Average outcomes with probability weights

Try

A reward system pays0,2 or5 tokens with probabilities1/2,1/4 and1/4. Before computing a mean, check that the probabilities are nonnegative and add to one. Predict whether its average payoff must be one of the three possible rewards. Ordinary averages already suggest that a center can fall between observed values.

Predict

Predict which has more risk: a guaranteed5 tokens or equal chances of0 and10. Both have average5, but one varies and the other does not. An expectation answers a specific average-outcome question; it does not summarize every feature relevant to a decision.

Experiment

Let X be the numerical reward. Make a weighted ledger.

x	P(X=x)	xP(X=x)	x²P(X=x)
0	1/2	0	0
2	1/4	1/2	1
5	1/4	5/4	25/4

Adding the third column gives E(X)=7/4. Adding the fourth gives E(X²)=29/4. Variance is29/4−(7/4)²=67/16. Check the alternative definition by weighting squared distances from7/4; it must give the same result. The standard deviation is√67/4 tokens, whereas variance has square-token units.

Understand

A random variable assigns a number to each outcome of a random process. Its distribution lists possible values and their probabilities. Discrete variables have a finite or countable set of possible values; several underlying outcomes can map to the same number. Aggregate their probabilities when constructing the distribution.

Expectation is the probability-weighted average E(X)=ΣxP(X=x), assuming the required sum exists. It can represent a long-run average under suitable repeated-trial assumptions, but need not be an attainable outcome or the most likely one. Here7/4 tokens is never paid on an individual play. The mode is0 because that reward has the largest probability.

Variance is E[(X−μ)²], where μ=E(X). Expanding gives E(X²)−μ² when the moments exist. Squaring prevents opposite deviations from canceling. A variance is nonnegative, so a negative calculated variance signals an error or numerical issue. Standard deviation returns to the original units by taking a square root.

For a fixed cost c, net payoff Y=X−c has expectation E(X)−c and the same variance as X. More generally, a+bX has expectation a+bE(X) and variance b²Var(X). Linearity of expectation also permits adding expected outcomes without requiring independence, though variance formulas for sums need covariance information unless independence or another condition removes it.

The guaranteed5-token option has variance0. The equally likely0-or10 option has variance25. Equal expectation does not mean equal desirability when losses, cash-flow constraints or attitudes to variability matter. A model should state the decision criterion rather than silently assuming everyone cares only about average payoff.

The probabilities themselves are assumptions or estimates. Do not treat an attractive expected return as a guarantee, and do not use a few favorable observations as proof of the underlying distribution. Model checking and data quality remain part of the investigation.

Apply

For the practice game that charges3 and pays10 with probability1/5, list the two net outcomes before averaging. They are7 and−3, with expected net−1. Explain why neither “you lose exactly1 each play” nor “a positive possible payoff makes the game profitable” follows. State which quantities a risk-aware comparison would also examine.

Interpret spread in the random variable’s units

An original package-delay model has X=0,2,4 delayed packages per dispatch with probabilities 1/4,1/2,1/4. The mean is 2 packages, and the variance is (0−2)²/4+(2−2)²/2+(4−2)²/4=2 squared packages. Thus SD is √2≈1.414 packages. It measures the root-mean-square distance of the count from its mean across repeated dispatches under this model; it is not a guaranteed maximum deviation, a confidence margin for an estimated mean, or a claim that a dispatch can contain 1.414 packages.

Changed-case check. A charge is Y=3X+2 currency units. Calculate and interpret its mean and SD. Does the fixed fee increase the SD? Would two models with the same mean have to have the same SD?

Feedback

E(Y)=3·2+2=8 currency units. Var(Y)=9·2=18 squared currency units, so SD(Y)=3√2≈4.243 currency units. The fee shifts all charges equally and therefore does not change deviations about their new mean; the multiplier triples their size. Equal means do not force equal spread: a charge always equal to 8 has SD zero but the same mean. A complete interpretation gives the variable and units, distinguishes an average from a guaranteed observation, and does not relabel distributional spread as uncertainty of a sample mean.

Continue

Next specialize to the number of successes in a fixed set of independent, equal-probability trials. The binomial formula is useful only when that structure fits.