Expected Value of a Discrete Random Variable

Why Expected Value Uses Probabilities

A discrete random variable can take particular values, each with an associated probability. A useful summary of its distribution should account for both: a value that occurs often should contribute more than a value that is rare. The expected value does this by taking a probability-weighted average of the possible values.

For an actuarial application, the random variable might represent a hypothetical claim payment, an underwriting result, or a count of events. The expected value is a measure of the distribution's center, expressed in the same units as the random variable. It is not necessarily a value that the variable can actually take, and it does not predict the outcome of any single observation.

Definition: Suppose a discrete random variable $X$ has a finite set of possible values $x_1,x_2,\ldots,x_n$, with probability mass function $p_X(x_j)=P(X=x_j)$. Its expected value is $$ E[X]=\sum_{j=1}^{n}x_jp_X(x_j). $$ Each possible value is multiplied by its probability, and the resulting products are added. The probabilities must be nonnegative and sum to one.

The support is the set of values that can occur with positive probability. In a finite-support problem, list each support value once and pair it with its probability. A probability of zero may be included in a table, but its contribution to the sum is zero. If an outcome is a dollar amount, the expected value is in dollars; if it is a count, the expected value is in counts.

Check the PMF Before Calculating

A probability mass function must assign a probability between zero and one to each possible value. Across all possible values, the probabilities must add to one. This check is important: if the listed probabilities do not sum to one, they do not describe a complete probability distribution as written. Do not calculate an expected value from an invalid PMF without first resolving the problem.

A convenient calculation table has three columns: the value $x_j$, its probability $p_X(x_j)$, and the product $x_jp_X(x_j)$. Summing the third column gives $E[X]$. Keeping the products visible makes it easier to detect a missing value, a mismatched probability, or an arithmetic error.

Worked Example: Expected Payment From a Hypothetical Benefit

A hypothetical one-period benefit has payment $X$ equal to $0, $250, $1,000, or $3,000, with respective probabilities 0.70, 0.20, 0.08, and 0.02. Find the expected payment. This is a single-time payment, so no interest rate or discounting is involved; no mortality assumption is involved. The distribution is the finite PMF stated above.

Method. First verify the PMF, then multiply each payment by its probability and add the products.

PMF check. Each probability lies between zero and one, and $0.70+0.20+0.08+0.02=1.00$. The probabilities therefore sum to one.

Weighted products. The products, in dollars, are: $0(0.70)=0$, $250(0.20)=50$, $1{,}000(0.08)=80$, and $3{,}000(0.02)=60$.

E[X]=0+50+80+60=\$190.

The expected payment is $190. The result is between the smallest and largest possible payments, $0 and $3,000, as it must be for a probability-weighted average. It is a distributional average, not a claim that the payment in a particular case will be $190.

Common mistake avoided. Taking the unweighted average of the four possible payments gives $(0+250+1{,}000+3{,}000)/4=\$1{,}062.50$. That calculation incorrectly treats all four outcomes as equally likely. The PMF assigns very different probabilities, so the correct calculation weights each payment by its own probability.

Why the Weighted-Sum Formula Holds

The formula can be justified by starting with the possible outcomes of the experiment. Suppose there are finitely many elementary outcomes $\omega$, each with probability $P(\{\omega\})$, and $X(\omega)$ is the value recorded at outcome $\omega$. The expected value can be calculated by multiplying each elementary outcome's recorded value by its probability and summing:

E[X]=\sum_{\omega}X(\omega)P(\{\omega\}).

Now group together all elementary outcomes that give the same value $x_j$. Within that group, the recorded value is always $x_j$, and the probabilities of the group's outcomes add to $P(X=x_j)$. Therefore that group's contribution is $x_jP(X=x_j)$. Adding the contributions from all possible values gives

E[X]=\sum_{j=1}^{n}x_jP(X=x_j).

This grouping argument explains why the PMF is sufficient for calculating the expected value: the identities of individual elementary outcomes do not matter once their probabilities have been combined by the value of $X$.

Work With Signed Values and Missing Probabilities

Possible values do not have to be nonnegative. If $X$ represents a net gain or loss, a negative value represents a loss and contributes a negative amount to the expected value. The same probability-weighting rule applies. Also, a problem may give an unknown probability; for a finite PMF, use the requirement that all probabilities sum to one before finding the expected value.

Worked Example: Expected Net Result With a Possible Loss

A hypothetical one-period project has net result $X$ of $-\$120$, $0, or $180, with respective probabilities 0.10, 0.50, and 0.40. Calculate the expected net result. The amounts are measured at the end of the same period, so no interest or discounting is involved; mortality is not part of the model. The distribution is the finite PMF stated here.

Method. Verify the probabilities and retain the negative sign on the loss when calculating its weighted contribution.

The probabilities are valid and sum to $0.10+0.50+0.40=1.00$. The weighted products are $(-120)(0.10)=-12$, $(0)(0.50)=0$, and $(180)(0.40)=72$, in dollars.

E[X]=-12+0+72=\$60.

The expected net result is a gain of $60 per project under the stated distribution. The answer lies between the possible results of $-\$120$ and $180. It does not mean every project makes $60.

Common mistake avoided. Dropping the minus sign gives $120(0.10)+0+180(0.40)=\$84$, which is not the expectation of the stated net result. A loss is a negative value, so its weighted product must be negative.

Worked Example: Completing a PMF Before Finding Its Mean

In a hypothetical inspection program, the discrete random variable $X$ is the number of repair items found on a randomly selected unit. The listed probabilities are $P(X=0)=0.25$, $P(X=1)=q$, $P(X=2)=0.35$, and $P(X=5)=0.10$. Find $q$, check the completed PMF, and calculate $E[X]$. This is a count per inspected unit; no interest rate or mortality assumption applies. The distribution is finite and has the stated support.

Method. Use the total-probability requirement to solve for the missing mass before computing the weighted sum.

The known probabilities sum to $0.25+0.35+0.10=0.70$. Thus the missing probability is $q=1-0.70=0.30$. It is between zero and one, and the completed probabilities sum to $0.25+0.30+0.35+0.10=1.00$.

The weighted products are $0(0.25)=0$, $1(0.30)=0.30$, $2(0.35)=0.70$, and $5(0.10)=0.50$, measured in repair items per unit.

$$ E[X]=0+0.30+0.70+0.50=1.50. $$

The expected number of repair items is 1.50 per inspected unit. A count for an individual unit must be an integer, but an average across repeated units need not be an integer. The expected value is also between the smallest and largest possible counts, 0 and 5.

Common mistake avoided. Assigning the missing probability as 0.70 confuses the sum of the known probabilities with the probability still needed. The remaining mass is $1-0.70=0.30$, because the complete PMF must total one.

Interpretation and Checks

For a finite PMF, the expected value is a center determined by the probability distribution. It may be fractional even when every possible value is an integer, and it may not appear in the support at all. For example, a distribution supported on 0 and 5 could have an expected value between 0 and 5 without ever producing that intermediate value.

Several checks can catch errors before an answer is finalized:

Check the probabilities: every probability must be in the interval from zero to one, and the total must be one.
Check the products: multiply each value by its matching probability, preserving signs and units.
Check the range: a finite probability-weighted average cannot be smaller than the least possible value or larger than the greatest possible value.
Check the interpretation: report the expected value in the units of $X$, and do not describe it as a guaranteed outcome.

The range check follows directly from the weights. Each probability is nonnegative and all probabilities sum to one, so the expected value is an average formed from the support values using nonnegative weights with total weight one. Such an average must remain within the smallest and largest values being averaged.

Expected value is useful when summarizing the average result represented by a distribution. In actuarial work, that summary can provide a starting point for considering a hypothetical payment or net result. It does not describe the spread of outcomes or the likelihood of a particular outcome; those require other distributional summaries and methods.

Key takeaway: For a finite discrete random variable, verify that the PMF is valid, then calculate $E[X]$ by summing every possible value multiplied by its own probability. Interpret the answer as a probability-weighted average, not as a certain or necessarily attainable result.

Check Your Understanding

For each problem, check the PMF before calculating the expected value. Unless stated otherwise, each scenario is hypothetical and involves a single-period amount or count, so no interest or mortality assumption is needed.

A hypothetical device-repair benefit $X$ pays $0, $400, or $1,200 with probabilities 0.65, 0.25, and 0.10, respectively. Verify the PMF and find the expected payment. (Answer: PMF total $=1.00$; $E[X]=\$220$.)
A hypothetical one-period trading result $X$ is $-\$50$, $30, or $90 with probabilities 0.20, 0.50, and 0.30, respectively. Find the expected result, keeping the sign of the loss. (Answer: $E[X]=\$32$.)
A hypothetical count $X$ takes values 1, 3, or 8, with probabilities 0.40, $q$, and 0.10. Find $q$, verify the PMF, and calculate $E[X]$. (Answer: $q=0.50$; PMF total $=1.00$; $E[X]=2.70$.)
A hypothetical audit count $X$ takes values 0, 2, 4, or 10 with respective probabilities 0.30, 0.40, 0.20, and 0.10. Check the probability total, calculate $E[X]$, and state whether the expected value must be a possible count. (Answer: PMF total $=1.00$; $E[X]=2.60$; it need not be a possible count.)
A hypothetical one-period payment $X$ is $0, $500, or $2,000 with probabilities 0.80, 0.15, and 0.05. Calculate the expected payment and explain why the unweighted average of the three payment amounts is not the answer. (Answer: $E[X]=\$175$; the unweighted average ignores that the payments have unequal probabilities.)

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