ProbabilityConcept reference

Expected values

A finite expectation is the sum of each value multiplied by its probability.

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01An expected value is a probability-weighted average#

An expected value is a probability-weighted average. Multiply each possible value by how likely it is, then add those contributions.

E[X]=\sum_x x\,p(x)

Let $X$ be $0$ with probability $0.75$ and $4$ with probability $0.25$ . Their weighted contributions are $0$ and $1$ , giving $E[X]=1$ .

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PMF: $p(0)=0.75$ , $p(8)=0.25$ . Find $E[X]$ .

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Weighted sum: $0(0.75)$ + $8(0.25)$ = $2$ .

Negative values keep their signs. A negative reward contributes a negative term; its probability stays nonnegative.

Equal probabilities justify averaging the distinct values equally. Otherwise, use the supplied weights.

Check your reasoning

A model assigns numerical rewards $-4$ , $6$ , with respective probabilities 0.7, 0.3. Find the expected reward $E[X]$ .

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-1

Weighted sum: $-4(0.7)$ + $6(0.3)$ = $-1$ .

The mean describes the probability model, not a guaranteed observation. In the example, only $0$ or $4$ can occur even though the mean is $1$ .

Keep the weighted result. Rounding it to a possible outcome changes the expected value.

Check your reasoning

PMF: $p(0)=0.5$ , $p(3)=0.5$ . A learner rounds the mean to a possible outcome. Find the unrounded $E[X]$ .

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1.5

Weighted sum: $0(0.5)$ + $3(0.5)$ = $1.5$ . Keep this mean.

Key takeaway