Variance of future loss

+1 vote
42 views
edited Aug 13, 2016

A whole life insurance with unit sum insured payable at the end of the year of death with a level annual premium is issued to $$x$$. Let $$L_0$$ be the net future loss random variable with the premium determined by the equivalence principle. Let $$L^∗_0$$ be the net future loss random variable if the premium is determined such that $$E[L^∗_0]=−0.5$$ . Given $$V[L_0] = 0.75$$, calculate $$V[L^∗_0]$$.

answered Aug 13, 2016 by (1,350 points)

Let $$P$$ be the equivalence principle premium that yields $$\mathbb{E}$$[$$L_0$$] $$= 0$$ and $$P^*$$ be the premium that yields $$\mathbb{E}$$[$$L_0$$] $$= -0.5$$.

The terms $$L_0$$ and $$L_0^*$$ can be written as $$(1 + \frac{P}{d}) v^{K_x +1} - \frac{P}{d}$$ and $$(1 + \frac{P^*}{d}) v^{K_x +1} - \frac{P^*}{d}$$ respectively.

From this follows:

$$Var[L_0] = (1 + \frac{P}{d})^2 Var[v^{K_x +1}]$$ and $$Var[L_0^*] = (1 + \frac{P^*}{d})^2 Var[v^{K_x +1}]$$.

From the equivalence principle: $$\mathbb{E}[L_0] = 0 \Rightarrow A_x - P \ddot{a_x} = 0 \Rightarrow P = \frac{A_x}{\ddot{a_x}}$$.

Also,  $$\mathbb{E}[L_0^*] = -0.5 \Rightarrow A_x - P^* \ddot{a_x} = -0.5 \Rightarrow P^* = \frac{A_x+0.5}{\ddot{a_x}} \Rightarrow P^* = P + \frac{0.5}{\ddot{a_x}}$$.

$$Var[L_0] = (1 + \frac{P}{d})^2 Var[v^{K_x +1}] \Rightarrow Var[v^{K_x +1}] = \frac{0.75}{(1 + \frac{P}{d})^2}$$.

Using this expression for $$Var[v^{K_x +1}]$$ in the equation for finding $$Var[L_0^*]$$, and in place of $$P^*$$ substituting in $$P + \frac{0.5}{\ddot{a_x}}$$, gives the answer for $$Var[L_0^*]$$  (after some simplification) as $$1.6875$$.