Definition: Suppose that two random variables, either continuous or discrete, X and Y have joint density


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1 HW MATH 461/561 Lecture Notes 15 1 Definition: Suppose that two random variables, either continuous or discrete, X and Y have joint density and marginal densities f(x, y), (x, y) Λ X,Y f X (x), x Λ X, f Y (y), y Λ Y. The conditional probability density that Y takes on the value y given that X is equal to x is denoted by f Y x (y) and is given by f Y x (y) = f(x, y) f X (x) (x, y) Λ X,Y, x Λ X and the conditional probability density that X takes on the value x given that Y is equal to y is denoted by f X y (x) and is given by f X y (x) = f(x, y) f Y (y) (x, y) Λ X,Y, y Λ Y Example: In a box, there are 4 white balls, 3 black balls, 2 red balls. We randomly select 2 balls without replacement. Define X = {number of white balls} Y = {number of black balls} (1) Find density function f(x, y) and marginal densities f X (x) and f Y (y). (2) Find the conditional probability P(Y 1 X = ). (3) Find the conditional probability P(X = 1 Y = 2).
2 HW MATH 461/561 Lecture Notes 15 2 Solution: (1) We can find P(X =, Y = ) = C2 2 C 9 2 = 1, P(X =, Y = 1) = C3 1C 2 1 C 9 2 = 6 P(X =, Y = 2) = C3 2 C 9 2 P(X = 1, Y = 1) = C4 1C 3 1 C 9 2 = 3, P(X = 1, Y = ) = C4 1C 2 1 C 9 2 = 12, P(X = 2, Y = ) = C4 2 C 9 2 Dist. and marginal dists table Y \X 1 2 f Y (y) = 8 = 6 (2) Thus, f X (x) 1 f Y (1) = f Y () = 2 6 f(, 1) f X () = 6/ 1/ = 3/5 f(, ) f X () = 1/ 1/ = 1/1 P(Y 1 X = ) = f Y () + f Y (1) = 3/5 + 1/1 = 7/1. P(X = 1 Y = 2) = f X 2 (1) = f(1, 2) f Y (2) = 3/ =
3 HW MATH 461/561 Lecture Notes 15 3 Example: Given the joint pdf Find (a) P(Y < 1 X = 1/2) (b) P(Y < 1 X < 1) Solution: (a) Since Thus, (b) Since f(x, y) = 2e (x+y) < x < y, f X (x) = x 2e x e y dy = 2e 2x x > f Y 1(y) = f(1 2, y) 2 f X ( 1 2 ) = 2e (y+1 2 ) = e y+1 2, 2e 2(1 2 ) P(Y < 1 X = 1/2) = P(X < 1) = < y e 1 2 y dy = ee y = 1 1 e 2e 2x dx = 1 e 2 = =.865 P(X < 1, Y < 1) = dy[ y 2e (x+y) dx] = dy{2e y [ e x y ]} = {2e y 2e 2y }dy = 1 2/e + e 2 = P(Y < 1 X < 1) = = Intuitively, a mean or an expected value of a rv X is X s central tendency or average. Definition:
4 HW MATH 461/561 Lecture Notes 15 4 Given a discrete random variable X with range space Λ = {x 1,, x n } and the distribution table. The distribution table of r.v X X x 1 x 2 x n p(x) p(x 1 ) p(x 2 ) p(x n ) The mean or expected value of X is defined by µ = E(X) = n x i p(x i ) Example: In a class A, there are total 2 students. The students scores of the final exam are divided into five categories. Let X be a rv which associates each student s score of the final exam with a specified category. X has following distribution table: i=1 X p(x) 3/2 6/2 3/2 3/2 5/2 Find the average category of the students scores of the final exam. Solution: E(X) = 1 3/ / / / /2 = 3.5 Example: A company has five applicants for two positions: two women and three men. Suppose that the five applicants are equally qualified and that no preference is given for choosing either gender. Let X equal the number of women chosen to fill the two positions.
5 HW MATH 461/561 Lecture Notes 15 5 (a) Find the probability distribution of X; (b) Find the average number of women who will be hired. Solution: (a). The sample space Ω = {ω 1 = {M 1 M 2 }, ω 2 = {M 1 M 3 }, ω 3 = {M 2 M 3 }, ω 4 = {W 1 M 1 }, ω 5 = {W 1 M 2 }, ω 6 = {W 1 M 3 }, ω 7 = {W 2 M 1 }, ω 8 = {W 2 M 2 }, ω 9 = {W 2 M 3 }, ω 1 = {W 1 W 2 }} and the range space Λ := {x 1 =, x 2 = 1, x 3 = 2}. p(x 1 ) = p() = P(X = ) = P({ω 1, ω 2, ω 3 }) = P({{M 1 M 2 }, {M 1 M 3 }, {M 2 M 3 }}) = 3 1, p(x 2 ) = p(1) = P(X = 1) = P({ω 4, ω 5, ω 6, ω 7, ω 8, ω 9 }) = P({{W 1 M 1 }, {W 1 M 2 }, {W 1 M 3 }, {W 2 M 1 }, {W 2 M 2 }, {W 2 M 3 }}) = 6 1, p(x 3 ) = p(2) = P(X = 2) = P({ω 1 = {W 1 W 2 }}) = 1 1.
6 HW MATH 461/561 Lecture Notes 15 6 Therefore, The distribution table of r.v X X 1 2 p(x) 3/1 6/1 1/1 (b) µ = E(X) = x i Λ x ip(x i ) = (3/1) + 1(6/1) + 2(1/1) = 8/1. Definition: Given a continuous random variable X with range space Λ and the density function: f(x), x Λ. The mean or expected value of X is defined by µ = E(X) = xf(x)dx Example: Let X be a rv which represents the lifetime of a bulb. X has pdf f(x) = λe λx, x >, where λ > is a constant. Find the average lifetime of a bulb. Solution: Using integration by part, we can get µ = E(X) = λxe λx dx Λ = 1/λ Theorem: Let X 1,, X n be rv s for which E(X i ) <, i = 1, 2,, n and a 1,, a n be constants. Then, E(a 1 X a n X n ) = a 1 E(X 1 ) + + a n E(X n ) Theorem: (1) Let X be a discrete rv with range space Λ = {x 1,, x n } and the distribution table.
7 HW MATH 461/561 Lecture Notes 15 7 The distribution table of r.v X X x 1 x 2 x n p(x) p(x 1 ) p(x 2 ) p(x n ) Let g be a function such that g(x i ) p(x i ) < x i Λ The mean of g(x) is given by Eg(X) = n g(x i )p(x i ) i=1 (2) Let Y be a continuous rv with range space Λ and the density function: h(y), y Λ. Let g be a function such that g(y) h(y)dy < y Λ The mean of g(y ) is given by Eg(Y ) = Λ g(y)h(y)dy
8 HW MATH 461/561 Lecture Notes 15 8 Theorem: If X and Y are two independent rv s and E(X) <,E(Y ) <, then, we have E(XY ) = E(X)E(Y ). Example: Let X, Y be two independent rv s with density functions : f X (x) = 2x, < x < 1 and f Y (y) = 1/2 < y < 2 Find E(2X 3Y ) 2 = E[(2X 3Y ) 2 ]. Solution: Since (2X 3Y ) 2 = 4X 2 12XY + 9Y 2, we have Since Thus, E(2X 3Y ) 2 = 4E(X 2 ) 12E(X)E(Y ) + 9E(Y 2 ). E(X) = E(Y ) = E(X 2 ) = E(Y 2 ) 2 2 x2xdx = 2/3 y(1/2)dy = 1 x 2 2xdx = 1/2 y 2 (1/2)dy = 4/3 E(2X 3Y ) 2 = 4(1/2) 12(2/3)1 + 9(4/3) = 6 Definition: Given a discrete random variable X with range space Λ = {x 1,, x n } and distribution table as follows:
9 HW MATH 461/561 Lecture Notes 15 9 The distribution table of r.v X X x 1 x 2 x n p(x) p(x 1 ) p(x 2 ) p(x n ) Let µ = E(X) = n i=1 x ip(x i ) be its mean. Then, the variance of X is denoted by σ 2 or Var(X), is defined by Var(X) = σ 2 = E(X µ) 2 = x i Λ(x i µ) 2 p(x i ) The standard deviation of X is defined as σ = σ 2. Remark: The variance of a rv X is a measurement to indicate the dispersion of the distribution of X. Example: A company has five applicants for two positions: two women and three men. Suppose that the five applicants are equally qualified and that no preference is given for choosing either gender. Let X equal the number of women chosen to fill the two positions. Find the mean and standard deviation of X. Solution: The sample space Ω = {ω 1 = {M 1 M 2 }, ω 2 = {M 1 M 3 }, ω 3 = {M 2 M 3 }, ω 4 = {W 1 M 1 }, ω 5 = {W 1 M 2 }, ω 6 = {W 1 M 3 }, ω 7 = {W 2 M 1 }, ω 8 = {W 2 M 2 }, ω 9 = {W 2 M 3 }, ω 1 = {W 1 W 2 }} and the range space Λ := {x 1 =, x 2 = 1, x 3 = 2}.
10 HW MATH 461/561 Lecture Notes 15 1 p(x 1 ) = p() = P(X = ) = P({ω 1, ω 2, ω 3 }) = P({{M 1 M 2 }, {M 1 M 3 }, {M 2 M 3 }}) = 3 1, p(x 2 ) = p(1) = P(X = 1) = P({ω 4, ω 5, ω 6, ω 7, ω 8, ω 9 }) = P({{W 1 M 1 }, {W 1 M 2 }, {W 1 M 3 }, {W 2 M 1 }, {W 2 M 2 }, {W 2 M 3 }}) = 6 1, p(x 3 ) = p(2) = P(X = 2) = P({ω 1 = {W 1 W 2 }}) = 1 1. Therefore, The distribution table of r.v X x 1 2 p(x) 3/1 6/1 1/1 The mean is µ = E(X) = x i Λ x ip(x i ) = (3/1) + 1(6/1) + 2(1/1) = 8/1. The variance of X is σ 2 = E[(X µ) 2 ] = x i Λ (x i µ) 2 p(x i ) = ( 8/1) 2 (3/1) + (1 8/1) 2 (6/1) + (2 8/1) 2 (1/1) =. and σ = σ 2 =.6
11 HW MATH 461/561 Lecture Notes Definition: Given a continuous random variable X with density function f(x), x Λ. Let µ = E(X) = Λ xf(x)dx be its mean. Then, the variance of X is denoted by σ 2 or Var(X), is defined by Var(X) = σ 2 = E(X µ) 2 = (x µ) 2 f(x)dx x Λ The standard deviation of X is defined as σ = σ 2. Theorem: Let X be a rv, discrete or continuous, have mean µ and finite E(X 2 ). Then, Var(X) = σ 2 = E(X µ) 2 = E(X 2 ) µ 2 Theorem: Let X be a rv, discrete or continuous, and a, b be two constants. Then, Var(aX + b) = a 2 Var(X) Theorem: Let X 1,, X n be independent rv s, discrete or continuous. Then, Var(X X n ) = Var(X 1 ) + + Var(X n ) Example: Let X,Y be two independent rv s with following pdf s: Find Var(7X + 8Y ). Solution: f X (x) = 2x, < x < 1 f Y (y) = 6y 5, < y < 1 E(X) = x2xdx = 2/3x 3 1 = 2/3
12 HW MATH 461/561 Lecture Notes E(Y ) = y6y 5 dy = 6/7x 7 1 = 6/7 Var(7X + 8Y ) = Var(7X) + Var(8Y ) = 7 2 Var(X) Var(Y ) Since Thus, = 49(EX 2 4/9) + 64(EY 2 /49) EX 2 = EY 2 = x 2 2xdx = 1/2 x 2 6x 5 dx = 3/4 Var(7X + 8Y ) = 49(EX 2 4/9) + 64(EY 2 /49) = 49(1/2 4/9) + 64(3/4 /49) = 3265/882 Theorem: Let X be a rv with binomial distribution b(n, p). Then, µ = E(X) = np and Var(X) = np(1 p). Example: Using Var(X X n ) = Var(X 1 ) + + Var(X n ) if X 1,, X n are independent, prove above theorem. Example: Let X 1,, X 1 be independent rv s and each with binomial distribution b(1, 1/3). Find and E{ i=1 X i } Var( i=1 X i)
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