University:
Rice University
Course:
STAT 310 | Probability and Statistics
Academic year:

2023
Views:

27

Pages:

6
Author:

MathMonk

Questions and Answers Sheet 4 Random Variables Question #1 Suppose that X and Y are continuous random variables with joint pdf 𝑓(𝑥, 𝑦) = 𝑒 −(𝑥+𝑦) 0 < 𝑥 < ∞ and 0 < 𝑦 < ∞ and zero otherwise. Find 𝑃(𝑋 > 𝑌) Answer: Step 1 Introduction: The joint density function of two random variables X and Y is given below: e−( x + y ) ; 0  x  ,0  y   f ( x, y) = { } 0; elsewhere The marginal density function of X is, ∞ 𝑓(𝑥) = ∫0 𝑒 −(𝑥+𝑦) 𝑑𝑦 ∞ −𝑦 = 𝑒 −𝑥 ∫0 𝑒 𝑑𝑦 = −𝑒 −𝑥 [−𝑒 −𝑦 ]∞ 0 = −𝑒 −𝑥 [𝑒 −∞ − 𝑒 −0 ] = −𝑒 −𝑥 [0,1] = 𝑒 −𝑥 Step 2 The probability of 𝑃(𝑋 > 𝑌) is obtained as 0.5 from the calculation given below: 𝑃(𝑋 ≥ 𝑌) = 𝑃(𝑌 ≤ 𝑋) ∞ 𝑥 = ∫0 ∫0 𝑒 −(𝑥+𝑦) 𝑑𝑥𝑑𝑦 ∞ 𝑥 = ∫0 ∫0 𝑒 −𝑥 𝑒 −𝑦 𝑑𝑥𝑑𝑦 ∞ 𝑥 = ∫0 𝑒 −𝑥 𝑑𝑥 ∫0 𝑒 −𝑦 𝑑𝑦 ∞ = ∫0 𝑒 −𝑥 𝑑𝑥[−𝑒 −𝑦 ]0𝑥 ∞ = − ∫0 𝑒 −𝑥 𝑑𝑥[𝑒 −𝑥 − 𝑒 −0 ] ∞ = − ∫0 𝑒 −𝑥 𝑑𝑥[𝑒 −𝑥 − 1] ∞ = ∫0 𝑒 −𝑥 𝑑𝑥[1 − 𝑒 −𝑥 ] ∞ = ∫0 𝑒 −𝑥 − 𝑒 −2𝑥 𝑑𝑥 = [𝑒 −𝑥 ]∞ 0 +[ = −[𝑒 =1− −∞ 𝑒 −2𝑥 −𝑒 2 −0 ] ∞ ] 0 𝑒 −2(∞) +[ 2 − 𝑒 −2(0) 2 ] 1 2 =0.5 Thus, the probability of 𝑃(𝑋 > 𝑌) is 0.5. Question #2 Suppose that X and Y are continuous random variables with joint pdf 𝑓(𝑥, 𝑦) = 𝑒 −(𝑥+𝑦) 0 < 𝑥 < ∞ and 0 < 𝑦 < ∞ and zero otherwise. Find 𝑃(𝑋 + 𝑌 > 3) Answer: Step 1 Introduction: The joint density function of two random variables X and Y is given below: e−( x + y ) ; 0  x  ,0  y   f ( x, y) = { } 0; elsewhere ∞ The marginal density function of X is, 𝑓(𝑥) = ∫0 𝑒 −(𝑥+𝑦) 𝑑𝑦 ∞ −𝑦 = 𝑒 −𝑥 ∫0 𝑒 𝑑𝑦 = −𝑒 −𝑥 [−𝑒 −𝑦 ]∞ 0 = −𝑒 −𝑥 [𝑒 −∞ − 𝑒 −0 ] = −𝑒 −𝑥 [0,1] = 𝑒 −𝑥 Step 2 The probability of 𝑃(𝑋 + 𝑌 > 3) is obtained as 0.1991 from the calculation given below: 𝑃(𝑋 + 𝑌 > 3) = 1 − 𝑃(𝑋 + 𝑌 ≤ 3) 3 3−𝑥 = 1 − ∫0 ∫0 =1− =1− =1− =1− =1− =1− 𝑒 −(𝑥+𝑦) 𝑑𝑥𝑑𝑦 3 3−𝑥 ∫0 ∫0 𝑒 −𝑥 𝑒 −𝑦 𝑑𝑥𝑑𝑦 3 3−𝑥 ∫0 𝑒 −𝑥 𝑑𝑥 ∫0 𝑒 −𝑦 𝑑𝑦 3 ∫0 𝑒 −𝑥 𝑑𝑥[−𝑒 −𝑦 ]3𝑥 0 3 −𝑥 −(3−𝑥) (− ∫0 𝑒 𝑑𝑥[𝑒 − 𝑒 −0 ]) 3 (− ∫0 𝑒 −𝑥 𝑑𝑥[𝑒 −3+𝑥 − 1]) 3 ∫0 𝑒 −𝑥 𝑑𝑥[1 − 𝑒 −3+𝑥 ] 3 ∫0 𝑒 −𝑥 − 𝑒 −3 𝑒 𝑥 𝑒 −𝑥 𝑑𝑥 ([−𝑒 −𝑥 ]30 − 𝑒 −3 [𝑥]30 ) −0 ] −3 [3 (−[𝑒 −3 =1− =1− =1− −𝑒 −𝑒 − 0]) = 1 − ([1 − 𝑒 −3 ] − 3𝑒 −3 ) = 1 − 1 + 𝑒 −3 + 3𝑒 −3 = 4𝑒 −3 =0.1991 Thus, the probability of 𝑃(𝑋 + 𝑌 > 3) is 0.1991. Question #3 A continuous random variable X that can has a density function given by .25 if 0  x  4 f ( x) =  0 otherwise Find 𝑃(𝑥 = .5) Answer: Step 1 From the given information, The density function is, f(x)=0.25, if 0