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solution08 - Solutions to Homework 08 1 Y = e X We denote...

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Unformatted text preview: Solutions to Homework 08 1. Y = e X . We denote the pdf of X by f X ( x ) and its cdf by F X ( x ). Also, Y = e X ⇒ X = ln Y . a. Note first that for all values of X , i.e.-∞ < x < ∞ , Y > 0. Thus, P [ Y ≤ 0] = 0. For Y > 0, F Y ( y ) = P [ Y ≤ y ] = P [ e X ≤ y ] = P [ X ≤ ln y ] = F X (ln y ) Therefore F Y ( y ) = braceleftBigg y ≤ F X (ln y ) y > Now, f Y ( y ) = d dy F Y ( y ) = d dy F X (ln y ) = F prime X (ln y ) d dy ln y = 1 y f X (ln y ) b. If X is a Gaussian random variable, its pdf is given by f X ( x ) = 1 √ 2 πσ 2 e- 1 2 ( x- m σ ) 2-∞ < x < ∞ Then, f Y ( y ) is given by f Y ( y ) = y ≤ 1 y √ 2 πσ 2 e- 1 2 ( ln y- m σ ) 2 y > 2. Y = A cos( ωt ) + c E [ Y ] = E [ A cos( ωt ) + c ] = integraldisplay ∞-∞ ( A cos( ωt ) + c ) f A ( a ) da = cos( ωt ) integraldisplay ∞-∞ Af A ( a ) da + c integraldisplay ∞-∞ f A ( a ) da = cos( ωt ) E [ A ] + c = m cos( ωt ) + c 1 V AR [ Y ] = E [( Y- E [ Y ]) 2 ] = integraldisplay ∞...
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solution08 - Solutions to Homework 08 1 Y = e X We denote...

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