University:
Stony Brook University
Course:
AMS 310 | Survey of Probability and Statistics
Academic year:

2024
Views:

397

Pages:

16
Author:

study_genius

Example 1 Joint Discrete Distributions The random variables X and Y have the following joint probability distribution f(x,y). Y f(x,y) X 1 2 3 1 0.10 0.20 0.15 2 0.05 0.05 0.05 3 0.04 0.15 0.10 4 0.01 0.04 0.06 a) Find P(X + Y = 5). b) Find the marginal probability distributions f1(x) and f2(y). c) Find P(X < 2  Y=3). d) Are X and Y independent? e) Find E(Y) and Var(Y) Solution a) To find P(X+Y=5) we find the combinations of X and Y that sum to 5 and add the corresponding probabilities. So P(X + Y = 5) = f(1,4) + f(2,3) + f(3,2) = 0.01 + 0.15 + 0.05 = 0.21 b) X f1(X) 1 0.20 2 0.44 3 0.36 Y f2(Y) 1 0.45 2 0.15 3 0.29 4 0.11 Solution continued c) P(X < 2  Y=3) = 0.04 0.04+0.15+0.10 = 0.04 0.29 = 0.138 d) X and Y are not independent. To see why notice that f(1,1) = 0.10 (from the table) but f1(1) f2(1) = (0.20)(0.45) = 0.09 e) E(Y) = 1(0.45) + 2(0.15) + 3(0.29) + 4(0.11) = 2.06 Y2 1 4 9 16 f2(y) 0.45 0.15 0.29 0.11 E(Y2) = 1(0.45) + 4(0.15) + 9(0.29) + 16(0.11) = 5.42 Var(Y) = E(Y2)  E(Y)2 = 5.42 – 2.062 = 1.176 Facts About Joint Continuous Distributions Joint pdf: 𝑓(𝑥1 , 𝑥2 ) 𝑃 𝑋1 , 𝑋2 ∈ 𝐴 = 𝑓 𝑥1 , 𝑥2 𝑑𝑥1 𝑑𝑥2 , 𝐴 𝑏 𝑑 𝑃 𝑎 ≤ 𝑋1 ≤ 𝑏, 𝑐 ≤ 𝑋2 ≤ 𝑑 = ∞ 𝑓 𝑥1 , 𝑥2 𝑑𝑥2 𝑑𝑥1 , 𝑎 ∞ 𝑓 𝑥1 , 𝑥2 𝑑𝑥1 𝑑𝑥2 = 1 −∞ −∞ Marginal pdf: ∞ 𝑓1 𝑥1 = 𝑓(𝑥1 , 𝑥2 ) 𝑑𝑥2 −∞ ∞ 𝑓2 𝑥2 = 𝑓 𝑥1 , 𝑥2 𝑑𝑥1 −∞ 𝑓 𝑥1 , 𝑥2 ≥ 0 for all 𝑥1 , 𝑥2 𝑐 Conditional Distributions and Expectation Conditional probability distribution pmf (discrete) or pdf (continuous): 𝑓2 𝑥2 𝑥1 = 𝑓(𝑥1 ,𝑥2 ) , 𝑓1 (𝑥1 ) 𝑓1 𝑥1 > 0 Given constants 𝑎 & 𝑏, 𝐸 𝑎𝑋 + 𝑏 = 𝑎𝐸 𝑋 + 𝑏 𝑉𝑎𝑟 𝑎𝑋 + 𝑏 = 𝑎2 𝑉𝑎𝑟(𝑋). Example 2 Joint Continuous Variables Let the random variables X and Y have joint pdf as follows: f(x,y) = 15xy2, 0 < y < x < 1 a) Find the marginal pdf f1(x) of X b) Find the conditional pdf f2(y  x) Example 2 Joint Continuous Variables Solution a) ∞ 𝑓1 𝑥 = 𝑥 15𝑥𝑦 2 𝑑𝑦 = 5𝑥𝑦 3 𝑓 𝑥, 𝑦 𝑑𝑦 = −∞ 0 = 5𝑥 4 , 0 < 𝑥 < 1 b) 𝑓(𝑥, 𝑦) 15𝑥𝑦 2 3𝑦 2 𝑓2 𝑦 𝑥 = = = 3 ,0 < 𝑦 < 𝑥 4 𝑓1 𝑥 5𝑥 𝑥 𝑥 𝑦=0 Example 2 Joint Continuous Variables (continued) Let the random variables X and Y have joint pdf as follows: f(x,y) = 15xy2, 0 < y < x < 1 c) Find P(Y > 1  3 1 3 X = x) for any < x < 1 d) Determine if X and Y are independent Example 2 Joint Continuous Variables Solution c) ∞ 𝑥 3𝑦 2 𝑃(𝑌 > 1/3|𝑋 = 𝑥) = 𝑓2 𝑦 𝑥 𝑑𝑦 = 𝑑𝑦 3 1/3 1/3 𝑥 3 𝑦 = 3 𝑥 𝑥 𝑦=1/3 From Part b 1 =1− , 𝑥 > 1/3 3 27𝑥 d) They are not independent because 𝑓2 𝑦 𝑥 includes 𝑥 from the answer to (b). Example 3 Joint Continuous Variables Let the joint pdf of X and Y be f(x,y) = 12e4x  3y, x > 0 and y > 0. a) Find the marginal pdf’s of X and Y. b) Determine if X and Y are independent? c) Find the conditional pdf f1(x  y). Solution a) ∞ 𝑓1 𝑥 = ∞ 12𝑒 −4𝑥−3𝑦 𝑑𝑦 𝑓 𝑥, 𝑦 𝑑𝑦 = −∞ = −4𝑒 −4𝑥−3𝑦 = −3𝑒 −4𝑥−3𝑦 0 ∞ 𝑦=0 = 4𝑒 −4𝑥 , 𝑥>0 ∞ 𝑓2 𝑦 = ∞ 12𝑒 −4𝑥−3𝑦 𝑑𝑥 𝑓 𝑥, 𝑦 𝑑𝑥 = −∞ 0 ∞ 𝑥=0 𝑦>0 b) 𝑋 and 𝑌 are independent because 𝑓 𝑥, 𝑦 = 12𝑒 −4𝑥−3𝑦 = 𝑓1 𝑥 𝑓2 𝑦 , c) By independence, 𝑓1 𝑥 𝑦 = 𝑓1 (𝑥) = 4𝑒 −4𝑥 , 𝑥 > 0, 𝑦 > 0. 𝑥 > 0. = 3𝑒 −3𝑦 , Example 3 Joint Continuous Variables continued Let the joint pdf of X and Y be f(x,y) = 12e4x  3y, x > 0 and y > 0. d) Find the marginal cdf’s of X and Y. e) Find P(1 < Y < 3) f) Find P(1 < Y < 3  X = 3). Solution d) By independence, 𝑥 𝐹1 𝑥 = 𝑥 4𝑒 −4𝑡 𝑑𝑡 𝑓1 𝑡 𝑑𝑡 = −∞ = 0 0, 𝑦 𝐹2 𝑦 = 𝑒 −4𝑡 𝑓2 𝑡 𝑑𝑡 = −∞ 0 0, 0 = 1 − 𝑒 −4𝑥 , 𝑥>0 𝑥≤0 𝑦 3𝑒 −3𝑡 𝑑𝑡 𝑥 = 𝑒 −3𝑡 𝑦 0 = 1 − 𝑒 −3𝑦 , 𝑦>0 𝑦≤0 e) 𝑃 1 < 𝑌 < 3 = 𝐹2 3 − 𝐹2 1 = 1 − 𝑒 −9 − 1 − 𝑒 −3 = 𝑒 −3 − 𝑒 −9 f) By independence, 𝑃 1 < 𝑌 < 3|𝑋 = 3 = 𝑃 1 < 𝑌 < 3 = 𝑒 −3 − 𝑒 −9 . Example 4 Joint Continuous Variables 1 Suppose the random variables X and Y have joint pdf f(x,y) = , 2 0<𝑦<𝑥<2 a) Find the marginal pdf of X and marginal pdf of Y b) Find the conditional pdf of Y given X = x. c) Find the conditional pdf of X given Y = y. d) Find E(X) and E(Y) Solution a) ∞ 𝑓1 𝑥 = 𝑥 𝑓 𝑥, 𝑦 𝑑𝑦 = −∞ 0 ∞ 𝑓2 𝑦 = 1 𝑦 𝑑𝑦 = 2 2 2 𝑓 𝑥, 𝑦 𝑑𝑥 = −∞ 𝑦 𝑥 0 1 𝑥 𝑑𝑥 = 2 2 𝑥 = , 2 2 𝑦 0<𝑥<2 2−𝑦 = , 2 0<𝑦<2 b) 𝑓(𝑥, 𝑦) 1/2 1 𝑓2 𝑦 𝑥 = = = , 𝑓1 𝑥 𝑥/2 𝑥 0<𝑦<𝑥<2 c) 𝑓(𝑥, 𝑦) 1/2 1 𝑓1 𝑥 𝑦 = = = , 𝑓2 𝑦 (2 − 𝑦)/2 2 − 𝑦 0<𝑦<𝑥<2 Solution continued d) ∞ 𝐸 𝑋 = 2 𝑥𝑓1 𝑥 𝑑𝑥 = −∞ ∞ 𝐸 𝑌 = 0 2 𝑥 𝑥 𝑑𝑥 = 2 6 2 𝑦𝑓2 𝑦 𝑑𝑦 = −∞ 4 2 =2− = 3 3 𝑦− 0 3 𝑦2 2 2 0 𝑑𝑦 = 4 = 3 𝑦2 2 − 2 3 𝑦 6 0