Lecture 4: Probability Distributions and Probability Densities - 2
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1 Lecture 4: Probability Distributions and Probability Densities - 2 Assist. Prof. Dr. Emel YAVUZ DUMAN MCB1007 Introduction to Probability and Statistics İstanbul Kültür University
2 Outline 1 Multivariate Distributions 2 Marginal Distributions 3 Conditional Distributions
3 Outline 1 Multivariate Distributions 2 Marginal Distributions 3 Conditional Distributions
4 In this section we shall concerned first with the bivariate case, that is, with situation where we are interested at the same time in a pair of random variables defined over a joint sample space that are both discrete. Later, we shall extend this discussions to the multivariate case, covering any finite number of random variables. If X and Y are discrete random variables, we write the probability that X will take on the value x and Y will take on the value y as P(X = x, Y = y). Thus, P(X = x, Y = y) is the probability of the intersection of the events X = x and Y = y. Asinthe univariate case, where we dealt with one random variable and could display the probabilities associated with all values of X by means of a table, we can now, in the bivariate case, display the probabilities associated with all pairs of the values of X and Y by mean of a table.
5 Example 1 Two caplets are selected at a random form a bottle containing three aspirin, two sedative, and four laxative caplets. If X and Y are, respectively, the numbers of the aspirin and sedative caplets included among the two caplets drawn from the bottle, find the probabilities associated with all possible pairs of values of X and Y. Solution. The possible pairs are (0, 0), (0, 1), (1, 0), (1, 1), (0, 2), and (2, 0). So we obtain the following probabilities: P(X =0, Y =0)= P(X =0, Y =1)= P(X =1, Y =0)= ( 3 )( 2 )( ( 2) 9 ) = 6 36, 2 ( 3 )( 2 )( ( 1) 9 ) = 8 36, 2 ( 3 )( 2 )( ( 1) 9 ) = 12 36, 2
6 ( 3 )( 2 )( P(X =1, Y =1)= ( 0) 9 ) = 6 36, 2 ( 3 )( 2 )( P(X =0, Y =2)= ( 0) 9 ) = 1 36, 2 ( 3 )( 2 )( P(X =2, Y =0)= ( 0) 9 ) = Therefore, we have the following table: x y 0 6/36 12/36 3/36 1 8/36 6/36 2 1/36
7 Definition 2 If X and Y are discrete random variables, the function given buy f (x, y) =P(X = x, Y = y) for each pair of values (x, y) within the range of X and Y is called the joint probability distribution of X and Y. Theorem 3 A bivariate function can serve as the joint probability distribution of a pair of discrete random variables X and Y if and only if its values f (x, y) satisfy the conditions 1 f (x, y) 0 for each pair of values (x, y) within its domain; 2 x y f (x, y) =1, where the double summation extends over all possible pairs (x, y) within its domain.
8 Suppose that X canassumeanyoneofm values x 1, x 2,, x m and Y canassumeanyoneofn values y 1, y 2,, y n. Then the probability of the event that X = x j and Y = y k is given by P(X = x j, Y = y k )=f (x j, y k ). A joint probability function for X and Y can be represented by a joint probability table as in the following: y x y 1 y 2 y n Totals x 1 f (x 1, y 1 ) f (x 1, y 2 ) f (x 1, y n ) g(x 1 ) x 2 f (x 2, y 1 ) f (x 2, y 2 ) f (x 2, y n ) g(x 2 )..... x m f (x m, y 1 ) f (x m, y 2 ) f (x m, y n ) g(x m ) Totals h(y 1 ) h(y 2 ) h(y n ) 1
9 Example 4 Determine the value of k for which the function given by f (x, y) =kxy for x =1, 2, 3; y =1, 2, 3 can serve as a joint probability distribution. Solution. Substituting the various values of x and y, weget f (1, 1) = k, f (1, 2) = 2k, f (1, 3) = 3k, f (2, 1) = 2k, f (2, 2) = 4k, f (2, 3) = 6k, f (3, 1) = 3k, f (3, 2) = 6k, f (3, 3) = 9k. To satisfy the first condition of Theorem 3, the constant k must be nonnegative, and to satisfy the second condition k +2k +3k +2k +4k +6k +3k +6k +9k =1 so that 36k =1andk =1/36.
10 f (x, y) =kxy for x =1, 2, 3; y =1, 2, 3 The joint probability function for X and Y can be represented by a joint probability table as in the following: y x Totals 1 k 2k 3k 6k 2 2k 4k 6k 12k 3 3k 6k 9k 18k Totals 6k 12k 18k 36k =1
11 Example 5 If the values of the joint probability distribution of X and Y are as shown in the table x y find (a) P(X =1, Y =2); (b) P(X =0, 1 Y < 3); (c) P(X + Y 1); (d) P(X > Y ) /12 1/6 1/24 1 1/4 1/4 1/40 2 1/8 1/20 3 1/120
12 x y /12 1/6 1/24 1 1/4 1/4 1/40 2 1/8 1/20 3 1/120 (a) P(X =1, Y =2)= 1 20 ; (b) P(X =0, 1 Y < 3) = f (0, 1) + f (0, 2) = = 3 8 ; (c) P(X + Y 1) = f (0, 0) + f (1, 0) + f (0, 1) = = 1 2 ; (d) P(X > Y )=f(1, 0) + f (2, 0) + f (2, 1) = = 7 30.
13 Example 6 If the joint probability distribution of X and Y is given by f (x, y) =c(x 2 + y 2 )forx = 1, 0, 1, 3, y = 1, 2, 3 find the value of c. Solution. Since x y Totals 1 2c 1c 2c 10c 15c 2 5c 4c 5c 13c 27c 3 10c 9c 10c 18c 47c Totals 17c 14c 17c 41c 89c =1 then we have that c = 1 89.
14 Example 7 Show that there is no value of k for which f (x, y) =ky(2y x) forx =0, 3, y =0, 1, 2 can serve as the joint probability distribution of two random variables. Solution. Since y x Totals 0 0 2k 8k 10k 3 0 k 2k k Totals 0 k 10k 11k =1 then we find that k =1/11. Butinthiscase,f (3, 1) differs in sign from all other terms.
15 Example 8 Suppose that we roll a pair of balanced dice and X is the number of dice that come up 1, and Y is the number of dice that come up 4, 5, or 6. (a) Construct a table showing the values X and Y associated with each of the 36 equally likely points of the sample space. (b) Construct a table showing the values the joint probability distribution of X and Y.
16 Solution. (a) If X is the number of dice that come up 1, and Y is the number of dice that come up 4, 5, or 6, then we have Roll 2 Roll (2,0) (1,0) (1,0) (1,1) (1,1) (1,1) 2 (1,0) (0,0) (0,0) (0,1) (0,1) (0,1) 3 (1,0) (0,0) (0,0) (0,1) (0,1) (0,1) 4 (1,1) (0,1) (0,1) (0,2) (0,2) (0,2) 5 (1,1) (0,1) (0,1) (0,2) (0,2) (0,2) 6 (1,1) (0,1) (0,1) (0,2) (0,2) (0,2) (b) Then, we simply count the number of times we have each of the possible (x, y) values, and divide by 36. x y Prob. 4/36 12/36 9/36 4/36 6/36 1/36
17 Definition 9 If X and Y are discrete random variables, the function given by F (x, y) =P(X x, Y y) = f (s, t) for < x, y < s x t y where f (s, t) is the value of the joint probability distribution of X and Y at (s, t), is called the joint distribution function, orthe joint cumulative distribution, of X and Y. Theorem 10 If F (x, y) is the value of the joint distribution function of two discrete random variables X and Y at (x, y), then (a) F (, ) =0; (b) F (, ) =1; (c) if a < bandc< d, then F(a, c) F (b, d).
18 Example 11 With reference to Example 1, find F (1, 1). x y Solution /36 12/36 3/36 1 8/36 6/36 2 1/36 F (1, 1) = P(X 1, Y 1) = f (0, 0) + f (0, 1) + f (1, 0) + f (1, 1) = = 32 36
19 Example 12 With reference to Example 5, find the following values of the joint distribution function of the two random variables: (a) F (1.2, 0.9) (b) F ( 3, 1.5) (c) F (2, 0) (d) F (4, 2.7). x y /12 1/6 1/24 1 1/4 1/4 1/40 2 1/8 1/20 3 1/120 Solution. (a) F (1.2, 0.9) = P(X 1.2, Y 0.9) = f (0, 0) + f (1, 0) = = 1 4 (b) F ( 3, 1.5) = P(X 3, Y 1.5) = 0
20 x y /12 1/6 1/24 1 1/4 1/4 1/40 2 1/8 1/20 3 1/120 (c) F (2, 0) = P(X 2, Y 0) = f (0, 0) + f (1, 0) + f (2, 0) = = 7 24 (d) F (4, 2.7) = P(X 4, Y 2.7) = 1 f (0, 3) = =
21 Example 13 If two cards are randomly drawn (without replacement) from an ordinary deck of 52 playing cards, Z is the number of aces obtained in the first draw and W is the total number of aces obtained in both draws, find F (1, 1). Solution. Let X be the number of aces obtained in the first draw, and Y be the number of aces obtained in the second draw. So, we have f (x, y), the joint probability distribution: f (0, 0) = = , f (1, 0) = = , f (0, 1) = = , f (1, 1) = =
22 Since w z f (0, 0) f (0, 1) 1 f (1, 0) f (1, 1) where z = x and w = x + y, thenwehave w z Therefore, we obtain that /221 16/ /221 1/221 F (1, 1) = = =
23 All the definitions in this section can be generalized to the multivariate case, where there are n random variables. Corresponding to Definition 2, the values of the joint probability distribution of n discrete random variables X 1, X 2,, and X n are given by f (x 1, x 2,, x n )=P(X 1 = x 1, X 2 = x 2,, X n = x n ) for each n-tuple (x 1, x 2,, x n ) within the range of the random variables; and corresponding to Definition 9, the values of their joint distribution function are given by F (x 1, x 2,, x n )=P(X 1 x 1, X 2 x 2,, X n x n ) for < x 1 <, < x 2 <,, < x n <.
24 Example 14 If the joint probability distribution of three discrete random variables X, Y,andZ is given by (x + y)z f (x, y, z) = 63 find P(X =2, Y + Z 3). for x =1, 2; y =1, 2, 3; z =1, 2 Solution. P(X =2, Y + Z 3) = f (2, 1, 1) + f (2, 1, 2) + f (2, 2, 1) = =
25 Example 15 Find c if the joint probability distribution of X, Y and Z is given by f (x, y, z) =kxyz for x =1, 2; y =1, 2, 3; z =1, 2 and determine F (2, 1, 2) and F (4, 4, 4) Solution. Since = x=1 y=1 z=1 =f (1, 1, 1) + f (1, 1, 2) + f (1, 2, 1) + f (1, 2, 2) + f (1, 3, 1) + f (1, 3, 2) +f (2, 1, 1) + f (2, 1, 2) + f (2, 2, 1) + f (2, 2, 2) + f (2, 3, 1) + f (2, 3, 2) =k( )=54k then we have that k =1/54.
26 f (x, y, z) =kxyz for x =1, 2; y =1, 2, 3; z =1, 2 Also, F (2, 1, 2) = P(X 2, Y 1, Z 2) = f (1, 1, 1) + f (1, 1, 2) + f (2, 1, 1) + f (2, 1, 2) = 1 54 ( )= 9 54 = 1 6 F (4, 4, 4) = P(X 4, Y 4, Z 4) = 1.
27 Outline 1 Multivariate Distributions 2 Marginal Distributions 3 Conditional Distributions
28 Marginal Distributions To introduce the concept of a marginal distribution, letus consider the following example. Example 16 In Example 1 we derived the joint probability distribution of two random variables X and Y, the number of aspirin and the number of sedative caplets included among two caplets drawn at random from a bottle containing three aspirin, two sedative, and four laxative caplets. Find the probability distribution of X alone and that of Y alone. Solution. The results of Example 1 are shown in the following table, together with the marginal totals, that is, the totals of the respective rows and columns:
29 x y h(y) 0 6/36 12/36 3/36 21/36 1 8/36 6/36 14/36 2 1/36 1/36 g(x) 15/36 18/36 3/36 1 The column totals are the probabilities that X will take on the values 0, 1, and 2. In other words, they are the values 2 g(x) = f (x, y) for x =0, 1, 2 y=0 of the probability distribution of X. 15/36 for x =0 g(x) = 18/36 for x =1 3/36 for x =2
30 x y h(y) 0 6/36 12/36 3/36 21/36 1 8/36 6/36 14/36 2 1/36 1/36 g(x) 15/36 18/36 3/36 1 By the same token, the row totals are the values h(y) = 2 f (x, y) for y =0, 1, 2 x=0 of the probability distribution of Y. 21/36 for y =0 h(y) = 14/36 for y =1 1/36 for y =2
31 We are thus led to the following definition: Definition 17 If X and Y are discrete random variables and f (x, y) is the value of their joint probability distribution at (x, y), the function given by g(x) = y f (x, y) for each x within the range of X is called the marginal distribution of X. Correspondingly, the function given by h(y) = x f (x, y) for each y within the range of Y is called the marginal distribution of Y.
32 Example 18 Two textbooks are selected at random from a shelf contains three statistics texts, two mathematics texts, and three physics texts. If X is the number of statistics texts and Y the number of mathematics texts actually chosen (a) construct a table showing the values of the joint probability distribution of X and Y. (b) Find the marginal distributions of X and Y.
33 Solution. (a) Since ( 3 f (0, 0) = ( 2) 8 = 2) 3 ( 3 3 ) 1)( 28, f (1, 0) = 1 ) = 9, f (0, 1) = 28 ( 8 2 ( 2 )( 3 1( 1) 8 ) = 6 28, 2 ( 3 )( 2 1 f (1, 1) = ( 1) 8 ) = 6 ( 3, f (2, 0) = ( 2) 28 8 ) = 3 ( 2, f (0, 2) = ( 2) 28 8 ) = 1 28, we have the following join probability distribution table: x h(y) y 0 3/28 9/28 3/28 15/28 1 6/28 6/28 12/28 2 1/28 1/28 g(x) 10/28 15/28 3/28 1
34 x y h(y) 0 3/28 9/28 3/28 15/28 1 6/28 6/28 12/28 2 1/28 1/28 g(x) 10/28 15/28 3/28 1 (b) The marginal distributions of X and Y 10/28 for x =0 g(x) = 15/28 for x =1 3/28 for x =2 respectively. 15/28 for y =0 h(y) = 12/28 for y =1 1/28 for y =2
35 Example 19 Given the values of the joint probability distribution of X and Y shown in the table x 1 1 y 1 1/8 1/ /4 1 1/8 0 find the marginal distribution of X and the marginal distribution of Y. Solution. g(x) = { = 1 4 for x = = 3 4 for x = = 5 8 for y = 1 h(y) = 1 4 for y =0 1 8 for y =1
36 Outline 1 Multivariate Distributions 2 Marginal Distributions 3 Conditional Distributions
37 Conditional Distributions We defined the conditional probability of an event A, given event B, as P(A B) P(A B) = P(B) provided P(B) 0. Suppose now that A and B are the events X = x and Y = y so that we can write P(X = x Y = y) = P(X = x, Y = y) P(Y = y) = f (x, y) h(y) provided P(Y = y) =h(y) 0,wheref (x, y) is the value of the joint probability distribution of X and Y at (x, y) andh(y) isthe value of the marginal distribution of Y at y. Denotingthe conditional probability f (x y) to indicate that x is a variable and y is fixed, let us make the following definition:
38 Definition 20 If f (x, y) is the value of the joint probability distribution of the discrete random variables X and Y at (x, y), and h(y) is the value of the marginal distribution of Y at y, the function given by f (x y) = f (x, y) h(y) h(y) 0 for each x within the range of X, is called the conditional distribution of X given Y = y. Correspondingly, if g(x) isthe value of the marginal distribution of X at x, the function w(y x) = f (x, y) g(x) g(x) 0 for each y within the range of Y, is called the conditional distribution of Y given X = x.
39 Example 21 With reference to Example 1 and Example 16, find the conditional distribution of X given Y =1. f (x 1) = f (x, 1)/h(1) x y h(y) 0 6/36 12/36 3/36 21/36 1 8/36 6/36 14/36 2 1/36 1/36 g(x) 15/36 18/36 3/36 1 Solution. Substituting the appropriate values from the table above, we get f (0 1) = f (0, 1) h(1) = 8/36 14/36 = 8 f (1, 1), f (1 1) = 14 h(1) f (2 1) = f (2, 1) h(1) = 0 14/36 =0. = 6/36 14/36 = 6 14,
40 Example 22 With reference to Example 18 find the conditional distribution of Y given X =0. w(y 0) = f (0, y)/g(0) x y h(y) 0 3/28 9/28 3/28 15/28 1 6/28 6/28 12/28 2 1/28 1/28 g(x) 10/28 15/28 3/28 1 Solution. Substituting the appropriate values from the table above, we get w(0 0) = f (0, 0) g(0) = 3/28 10/28 = 3 f (0, 1), w(1 0) = 10 g(0) w(2 0) = f (0, 2) g(0) = 1/28 10/28 = = 6/28 10/28 = 6 10,
41 Definition 23 Suppose that X and Y are discrete random variables. If the events X = x and Y = y are independent events for all x and y, thenwe say that X and Y are independent random variables. In such case, P(X = x, Y = y) =P(X = x) P(Y = y) or equivalently f (x, y) =f (x y) h(y) =g(x) h(y). Conversely, if for all x and y the joint probability function f (x, y) can be expressed as the product of a function of x alone and a function of y alone (which are then the marginal probability functions of X and Y ), X and Y are independent. If, however, f (x, y) cannot be so expressed, then X and Y are dependent.
42 Example 24 Determine whether the random variables of Example 1 are independent. Solution. With reference to Example 1, we have the following joint distribution table: x y h(y) 0 6/36 12/36 3/36 21/36 1 8/36 6/36 14/36 2 1/36 1/36 g(x) 15/36 18/36 3/36 1 Since f (0, 1) = we see that X and Y are dependent. = g(0) h(1)
43 Example 25 A box contains three balls labeled 1, 2 and 3. Two balls are randomly drawn from the box without replacement. Let X be the number on the first ball and Y the number on the second ball. (1) Find the joint probability distribution of X and Y. (2) Find the marginal distributions of X and Y. (3) Find the conditional distribution of X given Y =1. (4) Find the conditional distribution of Y given X =2. (5) Determine whether the random variables X and Y are independent.
44 Solution. (1) Since P(X =1, Y =2)=P(X =1, Y =3)=P(X =2, Y =1)= P(X =2, Y =3)=P(X =3, Y =1)=P(X =3, Y =2)= = 1 6, this joint probability distribution can be expressed by the following table: y g(x) x 1 1/6 1/6 2/6 2 1/6 1/6 2/6 3 1/6 1/6 2/6 h(y) 2/6 2/6 2/6 1
45 y x g(x) 1 1/6 1/6 2/6 2 1/6 1/6 2/6 3 1/6 1/6 2/6 h(y) 2/6 2/6 2/6 1 (2) 2/6, x =1, g(x) = 2/6, x =2, 2/6, x =3, 2/6, y =1, h(y) = 2/6, y =2, 2/6, y =3.
46 y x g(x) 1 1/6 1/6 2/6 2 1/6 1/6 2/6 3 1/6 1/6 2/6 h(y) 2/6 2/6 2/6 1 (3) The conditional distribution of X given Y =1: f (1 1) = f (2 1) = f (3 1) = f (x 1) = f (1, 1) h(1) f (2, 1) h(1) f (2, 1) h(1) f (x, 1) h(1) = 0 2/6 =0, = 1/6 2/6 = 1 2, = 1/6 2/6 = 1 2.
47 y x g(x) 1 1/6 1/6 2/6 2 1/6 1/6 2/6 3 1/6 1/6 2/6 h(y) 2/6 2/6 2/6 1 (4) The conditional distribution of Y given X =2: w(1 2) = w(y 2) = w(2 2) = w(3 2) = f (2, 1) g(2) f (2, 2) g(2) f (2, 3) g(2) f (2, y) g(2) = 1/6 2/6 = 1 2, = 0 2/6 =0, = 1/6 2/6 = 1 2.
48 y x g(x) 1 1/6 1/6 2/6 2 1/6 1/6 2/6 3 1/6 1/6 2/6 h(y) 2/6 2/6 2/6 1 (5) Since f (1, 1) = 0 g(1) h(1) = we see that X and Y are dependent.
49 Thank You!!!
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