这是一份southampton南安普敦大学MATH3066W1-01作业代写的成功案例
$Z$ is the present-value random variable for an $n$-year endowment insurance. Hence
$$
\mathrm{E}[Y]=\bar{a}{x: \bar{n}]}=\mathrm{E}\left[\frac{1-Z}{\delta}\right]=\frac{1-\bar{A}{x: \bar{n}}}{\delta}
$$
and
$$
\operatorname{Var}(Y)=\frac{\operatorname{Var}(Z)}{\delta^{2}}=\frac{{ }^{2} \bar{A}{x: \bar{n}}-A{x:\left.\bar{n}\right|^{2}}}{\delta^{2}}
$$
In terms of annuity values, becomes
$$
\begin{aligned}
\operatorname{Var}(Y) &=\frac{1-2 \delta^{2} \bar{a}{x: i n}-\left(1-\delta \bar{a}{x: i n}\right)^{2}}{\delta^{2}} \
&=\frac{2}{\delta}\left(\bar{a}{x: i n}-{ }^{2} \bar{a}{x: \bar{n}}\right)-\left(\bar{a}_{x: i n t}\right)^{2}
\end{aligned}
$$
MATH3066W1-01 COURSE NOTES ：
Since $Y=(1-Z) / d$, where
$$
Z= \begin{cases}v^{K+1} & 0 \leq K<n \ v^{n} & K \geq n\end{cases}
$$
is the present-value random variable for a unit of endowment insurance, payable at the end of the year of death or at maturity, we have
$$
\ddot{a}{x: \bar{n}}=\frac{1-\mathrm{E}[Z]}{d}=\frac{1-A{x \cdot \bar{n}}}{d} ;
$$