Page 141 - Elements of Distribution Theory

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4.7 Exercises 127

exponential distribution. That is, the conditional distribution of X given Y is Poisson with
mean Y and the marginal distribution of Y is a standard exponential distribution.
It follows from Example 4.13 that

2
3
2
3
2
E(X|Y) = Y, E(X |Y) = Y + Y , and E(X |Y) = Y + 3Y + Y .
r
Since E(Y ) = r!, it follows that
2
3
E(X) = 1, E(X ) = 3, and E(X ) = 13.
From Example 4.18, we know that all conditional cumulants of X given Y are equal to
Y. Hence,
Var(X) = E[Var(X|Y)] + Var[E(X|Y)] = 2.
To determine κ 3 , the third cumulant of X,we need ¯κ 3 , the third cumulant of E(X|Y) = Y,
¯ κ 11 , the covariance of E(X|Y) and Var(X|Y), that is, the variance of Y, and ¯κ 001 , the expected
value of κ 3 (Y) = Y.
Letting γ 1 ,γ 2 ,... denote the cumulants of the standard exponential distribution, it fol-
lows that
κ 3 = γ 3 + 2γ 2 + γ 1 .

According to Example 4.11, the cumulant-generating function of the standard exponential
distribution is − log(1 − t). Hence,

γ 1 = 1,γ 2 = 1, and γ 3 = 2.
It follows that κ 3 = 6.

4.7 Exercises
4.1 Let X denote a real-valued random variable with an absolutely continuous distribution with
density function
β α α−1
p(x) = x exp{−βx}, x > 0,
(α)
where α> 0 and β> 0; this is a gamma distribution. Find a general expression for the moments
of X.
4.2 Let X denote a real-valued random variable with an absolutely continuous distribution with
density function
(α + β) α−1 β−1
p(x) = x (1 − x) , 0 < x < 1,
(α) (β)
where α> 0 and β> 0; this is a beta distribution. Find a general expression for the moments
of X.
4.3 Prove Theorem 4.1.
4.4 Prove Theorem 4.2.
4.5 Prove Corollary 4.1.
4.6 Prove Theorem 4.4.