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Probability and Variability
Discrete Random Variables
Statistics Textbooks Boundless Statistics Probability and Variability Discrete Random Variables
Statistics Textbooks Boundless Statistics Probability and Variability
Statistics Textbooks Boundless Statistics
Statistics Textbooks
Statistics
Concept Version 7
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Expected Values of Discrete Random Variables

The expected value of a random variable is the weighted average of all possible values that this random variable can take on.

Learning Objective

  • Calculate the expected value of a discrete random variable


Key Points

    • The expected value of a random variable XXX is defined as: $E[X] = x_1p_1 + x_2p_2 + \dots + x_ip_i$, which can also be written as: E[X]=∑xipiE[X] = \sum x_ip_iE[X]=∑x​i​​p​i​​.
    • If all outcomes xix_ix​i​​ are equally likely (that is, $p_1=p_2=\dots = p_i$), then the weighted average turns into the simple average.
    • The expected value of XXX is what one expects to happen on average, even though sometimes it results in a number that is impossible (such as 2.5 children).

Terms

  • expected value

    of a discrete random variable, the sum of the probability of each possible outcome of the experiment multiplied by the value itself

  • discrete random variable

    obtained by counting values for which there are no in-between values, such as the integers 0, 1, 2, ….


Full Text

Discrete Random Variable

A discrete random variable XXX has a countable number of possible values. The probability distribution of a discrete random variable XXX lists the values and their probabilities, such that xix_ix​i​​ has a probability of pip_ip​i​​. The probabilities pip_ip​i​​ must satisfy two requirements:

  1. Every probability pip_ip​i​​ is a number between 0 and 1.
  2. The sum of the probabilities is 1: $p_1+p_2+\dots + p_i = 1$.

Expected Value Definition

In probability theory, the expected value (or expectation, mathematical expectation, EV, mean, or first moment) of a random variable is the weighted average of all possible values that this random variable can take on. The weights used in computing this average are probabilities in the case of a discrete random variable.

The expected value may be intuitively understood by the law of large numbers: the expected value, when it exists, is almost surely the limit of the sample mean as sample size grows to infinity. More informally, it can be interpreted as the long-run average of the results of many independent repetitions of an experiment (e.g. a dice roll). The value may not be expected in the ordinary sense—the "expected value" itself may be unlikely or even impossible (such as having 2.5 children), as is also the case with the sample mean.

How To Calculate Expected Value

Suppose random variable XXX can take value x1x_1x​1​​ with probability p1p_1p​1​​, value x2x_2x​2​​ with probability p2p_2p​2​​, and so on, up to value xix_ix​i​​ with probability pip_ip​i​​. Then the expectation value of a random variable XXX is defined as: $E[X] = x_1p_1 + x_2p_2 + \dots + x_ip_i$, which can also be written as: E[X]=∑xipiE[X] = \sum x_ip_iE[X]=∑x​i​​p​i​​.

If all outcomes xix_ix​i​​ are equally likely (that is, $p_1 = p_2 = \dots = p_i$), then the weighted average turns into the simple average. This is intuitive: the expected value of a random variable is the average of all values it can take; thus the expected value is what one expects to happen on average. If the outcomes xix_ix​i​​ are not equally probable, then the simple average must be replaced with the weighted average, which takes into account the fact that some outcomes are more likely than the others. The intuition, however, remains the same: the expected value of XXX is what one expects to happen on average.

For example, let XXX represent the outcome of a roll of a six-sided die. The possible values for XXX are 1, 2, 3, 4, 5, and 6, all equally likely (each having the probability of 16\frac{1}{6}​6​​1​​). The expectation of XXX is: E[X]=1x16+2x26+3x36+4x46+5x56+6x66=3.5E[X] = \frac{1x_1}{6} + \frac{2x_2}{6} + \frac{3x_3}{6} + \frac{4x_4}{6} + \frac{5x_5}{6} + \frac{6x_6}{6} = 3.5E[X]=​6​​1x​1​​​​+​6​​2x​2​​​​+​6​​3x​3​​​​+​6​​4x​4​​​​+​6​​5x​5​​​​+​6​​6x​6​​​​=3.5. In this case, since all outcomes are equally likely, we could have simply averaged the numbers together: 1+2+3+4+5+66=3.5\frac{1+2+3+4+5+6}{6} = 3.5​6​​1+2+3+4+5+6​​=3.5.

Average Dice Value Against Number of Rolls

An illustration of the convergence of sequence averages of rolls of a die to the expected value of 3.5 as the number of rolls (trials) grows.

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