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Joint Probability Distributions MATH30-6 Probability and Statistics Objectives At the end of the lesson, the students are expected to • Use joint probability mass functions and joint probability density functions to calculate probabilities; • Calculate marginal and conditional probability distributions from joint probability distributions; and • Interpret and calculate covariances and correlations between random variables. Joint Probability Mass Function The joint probability mass function of the discrete random variables X and Y, denoted as , , satisfies (1) , ≥ 0 (2) , = 1 (3) , = = , = (5-1) • Sometimes referred to as the bivariate probability distribution or bivariate distribution of the random variables • = and = is usually written as = Joint Probability Mass Function Examples: 5-1/153 Signal Bars Calls are made to check the airline schedule at your departure city. You monitor the number of bars of signal strength on your cell phone and the number of times you have to state the name of your departure city before the voice system recognizes the name. In the first four bits transmitted, let X denote the number of bars of signal strength on your cell phone Y denote the number of times you need to state your departure city Joint Probability Mass Function By specifying the probability of each of the points in Fig. 5-1, we specify the joint probability distribution of X and Y. Similarly to an individual random variable, we define the range of the random variables (X, Y) to be the set of points (x, y) in two-dimensional space for which the probability that X = x and Y = y is positive. Joint Probability Mass Function 3.14/95 Two refills for a ballpoint pen are selected at random from a box that contains 3 blue refills, 2 red refills, and 3 green refills. If is the number of blue refills and is the number of red refills selected, find (a) the joint probability function , , and (b) (b) , ∈ , where is the region , | + Joint Probability Density Function A joint probability density function for the continuous random variables X and Y, denoted as , , satisfies the following properties: (1) , ≥ 0 for all x, y (2) ∞ ∞ −∞ −∞ , = 1 (3) For any region R of two-dimensional space, , ∈ = (5-2) , Joint Probability Density Function Examples: 5-2/155 Server Access Time Let the random variable X denote the time until a computer server connects to your machine (in milliseconds), and let Y denote the time until the server authorizes you as a valid user (in milliseconds). Each of these random variables measures the wait from a common starting time and X < Y. Assume that the joint probability density function for X and Y is , = 6 × 10−6 −0.001−0.002 for < Joint Probability Density Function The region with nonzero probability is shaded in Fig. 5-4. The property that this joint probability density function integrates to 1 can be verified by the integral of , over this region as follows: ∞ ∞ , −∞ −∞ ∞ ∞ = 0 = 6 × 10−6 6 × 10−6 −0.001−0.002 ∞ 0 ∞ −0.002 −0.001 Joint Probability Density Function = 6 × 10−6 ∞ 0 −0.002 −0.001 0.002 ∞ = 0.003 ∞ 0 ∞ −∞ −∞ −0.003 1 , = 0.003 =1 0.003 Joint Probability Density Function Joint Probability Density Function The probability that X < 1000 and Y < 2000 is determined as the integral over the darkly shaded region in Fig 5-5. 1000 2000 ≤ 1000, ≤ 2000 = = 6 × 10−6 1000 0 = 6 × 10−6 1000 0 1000 = 0.003 0 2000 0 , −0.002 −0.001 −0.002 − −4 −0.001 0.002 −0.003 − −4 −0.001 Joint Probability Density Function −1 1 − −3 1 − = 0.003 − −4 0.003 0.001 ≤ 1000, ≤ 2000 = 0.003 316.738 − 11.578 = 0.915 Joint Probability Density Function 3.15/96 A privately owned business operates a drive-in facility and a walk-in facility. On a randomly selected day, let and , respectively, be the proportions of the time that the drive-in and the walk-in facilities are in use, and suppose that the joint density function of these random variables is 2 , = 5 2 + 3 , 0 ≤ ≤ 1, 0 ≤ ≤ 1, 0, elsewhere. (a) Verify condition 2. (b) Find , ∈ , where = 1 1 2 4 , |0 < < , < Marginal Probability Mass Function Marginal probability distribution • The individual probability distribution of a random variable The marginal probability mass functions of alone and of alone are = , and = , Marginal Probability Mass Function Examples: 5-3/157 Marginal Distribution The joint probability distribution of X and Y in Fig. 5-1 can be used to find the marginal probability distribution of X. For example, 3 = = 3 = = 3, = 1 + = 3, = 2 + = 3, = 3 + = 3, = 4 = 0.25 + 0.2 + 0.05 + 0.05 = 0.55 The marginal probability distribution for X is found by summing the probabilities in each column, whereas the marginal probability distribution for Y is found by summing the probabilities in each row. The results are shown in Fig. 5-6. Marginal Probability Mass Function Marginal Probability Mass Function 3.16/98 Show that the column and row totals of Table 3.1 give the marginal distribution of alone and of alone. Table 3.1: Joint Probability Distribution for Example 3.14 x Row Totals f(x, y) 0 1 2 0 3/28 3/14 1/28 1 9/28 3/14 0 2 3/28 0 0 15/28 3/7 1/28 Column Totals 5/14 15/28 3/28 1 y Marginal Probability Mass Function 3.50/106 Suppose that X and Y have the following joint probability distribution: f(x, y) y 1 3 5 x 2 0.10 4 0.15 0.20 0.10 0.30 0.15 (a) Find the marginal distribution of X. (b) Find the marginal distribution of Y. Marginal Probability Density Function If the joint probability density function of random variables X and Y is , , the marginal probability density functions of X and Y are = , and = , (5-3) where the first integral is over all points in the range of , for which X = x and the second integral is over all points in the range of , for which Y = y. Marginal Probability Density Function A probability for only one random variable, say, for example, < < , can be found from the marginal probability distribution of X or from the integral of the joint probability distribution of X and Y as << = = ∞ = , −∞ ∞ , −∞ Marginal Probability Density Function Examples: 5-4/157 Server Access Time For the random variables that denote times in Example 5-2, calculate the probability that Y exceeds 2000 milliseconds. This probability is determined as the integral of , over the darkly shaded region in Fig. 5-7. The region is partitioned into two parts and different limits of integration are determined for each part. Marginal Probability Density Function Marginal Probability Density Function > 2000 2000 ∞ 6 × 10−6 −0.001−0.002 = 0 ∞ 2000 ∞ 6 × 10−6 −0.001−0.002 + 2000 Marginal Probability Density Function The first integral is 2000 6 × 10−6 0 10−6 6× = −4 0.002 −0.002 −0.002 2000 ∞ −0,001 2000 −0.001 0 6 × 10−6 −4 1 − −2 = 0.002 0.001 = 0.0475 Marginal Probability Density Function The second integral is ∞ −0.002 6 × 10−6 2000 10−6 6× = 0.002 −0.002 ∞ ∞ −0.001 −0.003 2000 10−6 6× = 0.002 −6 = 0.0025 0.003 Therefore, > 2000 = 0.0475 + 0.0025 = 0.05 Marginal Probability Density Function Alternatively, the probability can be calculated from the marginal probability distribution of Y as follows. For y > 0, 6 × 10−6 −0.001−0.002 = 0 = 6 × 10−6 −0.002 −0.001 0 = 6 × 10−6 −0.002 =6× 10−6 −0.002 −0.001 −0.001 0 −0.001 1− 0.001 Marginal Probability Density Function = 6 × 10−3 −0.002 1 − −0.001 for y > 0 We have obtained the marginal probability density function of Y. Now, ∞ > 2000 = 6 × 10−3 = 6 × 10−3 −0.002 −0.002 = 6 × 10−3 −0.002 1 − −0.001 2000 ∞ 2000 −4 − −0.003 −0.003 −6 − = 0.05 0.002 0.003 ∞ 2000 Marginal Probability Density Function 3.40/105 A fast-food restaurant operates both a drive-through facility and a walk-in facility. On a randomly selected day, let X and Y, respectively, be the proportions of the time that the drive-through and walk-in facilities are in use, and suppose that the joint density function of these random variables is 2 , = 3 + 2 , 0 ≤ ≤ 1, 0 ≤ ≤ 1, 0, elsewhere. (a) Find the marginal density of X. (b) Find the marginal density of Y. (c) Find the probability that the drive-through facility is busy less than one-half of the time. Conditional Probability Mass Function Let X and Y be two random variables, discrete or continuous. The conditional distribution of the random variable Y, given that X = x, is , | = , > 0. Similarly, the conditional distribution of the random variable X, given that Y = y, is , | = , > 0. Conditional Probability Mass Function Examples: 5-5/159 Signal Bars For Example 5-1, X and Y denote the number of bars of signal strength and times you need to state your departure city received, respectively. Then, = 1| = 3 = = 3, = 1 = 3 = 3,1 3 = 0.25 0.55 = 0.454 The probability that = 2 given that = 3 is = 2| = 3 = = 3, = 2 = 3 = 3,2 3 = 0.2 0.55 = 0.364 Conditional Probability Mass Function Additional Conclusion: Further work shows that = 3| = 3 = 0.091 and = 4| = 3 = 0.091. Note that = 1| = 3 + = 2| = 3 + = 3| = 3 = = 4| = 3 = 1 . This set of probabilities defines the conditional probability distribution of Y given that X = 3. Conditional Probability Mass Function 3.18/99 Referring to Example 3.14, find the conditional distribution of X, given that Y = 1, and use it to determine = 0| = 1 . Table 3.1: Joint Probability Distribution for Example 3.14 x Row Totals f(x, y) 0 1 2 0 3/28 3/14 1/28 1 9/28 3/14 0 2 3/28 0 0 15/28 3/7 1/28 Column Totals 5/14 15/28 3/28 1 y Conditional Probability Mass Function 3.49/106 Let X denote the number of times a certain numerical control machine will malfunction: 1, 2, or 3 times on any given day. Let Y denote the number of times a technician is called on an emergency call. Their joint probability distribution is given as f(x, y) y 1 3 5 x 1 2 3 0.05 0.05 0.10 0.05 0.10 0.35 0.00 0.20 0.10 (a) Evaluate the marginal distribution of X. (b) Evaluate the marginal distribution of Y. (c) Find = 3| = 2 . Conditional Probability Density Function Given continuous random variables X and Y with joint probability density function , , the conditional probability density function of Y given X = x is | = , for > 0 Equivalently, | , = , > 0 (5-4) Conditional Probability Density Function Because the conditional probability density function | is a probability density function for all y in , the following properties are satisfied: (1) | ≥ 0 (2) | = 1 (3) ∈ | = = range of Y | for any set B in the (5-5) Conditional Probability Density Function Examples: 3.19/100 The joint density function for the random variables , , where X is the unit temperature change and Y is the proportion of spectrum shift that a certain atomic particle produces is 10 2 , 0 < < < 1 , = 0, elsewhere. a) Find the marginal densities , , and the conditional probability | . b) Find the probability that the spectrum shifts more than half of the total observations, given the temperature is increased to 0.25 units. Conditional Probability Density Function 3.20/100 Given the joint density function 1 + 3 2 , 0 < < 2, 0 < < 1 , = 4 0, elsewhere, find , , | , and evaluate 1 4 << Conditional Probability Density Function 3.53/106 Given the joint density function 6−− , 0 < < 2, 2 < < 4, , = 8 0, elsewhere, find 1 < < 3| = 1 . Independence For random variables X and Y, if any one of the following properties is true, the others are also true, and X and Y are independent. (1) , = for all x and y (2) | = for all x and y with > 0 (3) | = for all x and y with > 0 (4) ∈ , ∈ = ∈ ∈ for any sets A and B in the range of X and Y, respectively. (5-7) Independence Examples: 5-11/162 Independent Random Variables Suppose that Example 5-2 is modified so that the joint probability density function of X and Y is , = 2 × 10−6 −0.001 − 0.002 for ≥ 0 and ≥ 0. Show that X and Y are independent and determine > Independence The marginal probability density function of Y is ∞ 2 × 10−6 −0.001−0.002 = 0 = 0.002 −0.002 for y > 0 Therefore, , = for all x and y, and X and Y are independent. Independence To determine the probability requested, property (4) of Equation 5-7 and the fact that each random variable has an exponential distribution can be applied. > 1000, < 1000 = > 1000 < 1000 = −1 1 − −2 = 0.0318 Independence 3.21/102 Show that the random variables of Example 3.14 are not statistically independent. Table 3.1: Joint Probability Distribution for Example 3.14 f(x, y) 0 y 1 2 Column Totals 0 3/28 3/14 1/28 5/14 x 1 9/28 3/14 0 15/28 2 3/28 0 0 3/28 Row Totals 15/28 3/7 1/28 1 Examples 5-1/167 Show that the following function satisfies the properties of a joint probability mass function. x 1 1.5 y 1 2 fXY(x, y) 1/4 1/8 1.5 3 1/4 2.5 4 1/4 3 5 1/8 Determine the following: (a) < 2.5, < 3 (b) < 2.5 (c) < 3 (d) > 1.8, > 4.7 Examples (e) , , , and (f) Marginal probability distribution of the random variable X (g) Conditional probability distribution of Y given that X = 2.5 (h) Conditional probability distribution of X given that Y = 2. (i) | = 1.5 (j) Are X and Y independent? Examples 5-2/167 Determine the value of c that makes the function , = + a joint probability mass function over the nine points with x = 1, 2, 3 and y = 1, 2, 3. Determine the following: (a) = 1, < 4 (b) = 1 (c) = 2 (d) < 2, < 2 (e) , , , and (f) Marginal probability distribution of the random variable X (g) Conditional probability distribution of Y given that X = 1 Examples (h) Conditional probability distribution of X given that Y = 2 (i) | = 1 (j) Are X and Y independent? Examples 5-4/167 Four electronic printers are selected from a large lot of damaged printers. Each printer is inspected and classified as containing either a major or a minor defect. Let the random variables X and Y denote the number of printers with major and minor defects, respectively. Determine the range of the joint probability distribution of X and Y. Examples 5-6/167 A small-business Web site contains 100 pages and high graphic content, respectively. A sample of four pages is selected without replacement, and X and Y denote the number of pages with moderate and high graphics output in the sample. Determine: (a) , (b) (c) (d) |3 (e) | = 3 (f) | = 3 (g) Are X and Y independent? Expected Value of a Function of Two Random Variables ℎ , , ℎ , , discrete = ℎ , , (5-13) , continuous Expected Value of a Function of Two Random Variables Example: 5-19/171 For the joint probability distribution of the two random variables in Fig. 5-12, calculate − − . Covariance • A measure of linear relationship between the random variables • If the relationship between the random variables is nonlinear, the covariance might not be sensitive to the relationship, as illustrated in Fig. 5-13(d). The only points with nonzero probability are the points on the circle. cov , = = − − (5-14) Covariance Covariance Examples: 5-20/173 In Example 5-1, the random variables X and Y are the number of signal bars and the number of times you need to state your departure city, respectively. Is the covariance between X and Y positive or negative? Negative covariance Covariance 4.47/127 For the random variables X and Y whose joint density function is given in Exercise 3.40 on page 105, find the covariance. 2 , = 3 + 2 , 0 ≤ ≤ 1, 0 ≤ ≤ 1, 0, elsewhere Correlation The correlation between random variables X and Y, denoted as ρXY, is cov , σ = = (5-15) For any two random variables X and Y −1 ≤ ρXY ≤ +1 (5-16) If X and Y are independent random variables, σXY = ρXY = 0 (5-17) Covariance Examples: 5-21/173 Covariance For the discrete random variables X and Y with the joint distribution shown in Fig. 5-14, determine σXY and ρXY. Correlation 5-22/174 Correlation Suppose that the random variable X has the following distribution: = 1 = 0.2, Examples 5-29/175 Determine the covariance and correlation for the following joint probability distribution x y fXY(x, y) 2 3 1/8 1 4 1/4 2 5 1/2 4 6 1/8 Examples 5-31/175 Determine the value for c and the covariance and correlation for the joint probability mass function , = + for x = 1, 2, 3 and y = 1, 2, 3. 5-37/175 Determine the covariance and correlation for the joint probability density function , = −− over the range 0 < x and 0 < y. Examples 5-39/175 The joint probability distribution is x y fXY(x, y) −1 0 1/4 0 −1 1/4 0 1 1/4 1 0 1/4 Show that the correlation between X and Y is zero, but X and Y are not independent. Summary • A joint probability mass function is a function used to calculate probabilities for two or more discrete random variables. • A joint probability density function is a function used to calculate probabilities for two or more continuous random variables. • A marginal probability mass function is the probability mass function of a discrete random variable obtained from the joint probability distribution of two or more random variables. Summary • A marginal probability density function is the probability density function of a continuous random variable obtained from the joint probability distribution of two or more random variables. • A conditional probability mass function is the probability mass function of the conditional probability distribution of a discrete random variable. • A conditional probability density function is the probability density function of the conditional probability distribution of a continuous random variable. Summary • The covariance is a measure of association between two random variables obtained as the expected value of the product of the two random variables around their means; that is Cov , = = − References • Montgomery and Runger. Applied Statistics and Probability for Engineers, 5th Ed. © 2011 • Walpole, et al. Probability and Statistics for Engineers and Scientists 9th Ed. © 2012, 2007, 2002