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- The density function of two random variables X and Y is fxy (x, y) = 16 e-4 (x+y). u (x) u (y) Find the mean of the function (X, ¥)= 5 for 0 < X s; 1 = -1 for < X and lor< Y 2 = 0 for all other X and YLet X and Y be independent random variables with density f (x) = 3x² for 0 < x < 1. Then P (X+ Y < 1) is equal to3) The joint probability density function of the random variables X, Y, and Z is ², 0 }, 13. If X and Y are jointly continuous random variables with joint PDF fx.y (r, y) = exp(-y)I0.v)(x)I(0,0) (y) (a) Find the joint MGF of X and Y (b) Find the marginal MGFS of X and Y (c) Identify the marginal distribution of X and Y (d) Find the E(XY) using the joint MGF of X and Y4) The joint density function of the random variables X and Y is given by (&xy f(x, y) = 0SXS1,0 sysx otherwise Find (a) the marginal density of X, (b) the marginal density of Y, (c) the conditional density of X, (d) the conditional density of Y.4) The joint density function of the random variables X and Y is given by S8xy 0 s xs 1,0 s ys x f(x, y) = otherwise Find (a) the marginal density of X, (b) the marginal density of Y, (c) the conditional density of X, (d) the conditional density of Y.3. Let X be a continuous random variable. Let f(x) = c(x − 1)³ and Sx = [1,3]. Hint: (x - 1)³ = x³ + 3x − 3x² - 1 (a) What value of c will make f(x) a valid density? (b) What is P(X = 2)? (c) Find E(X). (d) What is P(1 < X < 2)?e) The density function of a certain random variable X is given by xa-'(1– x)®-1 f(x) = {B(a,0) 0, if 0 0. i) Derive the expected value of X. ii) Derive the variance of X.The random variable Y has probability density function f(V) = k(y + y³), 0 2. Hence find PG < Y <). iii) Find the variance of Y.Let Y1 and Y2 be two jointly continuous random variables with joint probability density function f (y1, Y2) = {10 y Y1 , 0 < y1 < y2 < 1, and 0 elsewhere. The marginal probability densities function of Y1 is fi (y1) = 5 yf, 0 < y1 <10 10 fi (y1) = y1 (1 – yf), 0Let Xand Y be two continuous random variables with joint probability density [3x function given by: f(x.y)%D 0sysxsl elsewhere with E(X) = ECX)- EC) - EC*)= ;and E(XY) = 10 3 E(Y*) = - and E(XY) =; %3D Then the value of the variance of 2X+Y is: O 3/80 O 91/320 43/320 7/20Let X and Y be independent normally distributed random variables with mean zero and variances og = 1 and of = 4. (a) Write the joint probability density function fx.y (r, y). • (b) Define new random variables U = aX + Y and V = X – Y, where a + -1 is a real number. Find the absolute value of the Jacobian of transform from X, Y to U, V. (c) Find the joint probability density function for U and V. Find a for which U and V are independent random variables. Write down fu,v (u, v) for this a in the answer.SEE MORE QUESTIONSRecommended textbooks for youCalculus For The Life SciencesCalculusISBN:9780321964038Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.Publisher:Pearson Addison Wesley,Linear Algebra: A Modern IntroductionAlgebraISBN:9781285463247Author:David PoolePublisher:Cengage LearningCalculus For The Life SciencesCalculusISBN:9780321964038Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.Publisher:Pearson Addison Wesley,Linear Algebra: A Modern IntroductionAlgebraISBN:9781285463247Author:David PoolePublisher:Cengage Learning