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[15] 5. (X, Y) have joint density (22 + y?) 0<*<1 0<y<1 f(x, y) else find the marginals fx(x) and fy (y).
The joint density function of the continuous variables X and Y is fX,Y(x,y) = (12/5)*x*(2-x-y) for 0<X<1 and 0<Y<1. a) Find the expected value of X+Y. (b) Find fX(x), and fY(y). (c) Find Cov(X,Y). (d) Find Corr(X,Y).
Find values of x and y such that both fx(x,y) = 0 and fy(x,y) = 0. f(x,y) = x² + xy + y2 - 3x +2 A. X=2, y= -1 OB. X = -2, y=1 O c. 1 x= 1, y = 2 OD. x= 0, y = 0
Find fx (x,y). f(x,y)= e - 4x + 3y O A. fx(x,y)= -4 e - 4x OB. fx(x,y) = -4 e - 4x + 3y O C. &x(x,y)= e = 4x+3 OD. &x(x,y)=3 € -4x+3y
given ellers. fx(z) = 0 ellers, 4(y-r) fr(u)o hvis 0 < y<1 ellers. Find P(X1/2 and P(1/3<Y < 1/2) Find E(Yl and EX Y) Find P(X+Y s 1/2)
5. Let fx(x) be a pdf given by fx(x) = (1/8)(e^(-x/8)) for x > 0. a) Find the CDF FX(x). b) Find P(X > 4) c) Find P(-2 ≤ X ≤ 12) d) Find P(X < 240) e) Find E(X) f) Find the standard deviation of X.
Suppose X and Y have joint probability density function fX,Y(x,y)=70e?3x?7y for 0<x<y; and fX,Y(x,y)=0 otherwise. Find E(X). (You may either use the joint density given here,
Consider fx (x)=e*, 0<x and joint probability density function fx (x, y) = e) for 0<x<y. Determine the following: (a) Conditional probability distribution of Y given X =1. (b) ECY X = 1) = (c) P(Y <2 X = 1) = (d) Conditional probability distribution of X given Y = 4.
Please explain The function below is a joint CDF of two continuous X, Y: iD else 1. Find the constant c and the marginal CDF Fx(u and Fy() 2. Are X, Y independent ? 3. Find the probabilities below: (a) p (X є (0.11, Ye (0,11 (b) p(X>0) (c) P(Y 1) (Hint: bound by a rectangle) (e) p(E), for the shaded area E in the figure. The function below is a joint CDF of two continuous X, Y: iD else...
Consider the joint density function fX,Y,Z(x,y,z)=(x+y)e−zfX,Y,Z(x,y,z)=(x+y)e−z where 0<x<1,0<y<1,z>0. b) Find the marginal density of (x,z) : fX,Z(x,z). For your spot check, please report fX,Z(1/2,1/4)+fX,Z(1/4,1/2)+fX,Z(1/2,2) rounded to 3 decimal places.