Show that the probability density function for the loss random variable is
begin{eqnarray*}
textrm{f}_{L_t}(y) &=& frac{1}{(65-t)(y delta + P^n)}.
end{eqnarray*}
Solution.
The distribution function of (L_t^n) is
begin{eqnarray*}
textrm{F}_{L_t}(y) &=& Pr(L_t^n leq y) = Pr(v^{T-t} – P
left(frac{1-v^{T-t}}{delta}
right) leq y) \
&=& Pr(v^{T-t} leq frac{y delta +P}{delta + P}) = Pr(T-t geq -frac{1}{delta} ln left(frac{y delta +P}{delta + P}right) )
end{eqnarray*}
Conditional on (T>t), (T-t) is uniform on ((0,65-t)), so
begin{eqnarray*}
textrm{F}_{L_t}(y) &=& frac{65-t -frac{1}{delta} ln left(frac{y delta +P}{delta + P}right) )}{65-t}
end{eqnarray*}
Taking a partial derivative with respect to (y) yields
begin{eqnarray*}
textrm{f}_{L_t}(y) &=& frac{partial}{partial y} frac{65-t -frac{1}{delta} ln left(frac{y delta +P}{delta + P}right) )}{65-t} \
&=& frac{-1}{delta (65-t)} frac{partial}{partial y} ln (y delta +P) \
&=& frac{1}{(65-t)(y delta + P)},
end{eqnarray*}
as required.