Discrete Random Variables

ST 370 Probability and Statistics for Engineers Discrete Random Variables

A random variable is a numerical value associated with the outcome of an experiment.

Discrete random variable When we can enumerate the possible values of the variable (such as 0, 1, 2, . . . ), the random variable is discrete.

Example: acceptance sampling Suppose that a sample of size 10 is drawn from a shipment of 200 items, of which some number are non-compliant; X is the number of non-compliant items in the sample. The possible values of X are 0, 1, 2, . . . , 10, so X is a discrete random variable.

1 / 15 Discrete Random Variables ST 370 Probability and Statistics for Engineers

Continuous random variable When the variable takes values in an entire interval, the random variable is continuous.

Example: ﬂash unit recharge time Suppose that a cell phone camera ﬂash is chosen randomly from a production line; the time X that it takes to recharge is a positive real number; X is a continuous random variable.

Presumably, there is some lower bound a > 0 that is the shortest possible recharge time, and similarly some upper bound b < ∞ that is the longest possible recharge time; however, we usually do not know these values, and we would just say that the possible values of X are {x : 0 < x < ∞}.

2 / 15 Discrete Random Variables ST 370 Probability and Statistics for Engineers Probability distribution

The probability distribution of a random variable X is a description of the probabilities associated with the possible values of X .

The representation of a probability distribution is diﬀerent for discrete and continuous random variables.

Probability mass function For a discrete random variable, the simplest representation is the probability mass function (pmf)

fX (x) = P(X = x) where x is any possible value of X .

3 / 15 Discrete Random Variables Probability distribution ST 370 Probability and Statistics for Engineers Example: acceptance sampling Suppose one item is chosen at random from a shipment of 200 items, of which 5 are non-compliant. Let ( 1 if the item is non-compliant, X = 0 if the item is compliant.

We could say that X is the number of non-compliant items seen.

The probability mass function of X is ( 0.975 x = 0 fX (x) = 0.025 x = 1

A random variable like X that takes only the values 0 and 1 is called a Bernoulli random variable.

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Example: Dice Suppose you roll a fair die, and the number of spots showing is X . Then X is a discrete random variable with probability mass function 1 f (x) = , x = 1, 2, 3, 4, 5, 6. X 6

Because the probability is the same for all the possible values of X , it is called the discrete uniform distribution.

Properties of the probability mass function

They are probabilities: fX (x) ≥ 0. P They cover all possibilities: x fX (x) = 1.

5 / 15 Discrete Random Variables Probability distribution ST 370 Probability and Statistics for Engineers Cumulative distribution function

As an alternative to the probability mass function, the probability distribution of a random variable X can be deﬁned by its cumulative distribution function (cdf)

FX (x) = P(X ≤ x), −∞ < x < ∞.

In terms of the probability mass function: X FX (x) = fx (xi ), −∞ < x < ∞.

xi ≤x

FX (·) has a jump at each possible value xi of X , and the jump equals the corresponding probability fX (xi ), so the probability mass function can be obtained from the cumulative distribution function.

6 / 15 Discrete Random Variables Cumulative distribution function ST 370 Probability and Statistics for Engineers

Example: acceptance sampling  0 x < 0  FX (x) = 0.975 0 ≤ x < 1 1 x ≥ 1 curve(pbinom(x, 1, .025), from = -1, to = 2)

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Example: dice  0 x < 1  1/6 1 ≤ x < 2  FX (x) = 2/6 2 ≤ x < 3 . .  1 x ≥ 6 curve(pmax(0, pmin(1, floor(x)/6)), from = 0, to = 7)

8 / 15 Discrete Random Variables Cumulative distribution function ST 370 Probability and Statistics for Engineers Mean and Variance

Mean value The mean value, or expected value, of a discrete random variable with probability mass function fX (·) is X µX = E(X ) = xfX (x). x E(X ) is a weighted average of the possible values of X , each weighted by the corresponding probability.

The expected value E(X ) is a typical value of the random variable X , in the same way that a sample meanx ¯ is a typical value of the sample x1, x2,..., xn.

9 / 15 Discrete Random Variables Mean and Variance ST 370 Probability and Statistics for Engineers

Example: acceptance sampling One item is chosen at random from a shipment of 200 items, of which 5 are non-compliant, and X is the number of non-compliant items seen:

E(X ) = 0 × fX (0) + 1 × fX (1) = 0.025.

For any Bernoulli random variable X ,

E(X ) = P(X = 1).

10 / 15 Discrete Random Variables Mean and Variance ST 370 Probability and Statistics for Engineers

Example: Dice Suppose you roll a fair die, and the number of spots showing is X :

E(X ) = 1 × fX (1) + 2 × fX (2) + ··· + 6 × fX (6) = (1 + 2 + 3 + 4 + 5 + 6)/6 = 3.5.

Note In these examples and in many others, the “expected” value is not one of the possible values of the random variable; this is not the paradox that it is sometimes made out to be!

11 / 15 Discrete Random Variables Mean and Variance ST 370 Probability and Statistics for Engineers

Variance

Suppose that X is a random variable with expected value µX . Then

2 Y = (X − µX ) is another random variable, and its expected value is X E(Y ) = yfY (y) y X 2 = (x − µX ) fX (x). x

12 / 15 Discrete Random Variables Mean and Variance ST 370 Probability and Statistics for Engineers

2 The variance of X is E(Y ) = E[(X − µX ) ]:

2 2 σX = V (X ) = E (X − µX ) .

The standard deviation of X is q 2 σX = σX .

13 / 15 Discrete Random Variables Mean and Variance ST 370 Probability and Statistics for Engineers

Example: acceptance sampling For any Bernoulli random variable X ,

µX = P(X = 1) = p, say, so

2 2 2 σX = (0 − p) × P(X = 0) + (1 − p) × P(X = 1) = p2(1 − p) + (1 − p)2p = p(1 − p) and p σX = p(1 − p).

14 / 15 Discrete Random Variables Mean and Variance ST 370 Probability and Statistics for Engineers

Example: Dice

2 2 x x − 3.5 (x − 3.5) fX (x)(x − 3.5) fX (x) 1 1 -2.5 6.25 6 1.0417 1 2 -1.5 2.25 6 0.3750 1 3 -0.5 0.25 6 0.0417 1 4 0.5 0.25 6 0.0417 1 5 1.5 2.25 6 0.3750 1 6 2.5 6.25 6 1.0417 Total: 2.9168

2 So σX = 2.917 and σX = 1.708.

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