Modes of Discrete and Continuous Laws

What a Mode Identifies

A mode identifies a value, or a set of values, where a distribution is most concentrated according to its probability mass function or density. For a discrete random variable, the comparison is between the probabilities of individual outcomes. For a continuous random variable, the comparison is between density values; a density value is not itself the probability of an exact outcome.

A distribution need not have just one mode. Several discrete outcomes may tie for the greatest probability, and a continuous density may attain its maximum at multiple points or across an entire interval. The calculations therefore identify a set of modes, not necessarily a single number.

Definition: Let $X$ be a discrete random variable with probability mass function $f(x)=P(X=x)$. Its mode set is $$ \mathcal{M}=\{x:f(x)=\sup_{y\in\mathbb{R}} f(y)\}. $$ For a random variable with density $f$, a mode of the specified density $f$ is any $x\in\mathbb{R}$ satisfying $f(x)=\sup_{y\in\mathbb{R}} f(y)$. The set of such points is its mode set. A mode exists under these definitions only if the relevant supremum is attained.

The distinction between probability mass and density is essential. At a discrete outcome $x$, $f(x)$ is the probability of that outcome. For a continuous variable, probabilities are obtained by integrating the density over sets; in particular, $P(X=x)=0$ for every individual point $x$. A density mode indicates where the density is highest, not where a positive point probability occurs.

Modes of Discrete Laws

To find discrete modes, compare the mass function at its possible values. A simple comparison may suffice when there are only a few outcomes. For a sequence of probabilities, ratios of consecutive masses often show where the mass function changes from increasing to decreasing.

Worked Example: A Bernoulli Mode

Let $X\sim\operatorname{Bernoulli}(0.3)$, so $P(X=1)=0.3$ and $P(X=0)=0.7$. The support is $\{0,1\}$, so the only candidates for a mode are 0 and 1.

The masses are $f(0)=0.7$ and $f(1)=0.3$. Since $0.7>0.3$, the unique mode is $0$. If the success probability were $0.5$, the two masses would be equal and both 0 and 1 would be modes.

For any discrete probability mass function, a largest mass is attained. This fact ensures that a discrete law has at least one mode, even if its support is infinite.

Theorem: If a discrete random variable has a probability mass function $f$ with total mass 1, then $f$ attains its supremum. Therefore its mode set is nonempty.

Proof. Let $M=\sup_x f(x)$. At least one mass is positive, since the masses sum to 1, so $M>0$. By the definition of supremum, there is an $x_0$ with $f(x_0)>M/2$. The set $A=\{x:f(x)>M/2\}$ is finite: if it contained more than $2/M$ points, their masses alone would sum to more than 1. It is nonempty because it contains $x_0$. Thus $f$ has a largest value on $A$, say at $x_1$. For $x\notin A$, $f(x)\leq M/2<f(x_0)\leq f(x_1)$. So $x_1$ maximizes $f$ over all possible values, and the supremum is attained. $\square$

A useful case is the binomial distribution, whose possible values form a finite sequence. Consecutive-mass ratios give its modes directly.

Theorem: Suppose $X\sim\operatorname{Binomial}(n,p)$, where $n$ is a positive integer and $0<p<1$. Put $t=(n+1)p$. If $t$ is not an integer, the unique mode is $\lfloor t\rfloor$. If $t$ is an integer, the two modes are $t-1$ and $t$.

Proof. For $k=0,\ldots,n$, the binomial mass is $f(k)=\binom{n}{k}p^k(1-p)^{n-k}$, and each of these masses is positive. For $k=0,\ldots,n-1$, divide consecutive masses:

\frac{f(k+1)}{f(k)} =\frac{n-k}{k+1}\frac{p}{1-p}.

This ratio is at least 1 exactly when $(n-k)p\geq(k+1)(1-p)$, or equivalently when $k+1\leq(n+1)p=t$. If $t$ is not an integer and $m=\lfloor t\rfloor$, the masses increase up to $f(m)$ and decrease after it, giving the unique mode $m$. If $t=r$ is an integer, the ratio equals 1 at $k=r-1$, so $f(r-1)=f(r)$; the masses increase before this tie and decrease after it. The modes are $r-1$ and $r$. Since $0<t<n+1$, these indices are within the support. $\square$

Worked Example: A Binomial Law With Two Modes

Let $X\sim\operatorname{Binomial}(5,0.5)$. Here $t=(5+1)(0.5)=3$, an integer, so the theorem gives modes 2 and 3. Direct calculation verifies the tie:

P(X=2)=\binom{5}{2}(0.5)^2(0.5)^3 =10(0.5)^5=\frac{10}{32}, \qquad P(X=3)=\binom{5}{3}(0.5)^3(0.5)^2 =10(0.5)^5=\frac{10}{32}.

The neighboring masses are $P(X=1)=5/32$ and $P(X=4)=5/32$; the endpoint masses are $P(X=0)=1/32$ and $P(X=5)=1/32$. Thus neither tied value is exceeded elsewhere, and the mode set is exactly $\{2,3\}$.

Modes of Continuous Laws

For a continuous law, examine a density formula over its domain. A density that rises to a peak and then falls has a mode at the peak. Derivatives can locate such a point, but a stationary point alone is not enough: it could be a minimum or another kind of stationary point. The sign of the derivative on either side, together with boundary values when relevant, establishes whether a candidate is a global maximum.

Theorem: Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. If there is a $c\in(a,b)$ such that $f'(x)>0$ for every $x\in(a,c)$ and $f'(x)<0$ for every $x\in(c,b)$, then $c$ is the unique maximizer of $f$ on $[a,b]$.

Proof. For any $x<y\leq c$, the mean value theorem applies on $[x,y]$: continuity holds on that interval and differentiability holds in its interior. It gives a point $\xi\in(x,y)$ with $f(y)-f(x)=f'(\xi)(y-x)>0$. Hence $f$ is strictly increasing up to $c$. Similarly, for any $c\leq x<y$, the theorem gives $f(y)-f(x)=f'(\xi)(y-x)<0$, so $f$ is strictly decreasing after $c$. Thus $f(c)$ exceeds every other value on $[a,b]$, proving that $c$ is the unique maximizer. $\square$

Worked Example: A Beta Density With a Unique Mode

Let $X\sim\operatorname{Beta}(2,3)$, with density $f(x)=12x(1-x)^2$ for $0\leq x\leq1$ and $f(x)=0$ otherwise. On $(0,1)$, differentiate:

f'(x)=12\bigl((1-x)^2-2x(1-x)\bigr) =12(1-x)(1-3x).

For $0<x<1/3$, both factors $1-x$ and $1-3x$ are positive, so $f'(x)>0$. For $1/3<x<1$, $1-x>0$ and $1-3x<0$, so $f'(x)<0$. The density is continuous on $[0,1]$, and the derivative-sign result shows that its unique maximum on that interval is at $x=1/3$. The endpoint values are $f(0)=f(1)=0$, and the density is zero outside the interval. Therefore the unique mode is $1/3$.

Worked Example: A Continuous Law With Many Modes

Let $X$ be uniform on $[0,1]$, using the density version $f(x)=1$ for $0\leq x\leq1$ and $f(x)=0$ elsewhere. Its largest density value is 1, and that value occurs at every point of $[0,1]$. Thus every point in $[0,1]$ is a mode for this specified density; there is no unique mode.

This example illustrates why mode sets matter. Reporting only one point would fail to describe all the maximizers. The density’s values at individual points do not give point probabilities: for instance, $P(X=1/2)=0$, even though $f(1/2)=1$.

Common Mistakes and Interpretation

A common mistake is to treat a density value as the probability of its argument. For a continuous variable, the probability of an interval is an integral, such as $P(a\leq X\leq b)=\int_a^b f(x)\,dx$; the value $f(x)$ is a height, not a point probability. A second mistake is to solve $f'(x)=0$ and stop. That equation only produces candidates; the derivative signs or other global comparisons are needed to determine whether a candidate is a maximum and whether it is unique.

There is also a qualification specific to densities: a density is defined only up to changes on sets of probability zero. Changing its value at a single point leaves all interval probabilities unchanged but can change its pointwise maximizer. Consequently, the definition above refers to a specified density version. In applications, one typically uses the natural formula for the density, such as the continuous formula in the Beta example. The maximizer of that formula is the mode being reported.

Modes can be useful as concise descriptions of the most probable outcome in a discrete model or the highest-density location in a continuous model. Their interpretation depends on the type of law, and a tie or flat maximum is a substantive feature rather than a calculation error. Always state the full mode set when more than one maximizer occurs.

Check Your Understanding

Use the definitions and methods in this tutorial to answer the following questions.

A discrete random variable has masses $0.2, 0.5, 0.3$ at $0,1,2$, respectively. What is its mode set?
For $X\sim\operatorname{Binomial}(7,0.5)$, calculate $(n+1)p$ and identify the mode or modes.
Why does a density value $f(x)=4$ not mean that $P(X=x)=4$, or even that $P(X=x)$ is positive?
A density is strictly increasing on $(a,c)$ and strictly decreasing on $(c,b)$, and is continuous on $[a,b]$. What does this imply about its mode on that interval?
Why should a report for a distribution with several maximizing points give a mode set rather than just one mode?