3.4 Joint Probability Distributions Joint Probability of two discrete random variables Joint probability of two continuous random variables Marginal distributions Conditional probability distributions Independence of two or more random variables

Reality However, there are many problems in which two or more random variables need to be studied simultaneously For example: 1. We might wish to study the number of available check-in counters at an airport in conjunction with the number of customers waiting in queue. 2. We might wish to study the yield of chemical reaction together with the temperature at which the reaction is run.

Typical questions to ask are: “What is average number of customers in the queue given that number of available counters is 5 ?”  “Is the yield independent of the temperature?” “What is the average yield if temperature is 40 C?”   To answer the questions of this type, we need to study what are called two-dimensional or multi-dimensional random variables of both discrete and continuous types.

Joint Probability Distribution of Discrete Random Variables Definition 3.8 Let X and Y be random variables The order the pair (X, Y) is called a two dimensional random variable. Ω ω X(s) S s y Y(s) x

For any region A in the xy plane, P[(X, Y)A] = Definition 3.8 A function f(x, y) is the joint probability distribution function or probability mass function for two-dimensional discrete random variable (X, Y) if: 1. f(x, y)  0 for all (x, y) 2. = 1 3. P(X = x, Y = y) = f(x, y) For any region A in the xy plane, P[(X, Y)A] = f(x, y) represents the probability distribution for the simultaneous occurrence of (X, Y) in any pair of (x, y) within the range of random variables X and Y.

Table for Joint Probability Distribution

Example 3.8, page 75 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 X is the number of blue refills and Y is the number of red refills selected, find (a) the joint probability function f(x,y), and (b) P((X,Y)A), where A is the region {(x, y) | x + y  1}

(a) find the joint probability function f(x,y)   f(0,0) = P(X=0,Y=0) = =3/28 f(0,1) = P(X=0,Y=1) = =3/14   In general,  f(x,y) = P(X =x, Y = y) = 1 2 Y X 3/28 9/28 3/14 1/28 (See Table 3.1, page 75.)

(b) P ((X,Y)A), where A is the region {(x, y) | x + y  1} because (0, 0), (0, 1) and (1, 0) are the ones , such that x + y  1, so: P((X,Y) A) = f(0, 0) + f(0, 1) + f(1, 0) = 3/28 + 3/14 + 9/28 = 9/14

2. Joint probability of two continuous random variables Definition 3.9 Let X and Y be continuous random variables. The order the pair (X,Y) is called a two dimensional continuous random variable. A function f(x, y) is the joint density function for (X, Y) if 1. f(x, y)  0 for all (x, y) 2. = 1 3. P((X, Y)A) = for any region A in the xy plane.

Example 3.9, page 76 A candy company distributes boxes of chocolates with a mixture of creams, toffees, and nuts coated in both light and dark chocolate. For a randomly selected box, let X and Y, respectively, be the proportions of the light and dark chocolates that are creams and suppose that the joint density function is (square) (may be circular, triangular or other regions) (a) Determine c. (b) Find P[(X, Y)A], A is the region A={(x, y) | 0 < x < ½, ¼ < y < ½ } (c) Find P[(X, Y)B], where B is the region B={(x, y) | 0 < x < y < 1 }

Solution (a) c = =2/5 (need to have ) (b) P[(X, Y)A] = =13/160 (c) P[(X, Y)B] = = x y 1 1/2 1/4

3.4 Joint Probability Distributions Joint Probability of two discrete random variables Joint probability of two continuous random variables Marginal distribution Conditional probability distributions Independence of two or more random variables

For any region A in the xy plane, P[(X, Y)A] = Definition 3.8 A function f(x, y) is the joint probability distribution function or probability mass function for two-dimensional discrete random variable (X, Y) if: 1. f(x, y)  0 for all (x, y) 2. = 1 3. P(X = x, Y = y) = f(x, y) For any region A in the xy plane, P[(X, Y)A] = f(x, y) represents the probability distribution for the simultaneous occurrence of (X, Y) in any pair of (x, y) within the range of random variables X and Y.

2. Joint probability of two continuous random variables Definition 3.9 Let X and Y be continuous random variables. The order the pair (X,Y) is called a two dimensional continuous random variable. A function f(x, y) is the joint density function for (X, Y) if 1. f(x, y)  0 for all (x, y) 2. = 1 3. P((X, Y)A) = for any region A in the xy plane.

3. Marginal distribution Definition 3.10 The marginal distributions of X alone and of Y alone are P(X = x) = g(x) = P(Y = y) = h(y) = for the discrete case, and by , for the continuous case.

Example 3.10, page 77 Use table 3.1, page 75 to find the marginal distributions for X and Y of Example 3.8. 1 2 Y X 3/28 9/28 3/14 1/28 h(y) g(x) 15/28 3/7 5/14 X g(x) 1 2 5/14 15/28 3/28 Y h(y) 1 2 15/28 3/7 1/28 g(0) = P(X = 0) = f(0, 0) + f(0, 1) + f(0, 2) = 3/28 + 3/14 + 1/ 28 = 5/14 g(1) = P(X = 1) = f(1, 0) + f(1, 1) + f(1, 2) = 9/28 + 3/14 + 0 = 15/28 g(2) = P(X = 2) = f(2, 0) + f(2, 1) + f(2, 2) = 3/28 + 0 + 0 = 3/28

Example Suppose (X, Y) has the joint density function Find P{X  x, Y  y} when x>0, y>0 (b) Find marginal distribution of X. x≥ 0 and g(x)=0 else where.

4.Conditional probability distributions Clearly, for discrete cases, P[X = x | Y = y] = = For continuous cases, it can be shown that P[X  x | Y = y] = Hence it is natural to define the conditional probability distributions as follows.

Definition Definition 3.11 Let X and Y be two random variables, discrete or continuous. The conditional distribution of the random variable X, given Y = y, , h(y) > 0 Similarly, the conditional distribution of the random variable Y, given X = x, is , g(x) > 0 P(a < X < b | Y = y) = for discrete cases. P(a < X < b | Y = y) = for continuous cases.

Example 3.12, page 79 f(x|1)=f(x,1) /h(1)=(7/3) f(x,1) x=0,1,2 Referring to Example 3.8, table 3.1, find the conditional distribution of X , given that Y=1 and use it to determine P(X = 0|Y= 1) f(x|1)=f(x,1) /h(1)=(7/3) f(x,1) x=0,1,2 f(0|1)=(7/3)f(0,1)=(7/3)(3/14)=1/2 f(1|1)=(7/3)f(1,1)=(7/3)(3/14)=1/2 f(2|1)=(7/3)f(2,1)=(7/3)(0)=0 Y g(x) 1 2 X 3/28 9/28 3/14 1/28 h(y) 15/28 3/7 5/14 X 1 2 f(x|1) 1/2 P(X = 0|Y= 1) =f(0|1) = f(0, 1)/h(1) = (3/14)/(3/7) = ½

Example 3.14 Page81 Given the joint density function: Find g(x), h(y), f(x|y), and evaluate P(1/4<X<1/2|Y=1/3). Solution: By definition and Therefore,

5.Independence of two or more random variables Definition 3.12 Let X and Y be two random variables, discrete or continuous, with joint probability distribution f(x,y) and marginal distribution g(x) and h(y), respectively. The random variable X and Y are said to be statistically independent if and only if f(x, y) = g(x)h(y) for all (x, y) within their range.

Example for discrete: ∵ f(x, y) = g(x)h(y) for ∀ x, y 1/2 1 2 -1 ∵ f(x, y) = g(x)h(y) for ∀ x, y ∴ X and Y are said to be statistically independent

Example 1.Example 3.15, page 82. From table 3.1, page 75, 1 2 Y X 3/28 9/28 3/14 1/28 h(y) g(x) 15/28 3/7 5/14 f(0, 1) = 3/14, g(0) = 5/14, h(1) = 3/7, f(0,1)  g(0)h(1) X and Y are not statistically independent. 2.Example 3.14, page 81. g(x)h(y)=f(x,y) X and Y are statistically independent.

Read page 82 – 83 Joint probability distribution for more than two random variables. Different marginal distributions and conditional distributions Definition 3.13 Let X1, X2, …, Xn be n random variables, discrete or continuous, with joint distribution f(x1, x2, …, xn) and marginal distributions f1(x1), f2(x2) , …, fn(xn), respectively. The random variables X1, X2, …, Xn are said to be mutually statistically independent if and only if f(x1, x2, …, xn) = f1(x1) f2(x2)··· fn(xn) for all x1, x2, …, xn. Remark: Distribution of X = X1 + X2 + ··· + Xn

Example 3. 16, page 83. Suppose that the shelf life, in years, of a certain perishable food product packaged in cardboard containers is a random variable whose probability density function is given by Let X1, X2, and X3 represent the shelf lives for three of these containers selected independently, find P(X1< 2, 1 < X2 < 3, X3 >2). P(X1< 2, 1 < X2 < 3, X3 >2)= =

Example 4 The joint density function of X and Y is given by Determine the conditional densities. Solution: The marginal densities for the given X and Y are Hence, from the formulas for conditional densities, we have Because , it is clear that X and Y are not independent.