Expected value and linear expectation of a discrete random variable

Confirm what the random variable measures

A finite random variable assigns one real value to each outcome. List its distinct possible values and their probabilities. Each probability must be nonnegative, and all probabilities sum to one. If several elementary outcomes produce one value, combine their probability masses first. Equal spacing of values does not imply equal probabilities

In the displayed distribution, what is Pr(X=1)?

If a finite distribution has probabilities 1/5, 3/10 and p, find p.

The mean weights values by probabilities

For values with corresponding probabilities, use the displayed formula to define the mean. It has the same units as X. Multiplying before adding matters: the unweighted average of listed values is generally a different number

For X taking −1,1,3 with probabilities 1/4,1/2,1/4, calculate E(X)

A random variable takes 0 with probability 3/4 and 4 with probability 1/4. Which interpretation of E(X)=1 is correct?

Long-run language requires a stable repeated model

Under independent repetitions from the same finite distribution, the sample average becomes concentrated near the expected value as the number of trials grows. Individual values remain variable, and a finite sample average need not equal E(X). The calculation itself does not establish that a real situation has a stable distribution or independent trials

A supplied mean can identify a missing value

Suppose X takes 0, 2 and u with probabilities one quarter, one half and one quarter, respectively, and has mean 2. The displayed equation gives u as 4. Check the value in the original equation and keep all probability assumptions

X takes 0, 2 and u with probabilities one quarter, one half and one quarter, respectively, and has mean 2. What is u?

Read probability masses from a graph

A filled point at (x,p) gives Pr(X=x)=p: read the horizontal value and its vertical probability together. A stem locates that mass; its area has no probability meaning. The following graphs show every positive mass and selected zero masses, with probability zero at every other value. A point on the horizontal axis has zero mass and contributes zero to the mean. The horizontal grid spacing is 1 and the vertical spacing is 0.25. Unequal spacing of horizontal values must be retained

Read the plotted probability masses for X and calculate E(X)

Read the plotted probability masses for Y and calculate E(Y)

Compare the means, not the shapes

Both graph calculations give a mean of 1/2, although the distributions differ: X puts probability 1/2 at 0, while Y puts probability zero there. A displayed zero-mass point does not create another possible result with positive probability. Equal means do not imply equal distributions; compare the weighted sums rather than the number of plotted points or the unweighted average of horizontal values

Prove the mean lies between the extremes

For a finite distribution with minimum value m and maximum value M, prove m≤E(X)≤M. Explain when an equality can hold if some displayed probabilities are zero. Independent written proof is ungraded

Model: use nonnegative weights

Every value lies between the minimum and maximum. Multiply by its nonnegative probability and sum to obtain the displayed bounds. The probabilities sum to one, giving the bounds on the mean. Equality at the minimum requires every value with positive probability to equal it: the displayed sum is zero only when each of its nonnegative terms is zero. Displayed values with zero probability do not affect equality. The upper equality is analogous