Pembahasan Ujian PAI: A70 – No. 6 – November 2017

• \(X\) ialah besarnya kerugian
  \(X \sim Lognormal(\mu = 7,\sigma = 2)\)
• Deductible (d) = 2.000

• Distribusi Lognormal
  \(E[{X^k}] = exp(k\mu + \frac{1}{2}{k^2}{\sigma ^2})\)
  \(E[{(X \wedge x)^k}] = \exp (k\mu + \frac{1}{2}{k^2}{\sigma ^2})\Phi \left( {\frac{{\ln x – \mu – k{\sigma ^2}}}{\sigma }} \right) + {x^k}[1 – F(x)]\)
  \(F(x) = \Phi \left( {\frac{{\ln x – \mu }}{\sigma }} \right)\)
• \(LER = \frac{{E[X \wedge d]}}{{E[X]}}\)

1. Mencari \(E[X]\)
   \(E[X] = \exp (7 + \frac{1}{2}{2^2})\)
   \(E[X] = 8.103,08\)
2. Mencari \(F(2.000)\)
   \(F(2.000) = \Phi \left( {\frac{{\ln 2.000 – 7}}{2}} \right)\)
   \(F(2.000) = \Phi \left( {0,3} \right)\)
   \(F(2.000) = 0,6179\)
3. Mencari \(E[X \wedge 2.000]\)
   \(E[X \wedge 2.000] = \exp (7 + \frac{1}{2}{2^2})\Phi \left( {\frac{{\ln 2.000 – 7 – {2^2}}}{2}} \right) + 2.000[1 – 0,6179]\)
   \(E[X \wedge 2.000] = \exp (9)\Phi \left( { – 1,7} \right) + 764,2\)
   \(E[X \wedge 2.000] = \exp (9)\left( {1 – \Phi \left( {1,7} \right)} \right) + 764,2\)
   \(E[X \wedge 2.000] = (8.103,08)0,0446 + 764,2\)
   \(E[X \wedge 2.000] = 1.125,59\)
4. Mencari LER
   \(LER = \frac{{1.125,59}}{{8.103,08}}\)
   \(LER = 0,1389\)

Paling sedikit 0,10 akan tetapi kurang dari 0,15