Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Besar kerugian pada suatu polis asuransi TPL mengikuti distribusi gabungan dari distribusi eksponensial dengan rata-rata 5 dan distribusi eksponensial dengan rata-rata . Rata-rata besar kerugian ialah 7,5 dan variansi dari besar kerugian ialah Tentukan koefisiensi skewness dari distribusi kerugian tersebut.
- 7,3210
- 3,3209
- 1,7923
- 1,4231
- 2,7063
| Diketahui |
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| Rumus yang digunakan | Skewness: \({\gamma _1} = \frac{{{\mu _3}}}{{{\sigma ^3}}} = \frac{{E\left[ {{X^3}} \right] – 3E\left[ {{X^2}} \right]\mu + 2{\mu ^3}}}{{Var{{\left( X \right)}^{\frac{3}{2}}}}}\)
\({\mu ‘_k} = E\left[ {{X^k}} \right] = \sum\nolimits_j {x_j^kp\left( {{x_j}} \right)} \)
\({\mu _3} = \sum\nolimits_j {{{\left( {{x_j} – \mu } \right)}^3}p\left( {{x_j}} \right)} = \sum\nolimits_j {\left( {{x_j}^3 – 3{x_j}^2\mu + 3{x_j}{\mu ^2} – {\mu ^3}} \right)p\left( {{x_j}} \right)} \)
\({\mu _3} = {{\mu ‘}_k} – 3{{\mu ‘}_2}\mu + 3{\mu ^3} – {\mu ^3}\)
\(Var\left( X \right) = E\left[ {{X^2}} \right] – {\left( {E\left[ X \right]} \right)^2}\)
Ekponensial: \(E\left[ X \right] = \theta \), \(E\left[ {{X^2}} \right] = 2{\theta ^2}\) dan \(E\left[ {{X^3}} \right] = {\theta ^3} \cdot 3! = 6{\theta ^3}\) Mixture function: \(f\left( x \right) = \sum\limits_{i = 1}^n {{w_i}f\left( {{x_i}} \right)} \) dengan \(\sum\limits_{i = 1}^n {{w_i}} = 1\) Mixture 2 Eksponensial: \(E\left[ X \right] = {w_1}{\theta _1} + {w_2}{\theta _2}\), \(E\left[ {{X^2}} \right] = 2\theta _1^2{w_1} + 2\theta _2^2{w_2}\), dst |
| Proses pengerjaan | Misalkan \(w\) adalah proporsi distribusi ekponensial dengan rata-rata \(\theta \), Maka \(E\left[ {{X^2}} \right]Var\left( X \right) + {\left( {E\left[ X \right]} \right)^2} = 75 + {7.5^2} = 131.25\) |
| \(E\left[ X \right] = {w_1}{\theta _1} + {w_2}{\theta _2}\) \(7.5 = \left( {1 – w} \right)5 + w\theta \) \(w\left( {\theta – 5} \right) = 2.5\) ——————– (1) | |
| \(E\left[ {{X^2}} \right] = 2\theta _1^2{w_1} + 2\theta _2^2{w_2}\) \(131.25 = 50\left( {1 – w} \right) + 2{\theta ^2}w\) \(2w\left( {{\theta ^2} – 25} \right) = 81.25\) ——————— (2) | |
| Dengan membagi (2) dengan (1) diperoleh \(\frac{{2w\left( {\theta – 5} \right)\left( {\theta + 5} \right)}}{{w\left( {\theta – 5} \right)}} = \frac{{81.25}}{{2.5}}\) \(2\left( {\theta + 5} \right) = 32.5\) \(\theta = \frac{{32.5}}{2} – 5 = 11.25\) |
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| Substitusikan nilai \(\theta = 11.25\) ke persamaan (1) \(w\left( {\theta – 5} \right) = 2.5\) \(w\left( {11.25 – 5} \right) = 2.5\) \(w = 0.4\) |
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| Diperoleh \(E\left[ {{X^3}} \right] = 6\theta _1^3{w_1} + 6\theta _2^3{w_2} = 6\theta _1^3\left( {1 – w} \right) + 6\theta _1^3w\) \(E\left[ {{X^3}} \right] = 6\left( {{5^3}} \right)\left( {0.6} \right) + 6\left( {{{11.25}^3}} \right)\left( {0.4} \right) = 3867.1875\) Jadi, |
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| Jawaban | e. 2,7063 |