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Asked Oct 17, 2019
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5. (Sec. 5.1) Suppose the marginal pdfs of two independent rvs X and Y are given by
5x4 for 0
|0 otherwise
1
fx(r)
2y3for 0y <1
|0 otherwise
fy(y)
(a) Find the joint pdf of X and Y
(b) What is the probability that both X and Y are greater than .5?
(c) What is the probability that at least one of X or Y is greater than .5?
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5. (Sec. 5.1) Suppose the marginal pdfs of two independent rvs X and Y are given by 5x4 for 0 |0 otherwise 1 fx(r) 2y3for 0y <1 |0 otherwise fy(y) (a) Find the joint pdf of X and Y (b) What is the probability that both X and Y are greater than .5? (c) What is the probability that at least one of X or Y is greater than .5?

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Expert Answer

Step 1

Given marginal p.d.f of random variable X and Y are given by

5x4for 0 x s
0 Otherwise
fx(x)
y for 0
y <
fy)=2y +yfor 0s ys 1)
0 Otherwise
a) The random variable X and Y are independent,
So, the joint pdf of X and Y is given by
(5x4(2y3 + y) 0 < x < 1
0 y s 1
fxy (x, y) fi(x)xf,y) =
0 Otherwise
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5x4for 0 x s 0 Otherwise fx(x) y for 0 y < fy)=2y +yfor 0s ys 1) 0 Otherwise a) The random variable X and Y are independent, So, the joint pdf of X and Y is given by (5x4(2y3 + y) 0 < x < 1 0 y s 1 fxy (x, y) fi(x)xf,y) = 0 Otherwise

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Step 2

b)
Probability that both X and Y are greater than 0.5 is given by

1
1
1
5x(2y3 y)dxdy = (2y3 y) dy
P(x >0.5, y > 0.5) =
5x4dx
11
0.5 0.5
0.5
0.5
1
(2y3 y) dy x 5
5 x 0.19375( 2y3 + y) dy
=>
5
0.5
0.5
0.5
1
y4
0.96875( 2y3 + y) dy 0.96875 x 2 x
4
E>
2
0.5
0.5
=> 0.96875 x (0.375 + 0.46875) = 0.81738 .... (1)
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1 1 1 5x(2y3 y)dxdy = (2y3 y) dy P(x >0.5, y > 0.5) = 5x4dx 11 0.5 0.5 0.5 0.5 1 (2y3 y) dy x 5 5 x 0.19375( 2y3 + y) dy => 5 0.5 0.5 0.5 1 y4 0.96875( 2y3 + y) dy 0.96875 x 2 x 4 E> 2 0.5 0.5 => 0.96875 x (0.375 + 0.46875) = 0.81738 .... (1)

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Step 3

c) Probability that atleast one of X or Y is gre...

P(X> 0.5 or Y > 0.5) P(X > 0.5) P(Y > 0.5) P(X> 0.5 and Y > 0.5)
1
1
1
(2y3 +y) dy
5х4dx +
5x*(2y3 + у)dxdy
=>
0.5 0.5
0.5
0.5
y4
X
4
0.81738(From eq 1)
=> 5
5 Jo.5
2 Jo.5
=> 0.96875+0.84375 - 0.81738 = 0.99512
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P(X> 0.5 or Y > 0.5) P(X > 0.5) P(Y > 0.5) P(X> 0.5 and Y > 0.5) 1 1 1 (2y3 +y) dy 5х4dx + 5x*(2y3 + у)dxdy => 0.5 0.5 0.5 0.5 y4 X 4 0.81738(From eq 1) => 5 5 Jo.5 2 Jo.5 => 0.96875+0.84375 - 0.81738 = 0.99512

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