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Let annual total loss be $L={\sum }_{i=1}^{N_1}{X}_i$, with an empty sum equal to zero. The assumptions give a compound Poisson distribution. Conditioning on the count,The tower property and total-variance formula therefore giveThe $\lambda {\mu }^2$ term is the extra uncertainty in the number of defaults. Replacing it by zero would incorrectly treat that count as deterministic.

For sufficiently many annual events and an adequate normal approximation, the value at risk at confidence level $p$ is the compound-Poisson annual loss quantileThe supplied quantiles give $z_{0.90}=1.282$, $z_{0.95}=1.645$ and $z_{0.99}=2.326$. In particular a one-sided 95% annual loss quantile uses $1.645$, not the $1.96$ from the two-sided 95% rate interval. If the expected loss is covered separately, the approximate unexpected-loss reserve is the second term, rather than the full quantile.

Estimate the rate from a suitable historical count/exposure or gap sample, then insert $\hat{\lambda }$ into the loss formula. The interval in part (d) measures rate-estimation uncertainty; for nonnegative loss mean it can be propagated to lower and upper approximate quantiles because the displayed expression is increasing in $\lambda$. Severity parameters also require estimation, so the rate interval alone is not a confidence interval accounting for every source of uncertainty.

Mean and variance do not determine the exact loss quantile. The actual distribution isFor nonnegative losses it has a zero atom at least $e^{-\lambda }$, so if that mass exceeds $p$ its exact value at risk is zero. With a small rate or a strongly skewed severity distribution, a normal approximation can therefore be poor. A loan book with dependent defaults, rating migration or varying exposures also needs more than the constant-intensity independent-severity model. These are model qualifications, not deductions from the credit rating alone.