Pembahasan Ujian PAI: A70 – No. 12 – Juni 2015

Banyaknya klaim mengikuti distribusi binomial negatif dengan:
\(\tau \) = 2
\(\beta \) = 1,5

Untuk setiap kerugian, berlaku deduktibel standard (“ordinary deductible“) ialah 50 dan loss limit dari besar klaim sebelum dipotong deduktibel ialah 175

• \(E[X \wedge d]{\rm{ }} = \frac{{\theta \Gamma (\alpha + 1)}}{{\Gamma (\alpha )}}\Gamma \left( {\alpha + 1;\frac{d}{\theta }} \right) + d\left[ {1 – \Gamma \left( {\alpha ;\frac{d}{\theta }} \right)} \right]\)
• Nilai rataan dari besar agregat pembayaran klaim “per loss-basis”
  \(E[aggregate\_payment]{\rm{ }} = E[{N^L}] \times E[{Y^L}]\)

\(\Gamma \left( {3;\frac{{50}}{{100}}} \right) = 0,0143877\)
\(\Gamma \left( {2;\frac{{50}}{{100}}} \right) = 0,090204\)

\(E[X \wedge 175]{\rm{ }} = 200\Gamma \left( {3;\frac{{175}}{{100}}} \right) + 175\left[ {1 – \Gamma \left( {2;\frac{{175}}{{100}}} \right)} \right] = 134,8347113\)
\(E[X \wedge 50]{\rm{ }} = 200\Gamma \left( {3;\frac{{50}}{{100}}} \right) + 50\left[ {1 – \Gamma \left( {2;\frac{{50}}{{100}}} \right)} \right] = 48,36734\)

Nilai rataan dari besar agregat pembayaran klaim “per loss-basis”
\(E[aggregate\_payment]{\rm{ }} = E[{N^L}] \times E[{Y^L}]\)
\({N^L}\) berdistribusi binomial negatif dengan parameter \(\tau = 2\) , \(\beta = 1,5\) dan \({Y^L}\) berdistribusi gamma
\(E[aggregate\_payment]{\rm{ }} = E[{N^L}] \times E[{Y^L}]\)
\(E[aggregate\_payment]{\rm{ }} = \tau \beta [E[X \wedge m] – E[X \wedge d]]\)
\(E[aggregate{\rm{\_ }}payment]{\rm{ }} = 2(1,{\rm{ }}5)[134,{\rm{ }}8347113 – 48,{\rm{ }}36734] = 259,401\)