1 0 1 2 2 1 0 1 2 2 1 0 1 2 c d f 2 1 0 y 1 2 figure

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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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