Expected Value, Mode, and Possible Outcomes

Introduction

An expected value is often called an average, which can make it sound like an outcome a random process will commonly produce. But an average can fall between the values that are possible, and it need not be the value with the greatest probability. These distinctions matter whenever a distribution is skewed: a small chance of a very large outcome can pull the expected value away from the outcome seen most often.

To keep these ideas separate, we will compare three features of a discrete random variable: its expected value, its mode, and its support. The expected value summarizes all outcomes using their probabilities; the mode identifies a most probable outcome; the support lists which outcomes can occur with positive probability.

Expected Value, Mode, and Support

Definition: For a discrete random variable $X$, its support is the set of values with positive probability: $S=\{x:P(X=x)>0\}$. If the support is finite, the expected value is $\operatorname{E}(X)=\sum_{x\in S}xP(X=x)$. A mode is a value in the support whose probability is at least as large as the probability of every other supported value. There may be more than one mode if the largest probability is tied.

The expected value weights each possible value by how likely it is. In contrast, the mode compares the probabilities of individual values and selects the largest. The support answers a different question: which outcomes have nonzero probability? A value can be in the support without being likely, while a value outside the support cannot occur under the model.

In this tutorial, examples use finite probability distributions. Before calculating, check that each listed probability is nonnegative and that all probabilities add to 1. Those conditions make the table a probability mass function: it assigns the full probability of the experiment across its possible outcomes.

Worked Example: A Mean Between Possible Prize Amounts

Consider a hypothetical game in which one play pays $0 with probability 0.80, $1 with probability 0.10, or $10 with probability 0.10. Let $X$ be the payout, measured in dollars. Find the expected payout, mode, and support.

The probabilities are nonnegative and sum to $0.80+0.10+0.10=1$, so they define a probability distribution. The support is $S=\{0,1,10\}$ dollars. The expected payout is

\operatorname{E}(X) =0(0.80)+1(0.10)+10(0.10) =0+0.10+1.00 =\$1.10.

The mode is $0, because $P(X=0)=0.80$ is greater than the probabilities of the other outcomes, each of which has probability 0.10. The expected payout, $1.10, is not in the support. Thus it is not a payout that one play of this game can produce. It is a probability-weighted average, not a prediction that a particular play will pay $1.10.

The relatively rare $10 payout raises the average above $1, even though most plays pay nothing. That pull from an occasional larger value is a feature of a right-skewed distribution.

Why an Expected Value Need Not Be an Outcome

For a finite distribution, the expected value is a weighted average of the supported values. The weights are probabilities: they are nonnegative and add to 1. A weighted average must lie between the smallest and largest values, but it does not have to equal one of the values being averaged.

Justification: Suppose the support has values $x_1,\ldots,x_k$, with probabilities $p_1,\ldots,p_k$, where $p_i\geq 0$ and $\sum_{i=1}^{k}p_i=1$. Let $m=\min_i x_i$ and $M=\max_i x_i$. Since $m\leq x_i\leq M$, multiplying by $p_i$ gives $mp_i\leq x_ip_i\leq Mp_i$. Adding these inequalities over all $i$ gives $\displaystyle m\sum_{i=1}^{k}p_i\leq \sum_{i=1}^{k}x_ip_i\leq M\sum_{i=1}^{k}p_i$, and therefore $m\leq \operatorname{E}(X)\leq M$. This establishes the bounds, not that the expected value must be one of $x_1,\ldots,x_k$.

In the prize example, the minimum possible payout is $0 and the maximum is $10, so an expected payout of $1.10 satisfies the bounds. But the only supported values are $0, $1, and $10. The fact that $1.10 lies between the minimum and maximum does not make it an attainable payout.

The weighted-average formula also explains the role of skew. A small probability attached to a sufficiently large value can have a substantial effect on the sum $xP(X=x)$. A mode, however, depends only on which individual probability is largest. It does not measure how far an outcome lies from the others.

Worked Example: A Rare Large Count Pulls Up the Mean

Suppose a hypothetical delivery route has $X$ late packages on a randomly selected day. A simple model assigns probability 0.70 to 0 late packages, 0.20 to 2, and 0.10 to 20. Find and interpret the expected count and mode.

The probabilities are nonnegative and $0.70+0.20+0.10=1$, so the model is a valid probability distribution. The support is $S=\{0,2,20\}$ packages. Compute the expected value by multiplying each count by its probability:

\operatorname{E}(X) =0(0.70)+2(0.20)+20(0.10) =0+0.40+2.00 =2.40\text{ packages}.

The mode is 0 late packages, because this outcome has the largest probability, 0.70. The expected count of 2.40 is not in the support, so no single day in this model has exactly 2.40 late packages. That causes no contradiction: the expected value summarizes probability across all modeled days, whereas each individual day has one of the listed integer counts.

Most of the probability is concentrated at 0, but the 10% chance of 20 late packages contributes 2 packages to the expected value by itself. Consequently, the average is greater than the mode. Calling 2.40 the “most likely count” would confuse two different summaries.

An Attainable Mean Can Still Differ From the Mode

The expected value does not have to be outside the support. It can be a possible outcome while still failing to be the most probable outcome. A separate example makes this distinction clear.

Worked Example: The Mean Is Possible but Not Most Likely

Let $X$ be the number of extra items in a hypothetical package. Suppose $X=0$ with probability 0.50, $X=1$ with probability 0.25, and $X=3$ with probability 0.25. Find the support, expected value, and mode.

The probabilities are nonnegative and sum to $0.50+0.25+0.25=1$. The support is $S=\{0,1,3\}$ items. The expectation is

\operatorname{E}(X) =0(0.50)+1(0.25)+3(0.25) =0+0.25+0.75 =1\text{ item}.

One item is in the support, so the expected value is attainable. But it is not the mode: $P(X=0)=0.50$, while $P(X=1)=0.25$ and $P(X=3)=0.25$. The unique mode is 0 extra items. In this distribution, the expected value is possible, but it is less likely than the modal outcome.

This example separates two questions that are easy to blend together. “Can the random variable equal its expected value?” asks whether the mean belongs to the support. “Is the expected value the mode?” asks whether its probability is maximal. One answer does not determine the other.

How to Read the Three Summaries

When describing a discrete distribution, name each summary precisely and attach it to its correct question:

Expected value: What is the probability-weighted average of the outcomes?
Mode: Which supported value has the highest probability?
Support: Which values have positive probability under the model?

A useful check is to compare the calculated expected value with the support. If the value is not listed in the support, it cannot be an outcome in the model. If it is listed, it is attainable, but its probability still needs to be compared with the probabilities of all other supported values before calling it a mode.

The mode may also be tied. For example, if two outcomes have the same largest probability, both are modes; it would be incorrect to report only one as uniquely most likely. The expected value can be different from both, equal to one of them, or outside the support. The probability table—not a rule equating “average” with “most likely”—decides which case applies.

Common Mistake

A common error is to say, “The expected value is 2.4 packages, so 2.4 packages is the most likely daily count.” This conclusion is wrong for the delivery-route model. Its possible counts are 0, 2, and 20, and the largest single-outcome probability is attached to 0. The correction is to report the summaries separately: the expected count is 2.40 packages, the mode is 0 packages, and the support is $\{0,2,20\}$ packages.

It is also a mistake to call an expected value impossible merely because it is fractional. A fractional expected value may be outside the support, as in the delivery example, but whether it is attainable depends on the support. Conversely, an integer expected value may still be absent from the support. Always check the actual possible values rather than relying on whether the mean looks like a plausible outcome.

Key takeaway: The expected value is a probability-weighted average, the mode is a most probable supported value, and the support contains the outcomes the model allows. These are distinct features: an expected value may be unattainable, and even when attainable it need not be a mode.

Check Your Understanding

Use the probability distribution in each question as stated.

A variable takes values 0, 4, and 12 with probabilities 0.60, 0.30, and 0.10. Find its support, expected value, and mode. Is the expected value attainable?
Explain why the expected value of a finite discrete random variable must be between its smallest and largest supported values.
Can an expected value be a supported value without being the mode? Give a reason or a counterexample.
If two supported values share the largest probability, how should the mode be reported?
In a right-skewed discrete distribution, how can an outcome with a small probability affect the expected value even if it is not the mode?