To make the Gaussian comparison precise, we use the standard Mills ratio bounds.

Asymptotic Decay Rates

For $z>0$, Mills ratio bounds give $$ \frac{z}{z^2+1}\,\phi(z)<\overline{\Phi}(z)<\frac{\phi(z)}{z}, \qquad \phi(z)=\frac{1}{\sqrt{2\pi}}\exp(-z^2/2). $$ The upper bound follows from $\int_z^\infty\phi(u)\,du\leq\int_z^\infty (u/z)\phi(u)\,du=\phi(z)/z$. Integration by parts gives $\overline{\Phi}(z)=\phi(z)/z-\int_z^\infty \phi(u)/u^2\,du$. The remaining integral is at most $\overline{\Phi}(z)/z^2$, which yields the lower bound. Dividing both bounds by $\phi(z)/z$ and letting $z$ tend to infinity shows that $\overline{\Phi}(z)\sim\phi(z)/z$.

Theorem: For fixed $\mu$, $\sigma>0$, $s>0$, and $b>0$, as $t=x-\mu$ tends to infinity, $$ \begin{aligned} P(N(\mu,\sigma^2)>x) &\sim \frac{\sigma}{t\sqrt{2\pi}}\exp\!\left(-\frac{t^2}{2\sigma^2}\right),\\ P(\text{logistic}(\mu,s)>x) &\sim \exp(-t/s),\\ P(\text{Laplace}(\mu,b)>x) &= \frac12\exp(-t/b). \end{aligned} $$ In particular, the Gaussian upper tail is eventually smaller than either exponential tail. The logistic and Laplace tail comparison depends on their scales: if $1/s>1/b$, the logistic tail is eventually smaller; if $1/s<1/b$, it is eventually larger; and if $s=b$, their ratio tends to $2$.

Proof. Set $z=t/\sigma$ in the Mills ratio asymptotic just established. Since $z\to\infty$, $$ \overline{\Phi}(t/\sigma) \sim \frac{\phi(t/\sigma)}{t/\sigma} =\frac{\sigma}{t\sqrt{2\pi}}\exp\!\left(-\frac{t^2}{2\sigma^2}\right). $$ For the logistic tail, factor $\exp(t/s)$ from the denominator: $$ \frac{1}{1+\exp(t/s)} =\frac{\exp(-t/s)}{1+\exp(-t/s)} \sim\exp(-t/s). $$ The Laplace survival formula is exact. The Gaussian-to-logistic ratio tends to zero because its exponential factor contains $-t^2/(2\sigma^2)+t/s$, which tends to negative infinity; the additional factor proportional to $1/t$ does not change this limit. The same argument applies to the Gaussian-to-Laplace ratio. Finally, dividing the logistic asymptotic by the Laplace survival gives a ratio asymptotic to $2\exp(-t(1/s-1/b))$. Its limit is zero if $1/s>1/b$, infinity if $1/s<1/b$, and $2$ if $s=b$. This proves the claims. $\square$

Equal-Variance Calibration

A convenient comparison holds location and variance fixed. The normal distribution $N(0,1)$ has variance one. The logistic distribution with location zero and scale $s$ has variance $\pi^2s^2/3$, so setting its variance to one gives $s=\sqrt{3}/\pi$. The Laplace distribution with location zero and scale $b$ has variance $2b^2$, so setting its variance to one gives $b=1/\sqrt{2}$.

Under this calibration, the exponential rates are $1/s=\pi/\sqrt{3}\approx1.8138$ for the logistic distribution and $1/b=\sqrt{2}\approx1.4142$ for the Laplace distribution. Thus the Laplace tail decays more slowly than the logistic tail, and both decay more slowly than the Gaussian tail. The theorem implies that this ordering holds for all sufficiently large thresholds. It does not say that the ordering holds at every positive threshold.

Worked Example: Comparing Tails at One Standard Deviation

Compare the probabilities above $x=1$ for three mean-zero, variance-one distributions: $N(0,1)$, logistic with scale $\sqrt{3}/\pi$, and Laplace with scale $1/\sqrt{2}$.

Normal. The standard normal tail is $P(X>1)=\overline{\Phi}(1)\approx0.158655$.

Logistic. Its exponent is $1/s=\pi/\sqrt{3}\approx1.8138$. Therefore $$ P(X>1)=\frac{1}{1+\exp(1.8138)} \approx\frac{1}{1+6.134} \approx0.1402. $$ Laplace. Its exponent is $1/b=\sqrt{2}$, so $$ P(X>1)=\frac12\exp(-\sqrt{2}) \approx\frac12(0.2431) \approx0.1216. $$ At this threshold the order is normal, then logistic, then Laplace: $0.158655>0.1402>0.1216$. This is not the eventual large-threshold ordering.

Worked Example: Comparing Tails at Two Standard Deviations

Use the same three mean-zero, variance-one distributions, now at $x=2$.

Normal. The standard normal probability is $P(X>2)=\overline{\Phi}(2)\approx0.022750$.

Logistic. The exponent is $2\pi/\sqrt{3}\approx3.6276$. Thus $$ P(X>2)=\frac{1}{1+\exp(3.6276)} \approx\frac{1}{1+37.623} \approx0.02589. $$ Laplace. The exponent is $2\sqrt{2}\approx2.8284$, giving $$ P(X>2)=\frac12\exp(-2.8284) \approx\frac12(0.05911) \approx0.02955. $$ The ordering is now Laplace, logistic, then normal: $0.02955>0.02589>0.022750$. In this example the eventual ordering has already appeared by $x=2$, although the result about asymptotic rates alone does not specify the threshold at which it must begin.

Worked Example: Comparing Tails at Three Standard Deviations

At $x=3$, the normal probability is $P(X>3)=\overline{\Phi}(3)\approx0.001350$. For the equal-variance logistic distribution, $3\pi/\sqrt{3}\approx5.4414$, so $$ P(X>3)=\frac{1}{1+\exp(5.4414)} \approx\frac{1}{1+230.9} \approx0.00431. $$ For the equal-variance Laplace distribution, the exponent is $3\sqrt{2}\approx4.2426$, and $$ P(X>3)=\frac12\exp(-4.2426) \approx\frac12(0.01437) \approx0.00718. $$ Thus $0.00718>0.00431>0.001350$. The Laplace probability is more than five times the normal probability at this threshold. Equal variances do not imply similar probabilities of large deviations.

Why Scale and Threshold Both Matter

Equal variance is a useful calibration, not a claim that these distributions must be compared that way in every application. A scale changes the tail probability, and at a fixed threshold the scale can outweigh the asymptotic family ordering. For example, consider $N(0,4)$, a logistic distribution with location zero and scale $1$, and a Laplace distribution with location zero and scale $1$. Their variances differ. At $x=4$, the normal standardized threshold is $4/2=2$, while the logistic and Laplace exponents are both $4$.

Worked Example: A Scale Choice Changes the Finite-Threshold Order

For $X\sim N(0,4)$, the standard deviation is $2$, and $P(X>4)=\overline{\Phi}(2)\approx0.022750$. For the logistic distribution with scale $1$, $$ P(X>4)=\frac{1}{1+\exp(4)} \approx\frac{1}{1+54.598} \approx0.01799. $$ For the Laplace distribution with scale $1$, $$ P(X>4)=\frac12\exp(-4) \approx\frac12(0.018316) \approx0.00916. $$ Here the normal tail is largest and the Laplace tail smallest. There is no contradiction with the equal-variance asymptotic result: these are different scale choices, and the calculation is at one finite threshold.

Common Pitfall

A common mistake is to compare only the words “Gaussian,” “logistic,” and “Laplace,” then claim that one family has the larger tail at every threshold. The asymptotic result concerns what happens as the threshold tends to infinity, with the location and scale parameters fixed. It gives no universal ordering at moderate thresholds, and it cannot be applied without accounting for the scale parameters. The equal-variance example illustrates this directly: at $x=1$, the Gaussian tail is largest, while by $x=2$ the Laplace tail is largest.

A second version of the same mistake is to call logistic and Laplace tails “exponential” and assume they decay at the same rate. Their rates are $1/s$ and $1/b$, respectively. Even with equal variance, these rates differ; with other scale choices, their eventual order can reverse. To make a numerical comparison, use the survival formulas with the actual parameters and threshold. To make an asymptotic comparison, compare the decay rates and keep the parameters fixed as the threshold grows.

Key takeaway: For fixed parameters, the Gaussian upper tail eventually falls faster than either the logistic or Laplace tail. The logistic-versus-Laplace comparison depends on their exponential rates. In the mean-zero, variance-one calibration, the eventual order from largest to smallest is Laplace, logistic, normal; a finite-threshold calculation is still needed to know whether that order holds at the threshold of interest.

Check Your Understanding

Use the survival formulas and the distinction between finite-threshold and asymptotic comparisons.

For fixed $\sigma$ and $s$, explain why the ratio of the $N(\mu,\sigma^2)$ upper tail to the logistic upper tail tends to zero as $x$ tends to infinity.
For logistic scale $s$ and Laplace scale $b$, which tail is eventually larger if $1/s<1/b$? What if $s=b$?
For the mean-zero, variance-one Laplace distribution, write the survival probability at $x=2$ and calculate its approximate value.
Why does observing that the normal tail is largest at $x=1$ in the equal-variance example not contradict the asymptotic result?
What information besides the family name is needed to compare two survival probabilities at a specified finite threshold?