When Modes and Medians Disagree

Two Different Questions About Location

A distribution can have more than one reasonable feature associated with “location.” The mode describes where probability mass or density is largest; the median divides probability into two halves. These answer different questions, so their locations need not agree. In a skewed distribution, the mode may sit near a high-density region while the median is shifted toward the longer tail. In a multimodal distribution, one or more peaks may lie far from a median that balances probability between them.

Neither feature is automatically the universal center of a distribution. A mode reports a peak, not a halfway point in probability. A median reports a halfway point in probability, not necessarily a peak or a typical-looking value. To compare them, identify the relevant mode or modes and then check how much probability lies on each side.

Definitions for the Comparison

Definition: For a discrete random variable \(X\) with probability mass function \(f\), a mode is any value \(x\) at which \(f(x)\) attains its largest value. For a continuous random variable with a specified density \(f\), a density mode is a point at which \(f\) attains its largest value. There may be multiple modes. A median is any real number \(m\) satisfying both \(P(X\le m)\ge 1/2\) and \(P(X\ge m)\ge 1/2\).

The density definition uses a specified density representative: changing a density at an isolated point does not change the distribution, although it can change the pointwise maximizer. This is one reason to interpret a density mode with care. Here, examples use the displayed density in its natural form. For a discrete law, modes are defined by comparing point probabilities; for a continuous law, they are defined by comparing density heights.

The median definition works for discrete, continuous, and mixed distributions. Its two inequalities allow for probability concentrated at a point and for a range of valid medians. In terms of the cumulative distribution function \(F\), the same conditions are \(F(m)\ge 1/2\) and \(F(m-)\le 1/2\), where \(F(m-)=P(X<m)\). A median is therefore a probability-balance location, not necessarily a value where the density or mass is largest.

A Probability-Balance Result

Theorem: Let \(X\) have a continuous cumulative distribution function \(F\). If \(c\) is a specified point, then:

If \(F(c)>1/2\), every median lies below \(c\).
If \(F(c)<1/2\), every median lies above \(c\).
If \(F(c)=1/2\), then \(c\) is a median.

In particular, this applies when \(c\) is a density mode.

Proof. Because \(F\) is continuous, \(P(X\ge m)=1-F(m)\). Thus \(m\) is a median exactly when \(F(m)=1/2\). If \(F(c)>1/2\), the limits of a CDF give values \(a<c\) with \(F(a)<1/2\): choose \(a\) sufficiently far to the left that this inequality holds. Continuity and the intermediate value theorem give some \(m\) between \(a\) and \(c\) with \(F(m)=1/2\). Moreover, no median can be at or above \(c\), since monotonicity would imply \(F(m)\ge F(c)>1/2\). The case \(F(c)<1/2\) follows in the same way using a point to the right with CDF above \(1/2\); no median can be at or below \(c\). Finally, when \(F(c)=1/2\), continuity gives \(P(X\ge c)=1-F(c)=1/2\), so both median inequalities hold at \(c\). \(\square\)

This result does not say that a mode determines the median from its location alone. It says that once a point \(c\) is identified, the CDF value there tells which side of \(c\) contains the median in a continuous distribution. The height of the density at \(c\) is not the probability to its left.

Worked Example: A Right-Skewed Exponential Distribution

Let \(X\sim\operatorname{Exp}(1)\), with density \(f(x)=e^{-x}\) for \(x\ge 0\) and \(f(x)=0\) otherwise. Find its mode and median, then compare their locations.

Mode. On the support, \(e^{-x}\) decreases as \(x\) increases, since its derivative is \(-e^{-x}<0\). Its largest value is \(f(0)=1\), so the mode is \(0\).

Median. For \(x\ge0\), integrating the density gives \(F(x)=1-e^{-x}\). Solve \(F(m)=1/2\):

1-e^{-m}=\frac12 \quad\Longrightarrow\quad e^{-m}=\frac12 \quad\Longrightarrow\quad m=\ln 2.

The value is valid because \(\ln 2>0\), and direct substitution gives \(F(\ln 2)=1-1/2=1/2\). Also \(P(X\ge \ln 2)=e^{-\ln 2}=1/2\), so the probability on each side is indeed one half. The median \(\ln 2\) lies to the right of the mode \(0\). Consistently, \(F(0)=0<1/2\), so the theorem places the median above the mode. The mode marks the density’s peak at the left endpoint; the median marks a probability split farther along the right-skewed distribution.

Several Modes, and a Median Between Them

A multimodal distribution has multiple peaks. Its mode set can have separated locations, while a median is determined by cumulative probability. The median may lie between peaks, and in some distributions an entire interval may satisfy the median conditions. The following discrete example makes the distinction explicit.

Worked Example: Two Discrete Modes and a Different Median

Suppose \(X\) has the probability mass function \(P(X=-4)=0.45\), \(P(X=0)=0.10\), and \(P(X=5)=0.45\). Find its modes and medians.

Modes. The largest point probability is \(0.45\). It occurs at both \(-4\) and \(5\), so the mode set is \(\{-4,5\}\).

Median. At \(m=0\), \(P(X\le0)=0.45+0.10=0.55\), and \(P(X\ge0)=0.10+0.45=0.55\). Both probabilities are at least \(1/2\), so \(0\) is a median.

To check that there is no other median, consider values below \(0\). If \(m<-4\), then \(P(X\le m)=0\); if \(-4\le m<0\), then \(P(X\le m)=0.45\). Neither reaches \(1/2\). For \(m>0\), if \(0<m\le5\), then \(P(X\ge m)=0.45\); if \(m>5\), then \(P(X\ge m)=0\). Neither reaches \(1/2\). Thus \(0\) is the unique median. It is not a mode: the mass at \(0\) is only \(0.10\), compared with \(0.45\) at each peak. The median balances the two sides even though it is not one of the most probable values.

Theorem: If \(m_1<m_2\) are both medians of a distribution, then every \(x\) with \(m_1\le x\le m_2\) is also a median.

Proof. Let \(m_1\le x\le m_2\). Since \(\{X\le m_1\}\) is contained in \(\{X\le x\}\), we have \(P(X\le x)\ge P(X\le m_1)\ge1/2\). Since \(\{X\ge m_2\}\) is contained in \(\{X\ge x\}\), we also have \(P(X\ge x)\ge P(X\ge m_2)\ge1/2\). Both median conditions hold at \(x\), proving the result. \(\square\)

Worked Example: A Median Interval in a Two-Peak Continuous Law

Consider the density \(f(x)=1/2\) on \([-3,-2]\) and \([2,3]\), and \(f(x)=0\) elsewhere. Each interval has length \(1\), so the total integral is \((1/2)(1)+(1/2)(1)=1\). Find the density modes and the median set.

Modes. The density reaches its maximum value \(1/2\) throughout both intervals, so every point in \([-3,-2]\cup[2,3]\) is a density mode for this displayed density. The law has two separated regions of maximum density.

Medians. The left interval carries probability \(1/2\), and the right interval carries probability \(1/2\). For every \(m\in[-2,2]\), all the left-interval probability is at or below \(m\), and all the right-interval probability is at or above \(m\). Therefore \(P(X\le m)=P(X\ge m)=1/2\). If \(m<-2\), the probability at or below \(m\) is less than \(1/2\); if \(m>2\), the probability at or above \(m\) is less than \(1/2\). Hence the median set is exactly \([-2,2]\). It includes points in the gap where the density is zero, illustrating that a median need not be a high-density value or even a value in the support.

When the Locations Agree

Agreement is possible, but it does not make mode and median interchangeable. For example, if \(X\sim N(\mu,\sigma^2)\) with \(\sigma^2>0\), its density is largest at \(\mu\), so \(\mu\) is its mode. The density is symmetric about \(\mu\), and a continuous distribution with that symmetry assigns probability \(1/2\) to each side of \(\mu\). Thus \(\mu\) is also a median. The match follows from the distribution’s symmetry, not from a general rule that modes and medians coincide.

In an asymmetric distribution, a mode gives the location of a largest mass or density, while a median gives a probability-balance location. In a multimodal distribution, there may be several modes but only one median, or several medians forming an interval. When describing a distribution, report the feature actually relevant to the question rather than calling either one “the center” without qualification.

Common Mistake

Wrong reasoning: “The mode is the most likely value, so it must have half the probability on either side and therefore be the median.” A peak’s height does not determine the total probability to its left or right.
Correction: For a continuous distribution, evaluate \(F(c)\) at the mode \(c\). If \(F(c)\) is above \(1/2\), the median lies below \(c\); if it is below \(1/2\), the median lies above \(c\). For a discrete or mixed distribution, check both median inequalities directly, allowing for jumps and nonunique medians.

Key takeaway: A mode locates a peak; a median balances probability. Skewness can separate them, and multiple peaks can leave the median between modes or across a whole interval. Neither feature is a universal center for every purpose.

Check Your Understanding

Use the definitions and probability-balance conditions to answer these questions.

For the exponential distribution in the worked example, what is \(F(0)\), and how does it help establish that the median lies to the right of the mode?
In the three-point discrete example, why is \(0\) a median even though it is not a mode?
For a continuous distribution with \(F(c)=0.7\), where must every median lie relative to \(c\)?
In the two-interval density example, why can a point in the gap be a median despite having density zero?
What is the difference between what a mode describes and what a median describes?