P302-01 (2)

Math 302 Final Examination, December 2015 Page 1 of 1 Instructions: The duration for this exam is 150 minutes (2 hours 30 minutes). There are a total of 8 questions worth 100 points in total. A scientific calculator is allowed and a table of common distributions is provided. 1. (15 points) (a) Define the moment generating function of a random variable X . (b) Define the covariance of two random variables X, Y . (c) Define what it means for two events A and B to be independent. (d) State the Chebyshev Inequality. (e) State the Central Limit Theorem. 2. (8 points) Assume that X, Y are independent random variables such that E ( X ) = 1 , Var( X ) = 2 , E ( Y ) = 3 and Var( Y ) = 4. (a) Calculate E (2 X + 4 XY - Y ). (b) Find Var(5 X + 3 Y ). 3. (14 points) A die is tossed seven times. (a) What is the probability that exactly two outcomes are 1? (b) What is the probability that all outcomes are odd, given that the first outcome was greater than 3? 4. (11 points) let X, Y be independent exponential random variables with the same parameter λ . Find the probability density function of X - Y . 5. (14 points) A random variable X has moment generating function g X ( t ) = (1 - 3 t ) - 2 , defined for t < 1 3 . (a) Find E ( X ). (b) Find Var( X ). 6. (16 points) Assume that the random variables X, Y have joint density function f ( x, y ) = braceleftBigg c (3 x 2 + 2 y ) 0 ≤ x ≤ 1 0 ≤ y ≤ 2 0 otherwise . (a) Find the constant c . (b) Find the marginal probability density function of Y . (c) Find the conditional expectation E ( X | Y = 1). 7. (11 points) Let X and Y be independent geometric random variables with the same parameter p . What is the probability that X + Y is odd? 8. (11 points) Let S 200 be the number of heads that turn up in 200 tosses of a fair coin. Use the Central Limit Theorem to estimate the probability P (90 ≤ S 200 ≤ 110).

Math 302 Final Examination, December 2015 Page 1 of 4 Instructions: The duration for this exam is 150 minutes (2 hours 30 minutes). There are a total of 8 questions worth 100 points in total. A scientific calculator is allowed and a table of common distributions is provided. 1. (15 points) (a) Define the moment generating function of a random variable X . Solution: The moment generating function M X ( t ) of X is defined by M X ( t ) = E ( e tX ) when- ever the expectation is finite. (b) Define the covariance of two random variables X, Y . Solution: The covariance of two random variables X, Y is defined by the formula Cov( X, Y ) = E (( X - E ( X ))( Y - E ( Y )) = E ( XY ) - E ( X ) E ( Y ) . (c) Define what it means for two events A and B to be independent. Solution: A and B are independent if their probabilities satisfy P ( A ∩ B ) = P ( A ) P ( B ). (d) State the Chebyshev Inequality. Solution: If X has mean μ and variance σ 2 , then for all k > 0, P ( | X - μ | ≥ kσ ) ≤ 1 k 2 . (e) State the Central Limit Theorem. Solution: Let { X i } be independent and identically distributed random variables each with mean μ and variance σ 2 . Suppose S n = ∑ n i =1 X i . Then S n − nμ σ √ n converges in distribution to Z , the standard normal random variable. 2. (8 points) Assume that X, Y are independent random variables such that E ( X ) = 1 , Var( X ) = 2 , E ( Y ) = 3 and Var( Y ) = 4. (a) Calculate E (2 X + 4 XY - Y ). Solution: Using linearity of expectation and since E ( XY ) = E ( X ) E ( Y ) if X, Y are indepen- dent, E (2 X + 4 XY - Y ) = 2 E ( X ) + 4 E ( X ) E ( Y ) - E ( Y ) = 2 + 12 - 3 = 11 . (b) Find Var(5 X + 3 Y ). Solution: Since X, Y are independent, Var( X + Y ) = Var( X ) + Var( Y ). Furthermore, since Var( aX ) = a 2 Var( X ), then Var(5 X + 3 Y ) = 25Var( X ) + 9Var( Y ) = 50 + 36 = 86 . 3. (14 points) A die is tossed seven times. (a) What is the probability that exactly two outcomes are 1?

Math 302 Final Examination, December 2015 Page 2 of 4 Solution: The number of 1s is distributed binomially with n = 7 , p = 1 6 . Hence the probability there are 2 1s is ( 7 2 ) ( 1 6 ) 2 ( 5 6 ) 5 = 21 ∗ 5 5 6 7 ≈ 0 . 23 . (b) What is the probability that all outcomes are odd, given that the first outcome was greater than 3? Solution: Let A be the event that all outcomes are odd and B be the event that the first outcome was greater than 3. We want to compute P ( A | B ). Since P ( B ) = 1 2 , and P ( A ∩ B ) = 1 6 ( 1 2 ) 6 (the event A ∩ B is the event that the first roll is 5 and all other rolls are odd.) Hence, P ( A | B ) = P ( A ∩ B ) P ( B ) = 1 6 ( 1 2 ) 5 = 1 192 ≈ 0 . 0052 . 4. (11 points) let X, Y be independent exponential random variables with the same parameter λ . Find the probability density function of X - Y . Solution: We will compute the cumultative density function of X - Y and then compute its prob- ability density function from the cumulative density function. The joint density function of ( X, Y ) is λ 2 e − λ ( x + y ) for x, y ≥ 0 since X, Y are independent and each exponentially distributed with pa- rameter λ . Let c ∈ R . Note that X - Y ≤ c is equivalent to the condition that Y ≥ X - c and furthermore each X, Y must be positive. Hence, if c is negative, then P ( X - Y ≤ c ) = integraldisplay ∞ 0 integraldisplay ∞ x − c λ 2 e − λ ( x + y ) dy dx = λ 2 integraldisplay ∞ 0 e − λx e − λy - λ vextendsingle vextendsingle vextendsingle ∞ x − c = λ integraldisplay ∞ 0 e − 2 λx + λc dx = λe λc e − 2 λx - 2 λ vextendsingle vextendsingle vextendsingle ∞ 0 = e λc 2 . (1) and if c is positive, then P ( X - Y ≤ c ) = λ 2 bracketleftbiggintegraldisplay c 0 integraldisplay ∞ 0 e − λ ( x + y ) dy dx + integraldisplay ∞ c integraldisplay ∞ x − c e − λ ( x + y ) dy dx bracketrightbigg . (2) Since integraldisplay ∞ c integraldisplay ∞ x − c λ 2 e − λ ( x + y ) = λe λc integraldisplay ∞ c e − 2 λx dx = λe λc e − 2 λc 2 λ = e − λc 2 and integraldisplay c 0 integraldisplay ∞ 0 λ 2 e − λ ( x + y ) dy dx = λ 2 integraldisplay c 0 e − λx dx integraldisplay ∞ 0 e − λy dy = λ 2 parenleftbigg 1 - e − λc λ parenrightbigg parenleftbigg 1 λ parenrightbigg = 1 - e − λc . Then P ( X - Y ≤ c ) = 1 - e − λc + e − λc 2 = 1 - e − λc 2