The fourier transform of the n 0 2 pdf is exp 2 2f 2

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Unformatted text preview: tor into a function of u times a function of v.1 Another reason is that the conditional distributions of u+v u+0.5 v ∈ [0, 1] for 0 ≤ u ≤ 1. So X and Y are not Y given X = u depend on u: fY |X (v |u) = 0, else independent by Proposition 4.4.2. 3u2 u ∈ [0, 1] (c) Yes, because fX,Y (u, v ) = fX (u)fY (v ) where fX (u) = and fY ≡ fX . 0, else, 4.5 Distribution of sums of random variables Section 3.8 describes a two or three step procedure for finding the distribution of a random variable that is is a function, g (X ), of another random variable. If g (X ) is a discrete random variable, Step 1 is to scope the problem, and Step 2 is to find the pmf of g (X ). If g (X ) is a continuous-type random variable, Step 1 is to scope the problem, Step 2 is to find the CDF, and Step 3 is to differentiate the CDF to find the pdf. The same procedures work to find the pmf or pdf of a random variable that is a function of two random variables, having the form g (X, Y ). An important function of (X, Y ) is the su...
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