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 \( 1.6875 \).