Random Variables. Numerical Outcomes Consider associating a numerical value with each sample point in a sample space. (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

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Presentation transcript:

Random Variables

Numerical Outcomes Consider associating a numerical value with each sample point in a sample space. (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) : The function relating each outcome from a roll of the die with their sum is considered a random variable. Refer to values of the random variable as events. For example, {Y = 9}, {Y = 10}, etc.

Probability Y = y The probability of an event, such as {Y = 9} is denoted P(Y = 9). In general, for a real number y, the probability of {Y = y} is denoted P(Y = y), or simply, p( y). P(Y = 10) or p(10) is the sum of probabilities for sample points which are assigned the value 10. When rolling two dice, P(Y = 10) = P({(4, 6)}) + P({(5, 5)}) + P({(6, 4)}) = 1/36 + 1/36 + 1/36 = 3/36

Discrete Random Variable A discrete random variable is a random variable that only assumes a finite (or countably infinite) number of distinct values. For an experiment whose sample points are associated with the integers or a subset of integers, the random variable is discrete.

Probability Distribution A probability distribution describes the probability for each value of the random variable. Presented as a table, formula, or graph. y p(y) 2 1/36 3 2/36 4 3/36 5 4/36 6 5/36 7 6/36 8 5/36 9 4/ / / /36

Probability Distribution For a probability distribution: y p(y) 2 1/36 3 2/36 4 3/36 5 4/36 6 5/36 7 6/36 8 5/36 9 4/ / / /36 = 1.0 Here we may take the sum just over those values of y for which p(y) is non-zero.

Expected Value The “long run theoretical average” For a discrete R.V. with probability function p(y), define the expected value of Y as: In a statistical context, E(Y) is referred to as the mean and so E(Y) and  are interchangeable.

For a constant multiple… Of course, a constant multiple may be factored out of the sum Thus, for our circles, E(C) = E(2  R) = 2  E(R).

For a constant function… In particular, if g(y) = c for all y in Y, then E[g(Y)] = E(c) = c.

Function of a Random Variable Suppose g(Y) is a real-valued function of a discrete random variable Y. It follows g(Y) is also a random variable with expected value In particular, for g(Y) = Y 2, we have

Try this! For the following distribution: y p(y)p(y) Compute the values E( Y ), E( 3Y ), E( Y 2 ), and E( Y 3 )

For sums of variables… Also, if g 1 (Y) and g 2 (Y) are both functions of the random variable Y, then

All together now… So, when working with expected values, we have Thus, for a linear combination Z = c g(Y) + b, where c and b are constants:

Try this! For the following distribution: y p(y)p(y) Compute the values E( Y ), E( 2Y + 5 ), and E( Y 2 - Y )

Variance, V(Y) For a discrete R.V. with probability function p(y), define the variance of Y as: Here, we use V(Y) and   interchangeably to denote the variance. The positive square root of the variance is the standard deviation of Y. It can be shown that Note the variance of a constant is zero.

Computing V(Y) And applying our rules for expected value, we find variance may be expressed as (as the mean is a constant) When computing the variance, it is often easier to use the formula

Try this! For the following distribution: y p(y)p(y) Compute the values V(Y), V(2Y), and V(2Y + 5). How would you compute V(Y 2 ) ?

“Moments and Mass” Note the probability function p(y) for a discrete random variable is also called a “probability mass” or “probability density” function. The expected values E(Y) and E(Y 2 ) are called the first and second moments, respectively.

Continuous Random Variables

For discrete random variables, we required that Y was limited to a finite (or countably infinite) set of values. Now, for continuous random variables, we allow Y to take on any value in some interval of real numbers. As a result, P(Y = y) = 0 for any given value y.

CDF For continuous random variables, define the cumulative distribution function F(y) such that Thus, we have

PDF For the continuous random variable Y, define the probability density function as for each y for which the derivative exists.

Integrating a PDF Based on the probability density function, we may write Remember the 2 nd Fundamental Theorem of Calc.?

Properties of a PDF For a density function f(y): 1). f(y) > 0 for any value of y. 2). Density function, f(y)Distribution function, F(y)

Try this! For what value of k is the following function a density function? We must satisfy the property

Try this! For what value of k is the following function a density function? Again, we must satisfy the property

P(a < Y < b) To compute the probability of the event a < Y < b ( or equivalently a < Y < b ), we just integrate the PDF:

Try this! For the previous density function Find the probability

Try this! Suppose Y is time to failure and Find the probability Determine the density function f (y)

Expected Value, E(Y) For a continuous random variable Y, define the expected value of Y as Note this parallels our earlier definition for the discrete random variable:

Expected Value, E[g(Y)] For a continuous random variable Y, define the expected value of a function of Y as Again, this parallels our earlier definition for the discrete case:

Properties of Expected Value In the continuous case, all of our earlier properties for working with expected value are still valid.

Properties of Variance In the continuous case, our earlier properties for variance also remain valid. and

Problem from MAT 332 Find the mean and variance of Y, given