a) for this to be valid: kxy dx dy = kx2y/2|10 dy = ky/2 dy =ky2/4 |10 =k/4 =1
k=4
b)
marginal pdf of x: f(x)= f(x,y) dy = 4xy dy =4xy2/2 |10
f(x=2x for 0 <x<1
marginal pdf of y: f(y)= f(x,y) dx = 4xy dy =4x2y/2 |10
f(y) =2y for 0 < y<1
c)Yes as f(x)*f(y) =f(x,y) , therefore x and y are independent
d)
P(X<0.5,Y>0.5) =P(X<0.5)*P(Y>0.5) = 2x dx * 2y dy =(x2) |0.50 *(y2) ||10.5 =0.25*0.75 =0.1875
For f(x, y) = k(x2 + y2), 0<x< 1 and 0 <y<1 and 0 elsewhere: a) Find k. b) Are X and Y independent? c) Find P(X<0.5, Y>0.5), P( X = 0.5, Y>0.5).
Let f(x, y) 2e-(x+y), x > 0, y > 0. Show that X, Y are independent. What are the marginal PDFS of each?
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22 If f(L,y) = 561, for 0 <<<y<1. find o elsewhere, (a) f(y) (b) f(z|y) (c) E[X | Y = y) (d) EX | Y = 0.5]
The joint density function for X and Y is given as: f(x, y) = kxy for 0 < x < 2y < 1. Find the value of the constant k for which the p.d.f is legitimate. If the video does not work, click here to go to YouTube directly.
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