Can expected value be negative?

Yes. If the random variable $X$ can take negative values (such as financial loss or downward displacement), the weighted sum can result in a negative expected value.

Does the expected value have to be a possible outcome?

No. The expected value is an average, not a specific draw. For example, rolling a fair die yields an expected value of 3.5, which cannot be rolled in a single trial.

How is expected value calculated for continuous random variables?

For continuous variables, the summation is replaced by an integral over the probability density function $f(x)$: $E[X] = \int_{-\infty}^{\infty} x f(x) dx$.

How to calculate expected value

The expected value $E[X]$ is the probability-weighted average of all possible values of a discrete random variable $X$ . It applies when you have a finite or countably infinite set of mutually exclusive outcomes, each with a known probability.

The setup

Identify the random variable $X$ , its possible values $x_i$ , and the corresponding probabilities $P(X = x_i)$ . Ensure that all probabilities are non-negative and that the sum of all probabilities equals exactly 1 ( $\sum P(X = x_i) = 1$ ).

The steps

List all mutually exclusive outcomes $x_1, x_2, \dots, x_n$ . 2. Determine the probability $p_i = P(X = x_i)$ for each outcome. 3. Multiply each outcome by its probability to compute $x_i p_i$ . 4. Sum these products to find the expected value: $E[X] = \sum_{i=1}^{n} x_i p_i$ .

Checking the result

Verify that the calculated expected value falls strictly within the inclusive range of the minimum and maximum possible outcomes: $\min(x_i) \le E[X] \le \max(x_i)$ . If $E[X]$ is outside this range, check your multiplication or probability sum.

Common errors

A frequent error is forgetting to verify that the probabilities sum to 1 before calculating. Another common mistake is treating the expected value as a guaranteed outcome for a single trial; it represents the long-term arithmetic mean as the number of trials approaches infinity.

Worked example

Calculate the expected value of rolling a standard, fair six-sided die.

Let $X$ be the outcome of the roll. The possible values are $x_i \in \{1, 2, 3, 4, 5, 6\}$ . The probability for each outcome is $P(X=x_i) = 1/6$ . The expected value formula is $E[X] = \sum_{i=1}^6 x_i P(X=x_i)$ . Substituting the values yields $E[X] = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6)$ . Factoring out the probability gives $E[X] = (1 + 2 + 3 + 4 + 5 + 6) / 6$ . Summing the numerator results in $E[X] = 21 / 6 = 3.5$ .

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References: OpenStax Introductory Statistics, Chapter 4 · Introduction to Probability by Bertsekas and Tsitsiklis