= Solution
Condition on the <claim count distribution>. If $\mu=\mathbb EX_1$ and $v=\operatorname{Var}(X_1)$, then independence of the count and the <independent and identically distributed random variables> gives
$$
\mathbb E[S\mid N]=N\mu,\qquad\operatorname{Var}(S\mid N)=Nv.
$$
The empty sum is zero. The <law of total expectation> and <law of total variance> therefore prove the <random sum of independent claims> formulas
$$
\boxed{\mathbb ES=\mathbb EN\,\mathbb EX_1,\qquad
\operatorname{Var}(S)=\mathbb EN\operatorname{Var}(X_1)+\operatorname{Var}(N)(\mathbb EX_1)^2.}
$$
For the variance calculation assume the count and severity have finite second moments; the expectation identity needs the corresponding integrability. Nonnegative claims also permit the expectation identity with extended nonnegative values.
Represent each policy payment as $S_i=B_iY_i$, where $B_i$ is a <Bernoulli random variable> of parameter $q_i$, independent of its severity. Equivalently, $Y_i$ specifies the payment distribution conditional on a claim. Then
$$
\mathbb ES_i=q_i\mu_i,\qquad
\mathbb ES_i^2=q_i(\sigma_i^2+\mu_i^2),\qquad
\operatorname{Var}(S_i)=q_i\sigma_i^2+q_i(1-q_i)\mu_i^2.
$$
Independence of the policies makes their <variances> add. Put
$$
Q=\sum_iq_i,\qquad M=\sum_iq_i\mu_i,\qquad
A=\sum_iq_i(\sigma_i^2+\mu_i^2),\qquad C=\sum_iq_i^2\mu_i^2.
$$
The <heterogeneous individual claims model> consequently has
$$
\boxed{\mathbb ET=M,\qquad\operatorname{Var}(T)=A-C
=\sum_i\bigl[q_i\sigma_i^2+q_i(1-q_i)\mu_i^2\bigr].}
$$
For the Poisson approximation, define the severity <mixture distribution>
$$
F=\sum_i\frac{q_i}{Q}F_i.
$$
Let $\varphi_i(t)=\mathbb E[e^{itY_i}]$ be the individual <characteristic functions>. Conditioning on each Poisson count and then multiplying transforms of the independent policy payments gives
$$
\begin{aligned}
\mathbb E[e^{it\widetilde T}]
&=\prod_i\exp\{q_i(\varphi_i(t)-1)\}\\
&=\exp\left\{Q\left[\sum_i\frac{q_i}{Q}\varphi_i(t)-1\right]\right\}.
\end{aligned}
$$
This is <Poisson superposition of insurance portfolios>. By the <uniqueness theorem for characteristic functions>,
$$
\boxed{\widetilde T\ \overset d=\ \sum_{j=1}^{K}Z_j,\qquad
K\sim\operatorname{Poisson}(Q),\quad Z_j\overset{\mathrm{iid}}\sim F,}
$$
with count and severities independent. Its <compound Poisson distribution> has distribution function
$$
\mathbb P(\widetilde T\leq x)=e^{-Q}\sum_{k=0}^{\infty}\frac{Q^k}{k!}F^{*k}(x),
$$
where $F^{*k}$ is the <convolution> distribution of $k$ independent severities and $F^{*0}$ is the distribution concentrated at zero. Characteristic functions avoid assuming that severity moment-generating functions exist. Since $\mathbb EZ=M/Q$ and $\mathbb EZ^2=A/Q$, the random-sum formulas give
$$
\boxed{\mathbb E\widetilde T=M,\qquad\operatorname{Var}(\widetilde T)=A.}
$$
Use $T_{\mathrm B}$ to distinguish the second, compound-binomial approximation from the Poisson one. It has $K_{\mathrm B}\sim\operatorname{Bin}(n,p)$ with $p=Q/n$ and the same severity law $F$. Thus $\mathbb EK_{\mathrm B}=Q$ and $\operatorname{Var}(K_{\mathrm B})=Q(1-p)$. Applying the <random sum of independent claims> identities again gives
$$
\begin{aligned}
\operatorname{Var}(T_{\mathrm B})
&=Q\left[\frac AQ-\left(\frac MQ\right)^2\right]
+Q(1-p)\left(\frac MQ\right)^2\\
&=A-\frac{M^2}{n}.
\end{aligned}
$$
Hence
$$
\boxed{\mathbb ET_{\mathrm B}=M,\qquad\operatorname{Var}(T_{\mathrm B})=A-M^2/n.}
$$
Both approximations match the exact <expected value>. The <Cauchy-Schwarz inequality> gives $M^2\leq nC$, proving the <variance comparison for compound portfolio approximations>:
$$
\boxed{\operatorname{Var}(T)\leq\operatorname{Var}(T_{\mathrm B})\leq\operatorname{Var}(\widetilde T).}
$$
The excess variances are respectively $C-M^2/n$ and $C$. The compound-binomial model is therefore closer in variance; equality with the exact variance occurs when all $q_i\mu_i$ are equal. This variance comparison is not a general ordering of approximation errors or tail probabilities.
The Poisson replacement is useful when all claim probabilities are small: it replaces the individual restriction of at most one claim by an unrestricted Poisson count, and the variance error is of second order in those probabilities. The compound-binomial replacement retains a count bounded by $n$, but pools both frequencies and severity laws and can lose policy-specific information. If all $q_i$ and all $F_i$ agree, it is exactly the original portfolio law. In a heterogeneous portfolio, agreement of means, or even variances, does not make either approximate law exact. Neither approximation is guaranteed accurate merely by having many policies.
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