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Expected value and variance of a discrete random variable

◈ 6 cards

The table method for μ, the shortcut Σx²p(x) − μ² for σ², and expected value as a decision criterion.

The long-run average

The expected value of a discrete random variable is its mean — the average value of over a very long run of repetitions, each value weighted by how often it occurs:

It is a weighted average with the probabilities as weights. The variance is the weighted average squared distance from , and the shortcut form is the one to compute by hand:

The standard deviation is , in the units of .

Worked example — the table method

Extend Maple Ledger's pmf with two columns and total each:

00.400.000.00
10.350.350.35
20.150.300.60
30.100.300.90
total1.000.951.85

On average a batch has 0.95 late invoices, give or take about one. No batch ever has 0.95 late invoices — the expected value is a long-run average, not a possible outcome, exactly as a family can have 1.8 children. The third column is , not : square the value, then weight it.

Forgetting the

The most common slip is to report the third column's total, 1.85, as the variance. That is ; the variance is smaller by . A variance larger than is impossible, so a quick check is that always.

Expected value as a decision rule

Maple Ledger's partner can install a review control for \$8,000 a year, or accept that without it there is a 12 % chance of a \$50,000 loss from an undetected billing error. The expected loss without the control is

which is less than \$8,000 — on expected value alone, skip the control. But expected value is an average over many years; in any single year the firm either loses nothing or loses \$50,000, and a firm that cannot absorb \$50,000 may rationally pay \$8,000 to avoid the chance. Expected value ranks decisions by their long-run average; the risk appetite — how much variance the decision-maker will carry — is a separate input the paper often asks you to name.

xp(x)x·p(x)x²·p(x)00.400.000.0010.350.350.3520.150.300.6030.100.300.90total1.000.951.85μ = 0.95 · σ² = 1.85 − 0.95² = 0.9475 · σ = 0.973.
The table method. μ is the total of the third column; E[X²] is the total of the fourth; σ² = 1.85 − 0.95² = 0.9475.
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