g 5. (Sec. 5.1) Suppose the marginal pdfs of two independent rvs X and Y are given byfX(x) = (5x4 for 0  x  10 otherwisefY (y) = (2y3 + y for 0  y  10 otherwise(a) Find the joint pdf of X and Y .

f(x,y) = 5x⁴(2y³ +y )           0<x<1 , 0<y<1 , 0 otherwise

Step-by-step explanation:

Given

F(x) = 5x⁴ for 0<x<1 and 0 otherwise

f(y)= 2y³ +y for 0<y<1 and 0 otherwise

the joint p.d.f is given by

f(x,y) = f(x) * f(y)= 5x⁴(2y³ +y ) 0<x<1 , 0<y<1 , 0 otherwise

The joint pdf is found by multiplying the marginal pdfs.

The two random variables are said to be statistically independent if the joint probability function can be expressed as the product of two marginal functions.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

it is b)

hope this : )

step-by-step explanation: