Pembahasan Ujian PAI: A70 – No. 13 – 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
Misal S = agregat pembayaran klaim “per-loss basis“

Distribusi gamma dengan parameter \(\alpha = 2\) dan \(\theta = 100\) , diketahui bawha:
\(E[{(X \wedge d)^2}]{\rm{ }} = \frac{{{\theta ^2}\Gamma (\alpha + 2)}}{{\Gamma (\alpha )}}\Gamma \left( {\alpha + 2;\frac{d}{\theta }} \right) + {d^2}\left[ {1 – \Gamma \left( {\alpha ;\frac{d}{\theta }} \right)} \right]\)
\(E[{(X \wedge d)^2}]{\rm{ }} = \frac{{{{100}^2}\Gamma (4)}}{{\Gamma (2)}}\Gamma \left( {4;\frac{d}{{100}}} \right) + {d^2}\left[ {1 – \Gamma \left( {2;\frac{d}{{100}}} \right)} \right]\)
\(E[{(X \wedge d)^2}]{\rm{ }} = 6{(100)^2}\Gamma \left( {4;\frac{d}{{100}}} \right) + {d^2}\left[ {1 – \Gamma \left( {2;\frac{d}{{100}}} \right)} \right]\)
\(\Gamma \left( {4;\frac{d}{{100}}} \right) = \frac{1}{{\Gamma (4)}}\int\limits_0^{\frac{d}{{100}}} {{t^3}{e^{ – t}}dt = } \frac{1}{6}\int\limits_0^{\frac{d}{{100}}} {{t^3}{e^{ – t}}dt} \)
\(\Gamma \left( {2;\frac{d}{{100}}} \right) = \frac{1}{{\Gamma (2)}}\int\limits_0^{\frac{d}{{100}}} {t{e^{ – t}}dt = } \int\limits_0^{\frac{d}{{100}}} {t{e^{ – t}}dt} \)
\(\Gamma \left( {3;\frac{{175}}{{100}}} \right) = 0,25603\)
\(\Gamma \left( {2;\frac{{175}}{{100}}} \right) = 0,52212165\)

\(\Gamma \left( {3;\frac{{50}}{{100}}} \right) = 0,0143877\)
\(\Gamma \left( {2;\frac{{50}}{{100}}} \right) = 0,090204\)
\(E[{\left( {X \wedge 175} \right)^2}]{\rm{ }} = 6{(100)^2}\Gamma \left( {4;\frac{{175}}{{100}}} \right) + {175^2}\left[ {1 – \Gamma \left( {2;\frac{{175}}{{100}}} \right)} \right] = 20.683,64547\)
\(E[{\left( {X \wedge 50} \right)^2}]{\rm{ }} = 6{(100)^2}\Gamma \left( {4;\frac{{50}}{{100}}} \right) + {50^2}\left[ {1 – \Gamma \left( {2;\frac{{50}}{{100}}} \right)} \right] = 2.379,58738\)

Dari soal nomor 12 diperoleh \(E[{Y^L}]{\rm{ }} = 134,{\rm{ }}8347113 – 48,{\rm{ }}36734 = 86,{\rm{ }}4673713\)
\(E[{({Y^L})^2}] = E[{(X \wedge 175)^2}] – E[{(X \wedge 50)^2}] – 2(50)[E[X \wedge 175] – E[X \wedge 50]]\)
\(E[{({Y^L})^2}] = 20.683,64547 – 2.379,58738 – 100(86,4673713) = 9.657,32096\)
\(Var[{Y^L}]{\rm{ }} = E[{({Y^L})^2}] – {(E[{Y^L}])^2} = 9.657,32096 – {(86,4673713)^2} = 2.180,71466\)
\(Var(S){\rm{ }} = E[{N^L}]Var[{Y^L}]{\rm{ }} + Var[{N^L}]{(E[{Y^L}])^2}\) \(= 3(2.180,71466){\rm{ }} + 7,{\rm{ }}5{(86,4673713)^2} = 62.616,69 \approx 62.616\)