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# soln4 - SS-1 1(d 2 Solution to this problem will be posted...

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SS-1.

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1.

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(d)

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2. Solution to this problem will be posted separately. SS-2. (a) P [ | X | < 5 , Y > 2 , Z 2 2] = P [ | X | < 5] P [ Y > 2] P [ Z 2 2] = P [ - 5 < X < 5](1 - P [ Y 2])(1 - P [ - 2 < Z < 2]) = [ F X (5 - ) - F X ( - 5)][1 - F Y (2)][1 - F Z ( 2 - ) + F Z ( - 2)] (b) P [ X > 5 , Y < 0 , Z = 1] = P [ X > 5] P [ Y < 0] P [ Z = 1] = [1 - F X (5)] F Y (0 - )[ F Z (1) - F Z (1 - )] (c) P [min( X, Y, Z ) > 2] = P [ X > 2] P [ Y > 2] P [ Z > 2] = [1 - F X (2)][1 - F Y (2)][1 - F Z (2)] (d) P [max( X, Y, Z ) < 6] = P [ X < 6] P [ Y < 6] P [ Z < 6] = F X (6 - ) F Y (6 - ) F Z (6 - )
SS-3. y -y y Figure 1: Event { Y y } shown in black. Mapping to function shown in blue. Equivalent event shown in orange. From Figure 1, { Y y } = {- y X y } , y 0 , , y < 0 F Y ( y ) = P ( Y y ) = P ( - y X y ) , y 0 0 , otherwise = F X ( y ) - F X ( - y ) , y 0 0 , otherwise f Y ( y ) = d dy F Y ( y ) = f X ( y ) + f X ( - y ) , y 0 0 , otherwise Note that you should be careful in taking the derivative of a function that is defined piecewise over some range. However, in this case the function is continuous at y = 0 , so y = 0 is not a special case.

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b) y+dy y+dy y y -y -y-dy Figure 2: Event { y Y y + dy } shown in black. Mapping to function shown in blue. Equiva- lent event shown in orange.
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soln4 - SS-1 1(d 2 Solution to this problem will be posted...

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