0.01 USD as an Expected Value
Consider a simple discrete random variable \(X\) (measured in USD) that represents a reward: \(X = 1\) USD with probability \(0.01\), and \(X = 0\) USD with probability \(0.99\). The expected value (mean) \(E[X]\) is the long-run average reward per trial.
The unit “USD” stays attached to the mean because expectation is a weighted average of outcomes measured in USD.
Step 1: Write the probability distribution
| Outcome \(x\) (USD) | Probability \(P(X=x)\) | Contribution \(x \cdot P(X=x)\) (USD) |
|---|---|---|
| \(1\) | \(0.01\) | \(1 \cdot 0.01 = 0.01\) |
| \(0\) | \(0.99\) | \(0 \cdot 0.99 = 0\) |
Step 2: Apply the definition of expected value
For a discrete random variable, the mean (expected value) is the probability-weighted sum:
\[ E[X] = \sum_x x \cdot P(X=x) \]
Substituting the two outcomes:
\[ E[X] = 1 \cdot 0.01 + 0 \cdot 0.99 = 0.01 \]
Since \(X\) is measured in USD, the expected value is 0.01 USD.
Step 3: Interpretation of 0.01 USD
The value 0.01 USD is not the most likely payout (the most likely payout is 0 USD). Instead, it is the long-run average payoff per trial: over many independent repetitions, the average reward approaches 0.01 USD per attempt.
Visualization: Probability mass function (PMF)
Quick check: If the reward is 1 USD only 1% of the time, then the average reward per attempt is 1% of 1 USD, which is 0.01 USD.