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