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Variance of future loss

+1 vote
41 views
asked Aug 13, 2016 in BUS 3024S - Contingencies by anonymous
edited Aug 13, 2016 by simon_rigby

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 \(V[L^∗_0]\).

1 Answer

+2 votes
answered Aug 13, 2016 by Feroz (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 \( 1.6875 \).

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