Questions and Answers #2 Random Variables

Questions and Answers Sheet 2 Random Variables Question #1 Random variables X and Y have joint PDF 12e− (3 x + 4 y ) , x  0, y  0 f X ,Y ( x, y ) =  0, otherwise Find 𝑃[𝑋 + 𝑌 ≤ 1] Answer: The value of 𝑃[𝑋 + 𝑌 ≤ 1] is obtained as given below: 𝑓(𝑥, 𝑦) = 12𝑒 −(3𝑥+4𝑦) ; 𝑥 ≥ 0 and 𝑦 ≥ 0 𝑓(𝑥) = ∫𝑓(𝑥, 𝑦)𝑑𝑦 = 𝑦 ∞ ∫0 12𝑒 −(3𝑥+4𝑦) 𝑑𝑦 ∞ 𝑒 −4𝑦 −3𝑥 = 12𝑒 = 12𝑒 [− −3𝑥 4 1 × ] 0 4 𝑓(𝑥) = 3𝑒 −3𝑥 𝑃(𝑋 + 𝑌 ≤ 1) = 𝑃(𝑋 ≤ 1 − 𝑌) 1−𝑦 = ∫0 = 𝑓(𝑥)𝑑𝑥 1−𝑦 ∫0 3𝑒 −3𝑥 𝑑𝑥 1−𝑦 𝑒 −3𝑥 = 3 [− ] 3 0 −3(1−𝑦) = −[𝑒 − 1] −3(1−𝑦) = [1 − 𝑒 ] Question #2 Random variables X and Y have joint PDF 12e− (3 x + 4 y ) , x  0, y  0 f X ,Y ( x, y ) =  0, otherwise Find 𝑃[𝑚𝑖𝑛(𝑋, 𝑌) ≥ 2] Answer: The value of 𝑃[𝑚𝑖𝑛(𝑋, 𝑌) ≥ 2] is obtained as given below: 𝑃[𝑚𝑖𝑛(𝑋, 𝑌) ≥ 2] = 𝑃[2 ≤ 𝑚𝑖𝑛(𝑋, 𝑌)] = 𝑃(2 ≤ 𝑋, 2 ≤ 𝑌) = 𝑃(𝑋 ≥ 2, 𝑌 ≥ 2) = [1 − 𝑃(𝑋 < 2)][1 − 𝑃(𝑌 < 2)] 2 𝑃(𝑋 < 2) = ∫0 3𝑒 −3𝑥 𝑑𝑥 = 3 [− 𝑒 −3𝑥 3 2 ] 0 = 1 − 𝑒 −6 =0.9975 1 − 𝑃(𝑋 < 2) = 0.0025 𝑓(𝑦) = ∫𝑥 𝑓(𝑥, 𝑦)𝑑𝑥 ∞ = ∫0 12𝑒 −(3𝑥+4𝑦) 𝑑𝑥 = 12𝑒 −4𝑦 [ −𝑒 −3𝑥 3 𝑓(𝑦) = 4𝑒 −4𝑦 ∞ ] 0 2 𝑃(𝑌 < 2) = ∫0 4𝑒 −4𝑦 𝑑𝑦 = 4 [− 𝑒 −4𝑦 4 2 ] 0 = 1 − 𝑒 −8 =0.9997 1 − 𝑃(𝑌 < 2) = 0.0003 𝑃[𝑚𝑖𝑛(𝑋, 𝑌) ≥ 2] = [1 − 𝑃(𝑋 < 2)][1 − 𝑃(𝑦 < 2)] = 0.0025 × 0.0003 = 0.0000075 Question #3 Random variables X and Y have joint PDF 12e− (3 x + 4 y ) , x  0, y  0 f X ,Y ( x, y ) =  0, otherwise Find 𝑃[𝑚𝑎𝑥(𝑋, 𝑌) ≤ 0.5] Answer: The value of 𝑃[𝑚𝑎𝑥(𝑋, 𝑌) ≤ 0.5] is obtained as given below: 𝑃[𝑚𝑎𝑥(𝑋, 𝑌) ≤ 0.5] = 𝑃(𝑋 ≤ 0.5, 𝑌 ≤ 0.5) = 𝑃(𝑋 ≤ 0.5) × 𝑃(𝑌 ≤ 0.5) 0.5 𝑃(𝑋 < 0.5) = ∫0 3𝑒 −3𝑥 𝑑𝑥 = 3 [− 𝑒 −3𝑥 3 0.5 ] 0 = 1 − 𝑒 −1.5 =0.7769 0.5 𝑃(𝑌 < 0.5) = ∫0 4𝑒 −4𝑦 𝑑𝑦 = 4 [− 𝑒 −4𝑦 4 0.5 ] 0 = 1 − 𝑒 −2 =0.8647 𝑃[𝑚𝑎𝑥(𝑋, 𝑌) ≤ 0.5] = 𝑃(𝑋 ≤ 0.5) × 𝑃(𝑌 ≤ 0.5) = 07769 × 0.8647 =0.6718 Question #4 Let 𝑋1 … . , 𝑋𝑛 and 𝑌1 , … , 𝑌𝑚 be two sets of random variables. Let 𝑎𝑖 , 𝑏𝑗 be arbitrary constant. Show that 𝑛 𝑚 𝐶𝑜𝑣(∑𝑛𝑖=1 𝑎𝑖 𝑋𝑖 , ∑𝑚 𝑗=1 𝑏𝑗 𝑌𝑗 ) = ∑𝑖=1 ∑𝑗=1 𝑎𝑖 𝑏𝑗 𝐶𝑜𝑣(𝑋𝑖 , 𝑌𝑗 ) Answer: Step 1 It is given that 𝑋1 , … , 𝑋𝑛 and 𝑌1 , … , 𝑌𝑚 be the two sets of random variables. Let 𝑎𝑖 , 𝑏𝑗 be the arbitrary constant. It is asked to show, 𝑛 𝑚 𝐶𝑜𝑣(∑𝑛𝑖=1 𝑎𝑖 𝑋𝑖 , ∑𝑚 𝑗=1 𝑏𝑗 𝑌𝑗 ) = ∑𝑖=1 ∑𝑗=1 𝑎𝑖 𝑏𝑗 𝐶𝑜𝑣(𝑋𝑖 , 𝑌𝑗 ) Step 2 Proof : 𝐶𝑜𝑣(∑𝑛𝑖=1 𝑎𝑖 𝑋𝑖 , ∑𝑚 𝑗=1 𝑏𝑗 𝑌𝑗 ) 𝑛 𝑚 = 𝐸(∑𝑖=1 𝑎𝑖 𝑋𝑖 , ∑𝑗=1 𝑏𝑗 𝑌𝑗 ) − 𝐸(∑𝑛𝑖=1 𝑎𝑖 𝑋𝑖 )𝐸(∑𝑚 𝑗=1 𝑏𝑗 𝑌𝑗 ) 𝑛 𝑚 = 𝐸 ∑𝑖=1 ∑𝑗=1 𝑎𝑖 𝑏𝑗 [𝐸(𝑋𝑖𝑌𝑗) = 𝐸(𝑋𝑖)𝐸(𝑌𝑗)] = 𝐸 ∑𝑛𝑖=1 ∑𝑚 𝑗=1 𝑎𝑖 𝑏𝑗 𝐶𝑜𝑥(𝑋𝑖, 𝑌𝑗) .... by covariance property 𝐶𝑜𝑣(𝑋, 𝑌) = 𝐸(𝑋𝑌) − 𝐸(𝑋)𝐸(𝑌). Hence, It is proved. Question #5 Let the continuous random variables X and Y have joint pdf 𝑓(𝑥, 𝑦) = 𝑒 −𝑥−𝑦 , 0 < 𝑥 < ∞, 0 < 𝑦 < ∞, then 𝑦 𝑓( ) = 𝑥 a) 𝑒 −𝑦 b) 𝑒 −𝑥 c) 𝑒𝑦 𝑒 −𝑥 Answer: Step 1 The joint pdf of the continuous random variables X and Y is given by: 𝑓(𝑥, 𝑦) = 𝑒 −𝑥−𝑦 , 0 < 𝑥 < ∞, 0 < 𝑦 < ∞. 𝑦 Here we have to find the value of 𝑓 ( ). 𝑥 𝑦 As we know that 𝑓 ( ) = And 𝑓𝑋 (𝑥) = 𝑓(𝑥,𝑦) 𝑥 𝑓𝑋 (𝑥) 𝑏 ∫𝑎 𝑓(𝑥, 𝑦)𝑑𝑦 Step 2 Let us finding the required value: 𝑦 𝑓(𝑥, 𝑦) 𝑓( ) = 𝑏 𝑥 ∫ 𝑓(𝑥, 𝑦)𝑑𝑦 = 𝑎 𝑒 −𝑥−𝑦 ∞ ∫0 𝑒 −𝑥−𝑦 𝑑𝑦 𝑒 −𝑥−𝑦 = [−𝑒 −𝑥−𝑦 ]∞ 𝑒 −𝑥−𝑦 = [−𝑒 −∞ = 0 𝑒 −𝑥 ] 𝑒 −𝑥−𝑦 𝑒 −𝑥 = 𝑒 −𝑥−𝑦+𝑥 = 𝑒 −𝑦 𝑦 Thus, the required value of 𝑓 ( ) = 𝑒 −𝑦 . 𝑥 Hence, option (a) is correct. Question #6 Find the expected values and variances of the exponential random variables with the density functions given below 3𝑒 −3𝑥 Answer: Step 1 It is given that x is a exponential random variables with pdf, 𝑓(𝑥) = 3𝑒 −3𝑥 Step 2 Pdf of exponential distribution 𝑓(𝑥) = 0𝑒 −0𝑥 , 𝑥 ≥ 0 1 Mean of exponential distribution = 0 Variance of exponential distribution = 1 02 Given pdf, f(𝑥) = 3𝑒 −3𝑥 Comparing it with pdf of exponential distribution, 0 = 3 1 Mean of exponential distribution = 3 Variance of exponential distribution = 1 = 32 1 9 Question #7 Find the expected values and variances of the exponential random variables with the density functions given below 1 4 𝑥 𝑒 −4 Answer: Step 1 1 𝑥 it is given that x is a exponential random variables with pdf, 𝑓(𝑥) = 𝑒 −4 4 Step 2 Pdf of exponential distribution 𝑓(𝑥) = 0𝑒 −0𝑥 , 𝑥 ≥ 0 1 Mean of exponential distribution = 0 Variance of exponential distribution = 1 𝑥 1 02 Given pdf, 𝑓(𝑥) = 𝑒 −4 4 Comparing it with pdf of exponential distribution, 0 = Mean of exponential distribution = 1 1 4 1 4 =4 Variance of exponential distribution = 1 1 2 4 = 16 ( ) Question #8 Assume that X and Y are jointly continuous random variables with joint probability density function given by 1  (3x − xy + 4 y ) if 0  x  2 and 1  y  3 f ( x, y ) =  36 0 othrewise Find the marginal density functions for X and Y. Answer: The joint density function for X and Y is given below: 1  (3x − xy + 4 y ) if 0  x  2 and 1  y  3 f ( x, y ) =  36 0 othrewise The marginal density function of X, g(x) is obtained as given below: 3 1 𝑔(𝑥) = ∫1 (3𝑥 − 𝑥𝑦 + 4𝑦)𝑑𝑦 36 = = = = 1 36 1 36 2 36 1 18 𝑦2 (3𝑥𝑦 − 𝑥 2 +4 𝑦2 2 3 ) 1 (2𝑥 + 16) (𝑥 + 8) (𝑥 + 8) 1  ( x + 8), for 0  x  2 = 18 0, otherwise The marginal density function of y, h(y) is obtained as given below: 2 1 ℎ(𝑦) = ∫0 (3𝑥 − 𝑥𝑦 + 4𝑦)𝑑𝑥 36 = = 1 36 6 36 1 (3 𝑥2 2 𝑥2 − 2 2 𝑦 + 4𝑥𝑦) 0 (1 + 𝑦) = (𝑦 + 1) 6 1  ( y + 1), for 1  y  3 = 6 0, otherwise Question #9 Assume that X and Y are jointly continuous random variables with joint probability density function given by 1  (3x − xy + 4 y ) if 0  x  2 and 1  y  3 f ( x, y ) =  36 0 othrewise Find E[X] and E[Y]. Answer: Step 1 𝐸[𝑋] = 𝜇𝑋 = ∫𝑥 𝑥𝑔(𝑥)𝑑𝑥 2 1 = ∫0 𝑥 ( (𝑥 + 8)) 𝑑𝑥 18 = = = = = = 1 2 2 ∫ (𝑥 18 0 1 𝑥3 [ 18 3 1 8 + 8𝑥)𝑑𝑥 +8 𝑥2 2 ] 2 0 [ + 8(2)] 18 3 8 7 [ ] 18 3 56 54 28 27 Step 2 E[Y]: 𝜇𝑌 = ∫𝑦 𝑦ℎ(𝑦)𝑑𝑦 1 1 = 𝑦 ( (𝑦 + 1)) 𝑑𝑦 6 6 1 52+24 = ( 6 6 ) 1 16+24 = [ = = 6 76 6 ] 36 19 9 Question #10 Assume that X and Y are jointly continuous random variables with joint probability density function given by 1  (3x − xy + 4 y ) if 0  x  2 and 1  y  3 f ( x, y ) =  36 0 othrewise Find Cov(X,Y). Answer: Step 1 Covariance of two random variables X and Y with means E[X] and E[Y] is given below: 𝜎𝑋𝑌 = 𝐸(𝑋𝑌) − 𝜇𝑋 𝜇𝑌 = ∑𝑥 ∑𝑦 𝑥𝑦𝑓(𝑥, 𝑦) − {∑𝑥 𝑥𝑓(𝑥, 𝑦)} × {∑𝑦 𝑦𝑓(𝑥, 𝑦)} (∵ For discrete) = ∫𝑥 ∫𝑣 𝑥𝑦𝑓(𝑥, 𝑦)𝑑𝑦𝑑𝑥 − {∫𝑥 𝑥𝑔(𝑥)𝑑𝑥} × {∫𝑣 𝑦ℎ(𝑦)𝑑𝑦} (∵ For\ continuous) The covariance of the random variable X and Y is obtained as shown below: For E[XY], 𝐸(𝑋𝑌) = ∫𝑥 ∫𝑦 𝑥𝑦𝑓(𝑥, 𝑦)𝑑𝑦𝑑𝑥 2 3 1 = ∫0 ∫1 = = = 𝑥𝑦(3𝑥 − 𝑥𝑦 + 4𝑦)𝑑𝑦𝑑𝑥 36 2 3 ∫ ∫ (3𝑥 2 𝑦 − 36 0 1 2 5𝑥 2 +52𝑥 ∫0 ( 54 ) 𝑑𝑥 176 1 𝑥 2 𝑦 2 + 4𝑥𝑦 2 )𝑑𝑦𝑑𝑥 81 Step 2 Therefore, the covariance of X and Y will be obtained as given below: 𝜎𝑋𝑌 = 𝐸(𝑋𝑌) − 𝜇𝑋 𝜇𝑌 = = 176 81 1 81 28 19 27 9 − ( )( ) (176 − 177.33) = −0.01646