= Solution
To distinguish the random intensity from its possible values, write it as $\Lambda$. Its law is a <gamma distribution> with shape $2$ and rate $p/q$. Hence
$$
\boxed{\lambda_0=\mathbb E\Lambda=\frac{2q}{p}},\qquad
\operatorname{Var}(\Lambda)=\frac{2q^2}{p^2}.
$$
For the <Poisson mixture>, <conditional expectation> and <conditional variance> both equal $\Lambda$. The <law of total expectation> and the <law of total variance> yield
$$
\mathbb EN=\frac{2q}{p},\qquad
\operatorname{Var}(N)=\mathbb E\Lambda+\operatorname{Var}(\Lambda)
=\frac{2q}{p^2},
$$
where $p+q=1$. Applying the <random sum of independent claims> formulas with the <exponential distribution> of the claim sizes gives \b[the portfolio B moments]
$$
\boxed{\mathbb ES_B=\frac{2q\mu}{p},\qquad
\operatorname{Var}(S_B)=\frac{2q(1+p)\mu^2}{p^2}.}
$$
At the matched intensity $\lambda_0$, the <expected value> for portfolio A is also $2q\mu/p$, whereas its <variance> is $4q\mu^2/p$. Thus \b[the expected totals agree, but mixing increases the variance]:
$$
\boxed{\operatorname{Var}(S_B)-\operatorname{Var}(S_A)
=\frac{2q^2\mu^2}{p^2}>0.}
$$
The extra term is precisely $\mu^2\operatorname{Var}(\Lambda)$.
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