Questions and Answers #1 Random Variables

Questions and Answers Sheet 1 Random Variables Question #1 The joint density function of two continuous random variables X and Y is: −y  cx 2e 3 f ( x, y) = f ( x, y) =   0 1  x  6, 2  y  4 = 0 otherwise Determine the value of the constant c. Answer: From the given information, the joint density function of two continuous random variables X and Y is, −𝑦 𝑓(𝑥, 𝑦) = 𝑐𝑥 2 𝑒 3 , 1 < 𝑥 < 6,2 < 𝑦 < 4 The value of constant c is obtained below: ∫𝑥 ∫𝑦 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑦 = 1 6 4 −𝑦 ∫1 ∫2 𝑐𝑥 2 𝑒 3 𝑑𝑥𝑑𝑦 = 1 4 𝑐 ∫2 𝑒 6 −𝑦 𝑥3 3 [ ] 𝑑𝑦 = 1 3 1 4 −𝑦 𝑐 ∫2 𝑒 3 𝑑𝑦 3 215 215 3 =1 𝑐 ∗ 0.7495 = 1 𝑐 = 0.0186 Question #2 The joint density function of two continuous random variables X and Y is: −y  cx 2e 3 f ( x, y) = f ( x, y) =   0 1  x  6, 2  y  4 = 0 otherwise Find the marginal distribution functions of X. Answer: Marginal distribution function of X: 𝑓(𝑥) = ∫𝑦 𝑓(𝑥, 𝑦)𝑑𝑦 −𝑦 4 = ∫2 0.0186𝑥 2 𝑒 3 𝑑𝑦 4 −𝑦 3 = 0.0186𝑥 2 ∫2 𝑒 𝑑𝑦 = 0.0139𝑥 2 , 1 < 𝑥 < 6 Question #3 The joint density function of two continuous random variables X and Y is: −y  cx 2e 3 f ( x, y) = f ( x, y) =   0 1  x  6, 2  y  4 = 0 otherwise Draw the integration boundaries and write the integration only for 𝑃(𝑋 + 𝑌 ≤ 6) Answer: Integration boundaries for 𝑋 + 𝑌 ≤ 6 ∶ X: 1 to 6-y Y: 2 to 4 Integration: 4 6−𝑦 −𝑦 𝑃(𝑋 + 𝑌 ≤ 6) = ∫𝑦=2 ∫𝑥=1 0.0186𝑥 2 𝑒 3 𝑑𝑥𝑑𝑦 4 = 0.0186 ∫𝑦=2 𝑒 4 = 0.0186 ∫2 𝑒 6−𝑦 −𝑦 𝑥3 3 [ ] 3 1 −𝑦 (6−𝑦)3 3 [ 4 −𝑦 0.0062 ∫2 𝑒 3 [(6 3 𝑑𝑦 1 − ] 𝑑𝑦 3 3 = − 𝑦) − 1]𝑑𝑦 = 0.0062 ∗ 24.0489 = 0.1491 Question #4 Suppose X and Y are independent continuous random variables uniformly distributed on the intervals 1 0 ≤ 𝑥 ≤ 2 and 0 ≤ 𝑦 ≤ 4 , respectively. Compute the variance of √2𝑋 − 𝑌. Hint: First find the variance 2 of X and the variance of Y. Answer: Step 1 Given information X and Y independent continuous random variables follows uniform distribution 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 4 Step 2 𝑉𝑎𝑟(𝑋) = 𝑉𝑎𝑟(𝑌) = (2−0)2 12 (4−0)2 12 = = 4 12 16 12 = = 1 3 4 3 Variance of 𝑎𝑥 + 𝑏𝑦 = 𝑎2 𝑣𝑎𝑟(𝑥) + 𝑏 2 𝑣𝑎𝑟(𝑦) 1 2 1 2 1 1 4 2 3 4 3 𝑉𝑎𝑟 (√2𝑋 − 𝑌) = (√2) 𝑉𝑎𝑟(𝑋) + (− ) 𝑉𝑎𝑟(𝑌) = 2 × + × = 1 2 Question #5 If two random variables X and Y are independent with marginal pdfs 𝑓𝑥(𝑥) = 2𝑥, 0 ≤ 𝑥 ≤ 1 and 𝑓𝑦(𝑦) = 1,0 ≤ 𝑦 ≤ 1 Calculate 𝑃(𝑌|𝑋 > 2) Answer: Step 1 From the provided information, the marginal probability distribution of both the random variables X and Y are given. Since both the variables are independent, the condition probability distribution of Y given X is equal to the marginal probability distribution of Y that is, 𝑃(𝑌|𝑋) = 𝑃(𝑌) = 𝑓𝑦 (𝑦), 0 ≤ 𝑦 ≤ 1 Step 2 The required probability can be calculated as: 𝑃(𝑌|𝑋 > 2) = 𝑃(𝑌 > 2) = 0 Thus, the probability is 0 because the range of Y is from 0 to 1 only. Question #6 Suppose the random variables X and Y have a joint pdf  x + y 0  x, y  1 f XY ( x, y) =  otherwise 0 Find 𝑃𝑟(𝑋 > √𝑌) Answer: Step 1 Given information: The joint probability density function of the random variables X and Y is as given below:  x + y; 0  x, y  1 f XY ( x, y) =  0; otherwise Step 2 Compute the value of 𝑃(𝑋 > √𝑌): The marginal probability density function of X is obtained as given below: 𝑓𝑋 (𝑥) = ∫𝑦 𝑓𝑋𝑌 (𝑥, 𝑦)𝑑𝑦 1 = ∫0 (𝑥 + 𝑦)𝑑𝑦 𝑦2 1 = 𝑥[𝑦]10 + [ ] 2 0 1 2 The value of 𝑃(𝑋 > √𝑌) is obtained as given below: 𝑓𝑋 (𝑥) = 𝑥 + 1 1 𝑃(𝑋 > √𝑌) = ∫ 𝑦 (𝑥 + ) 𝑑𝑥 = 1 ∫√𝑦 𝑥𝑑𝑥 1 𝑥2 √ 1 1 + ∫ 𝑦 𝑑𝑥 √ 2 𝑥 1 2 √𝑦 1 𝑦 1 =[ ] +[ ] = − + 2 2 𝑦 2 =1− − 2 √𝑦 𝑦 −√ 2 2 √𝑦 2 2 Question #7 Let X and Y be jointly continuous random variables with joint PDF is given by: 𝑓𝑋,𝑌 (𝑥, 𝑦) = 2 where 0 ≤ 𝑦 ≤ 𝑥 ≤ 1 Show the marginal PDF of X and Y. Answer: It is given that X and Y be jointly continuous random variables with joint PDF is given by: 𝑓𝑋,𝑌 (𝑥, 𝑦) = 2 where 0 ≤ 𝑦 ≤ 𝑥 ≤ 1 We have to find marginal PDF of X and Y. 𝑏 𝑏 𝑓𝑋 (𝑥) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑦 𝑎𝑛𝑑 𝑓𝑌 (𝑦) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑎 1 1 𝑓𝑋 (𝑥) = ∫0 2𝑑𝑦 𝑎𝑛𝑑 𝑓𝑌 (𝑦) = ∫0 2𝑑𝑥 𝑎 𝑓𝑋 (𝑥) = 2 and 𝑓𝑌 (𝑦) = 2 Thus, the marginal PDF of X and Y are: 2, 0  x  1 f X ( x) =  0, otherwise 2, 0  y  1 fY ( y ) =  0, otherwise Question #8 Let X and Y be jointly continuous random variables wth joint PDF is given by: 𝑓𝑋,𝑌 (𝑥, 𝑦) = 2 where 0 ≤ 𝑦 ≤ 𝑥 ≤ 1 1 Find 𝑃 (𝑋 ≥ ). 2 Answer: 1 Let us find the value of 𝑃 (𝑋 ≥ ) . 2 1 1 𝑃 (𝑋 ≥ ) = ∫1 2𝑑𝑥 2 = [2𝑥]11 2 2 =2-1 =1 1 Hence, the required value of 𝑃 (𝑋 ≥ ) = 1 2 Question #9 Let X and Y be independent, continuous random variables with the same marginal probability density function, defined as 2  ,t 2 f X (t ) = fY (t ) =  t 2 0, otherwise (a) What is the joint probability density function f(x,y)? (b) Find the probability density of W=XY. Hind: Determine the cdf of Z. Answer: Step 1 Here, X and Y are independent, continuous random variables. So, it can be written as: 𝑓(𝑥, 𝑦) = 𝑓(𝑥)𝑓(𝑦) It is given that, 2  ,x2 f X ( x) =  x 2 0, otherwise 2  , y2 fY ( y ) =  y 2 0, otherwise  Step 2 (a) The joint probability density function can be obtained as: 𝑓𝑋𝑌 (𝑥, 𝑦) = 𝑓𝑋 (𝑥)𝑓𝑌 (𝑦) = 2 × 𝑥2 4 2 𝑦2 = (𝑥𝑌)2 Thus,  4 , x  2, y  2  f XY ( x, y ) =  ( xy )2 0, otherwise  Step 3 (b)The probability density function of W=XY is same as probability density function of f(x,y).  4 , x  2, y  2  fW ( w) = f XY ( x, y ) =  ( xy )2 0, otherwise  Question #10 The continuous random variables X and Y are statistically independent and have marginal density 1 functions 𝑓𝑋 (𝑥) = 2𝑥, 0 ≤ 𝑥 ≤ 1, 𝑓𝑌 (𝑦) = 2 , 𝑦 ≥ 1. 𝑦 Calculate the probability 𝑃(𝑋 ≤ 0.5, 𝑌 ≤ 2) =? Answer: Step 1 If two random variables X and Y, are stochastically independent then the joint probability distribution of X and Y can be expressed as 𝑓(𝑥, 𝑦) = 𝑓𝑋 (𝑥)𝑓𝑋 (𝑦). Step 2 The probability density function of X is 𝑓𝑋 = 2𝑥, for 0 ≤ 𝑥 ≤ 1 . 1 The probability density function of Y is, 𝑓𝑋 = 2, for 1 ≤ 𝑦 . 𝑦 Thus, 0.5 2 1 𝑃(𝑋 ≤ 0.5, 𝑌 ≤ 2) = (∫0 2𝑥𝑑𝑥 ) (∫1 𝑥2 0.5 = (2 [ ] = 2 0 2 (0.5 𝑦 −2+1 ) ([ 𝑦2 𝑑𝑦) 2 ] ) −2+1 1 1 1 − 0) ( − ) 1 2 =0.125 Thus, the required probability is 0.125.