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Unformatted text preview: ) = 0.
To ﬁnd the mean and variance of Y we could use the pdf just found. However, instead of that,
∞
we will use LOTUS and the fact 0 un e−u du = n! for all integers n ≥ 03 to get
∞ E [Y ] = g (u)fX (u)du
−∞
∞ =
−∞
∞ = u2 e−u
du
2
u2 e−u du = 2! = 2. 0 and
∞ E [Y 2 ] = (g (u))2 fX (u)du
−∞
∞ =
−∞
∞ = u4 e−u
du
2
u4 e−u du = 4! = 24. 0 Therefore, Var(Y ) = E [Y 2 ] − E [Y ]2 = 20.
3 This identity can be proved by induction on n, using integration by parts for the induction step. 3.8. FUNCTIONS OF A RANDOM VARIABLE 101 Example 3.8.2 Suppose Y = X 2 , where X is a N (µ, σ 2 ) random variable with µ = 2 and σ 2 = 3.
Find the pdf of Y .
Solution: Note that Y = g (X ) where g (u) = u2 . The support of the distribution of X is the
whole real line, and the range of g over this support is R+ . Next we ﬁnd the CDF, FY . Since
P {Y ≥ 0} = 1, FY (c) = 0 for c < 0. For c ≥ 0,
√
√
FY (c) = P {X 2 ≤ c} = P {− c ≤ X ≤ c}
√
√
− c−2
c−2
X −2
√
=P
≤√
≤√
3
3
3
√
√
c−2
− c−2
√
√
−Φ
.
=Φ
3
3
Diﬀerentiate...
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 Spring '08
 Zahrn
 The Land

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