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into a function of u times a function of v.1 Another reason is that the conditional distributions of
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) =
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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- Spring '08
- The Land