Pembahasan Ujian PAI: A50 – No. 7 – April 2019

Pembahasan Soal Ujian Profesi Aktuaris

SOAL

Diberikan sebuah persamaan dalam proses ARMA (3,2) sebagai berikut

\({y_t} = 0,5{y_{t – 1}} + 0,3{y_{t – 2}} + 0,2{y_{t – 3}} + 4,2 + {\varepsilon _t} – 0,4{\varepsilon _{t – 1}} + 0,1{\varepsilon _{t – 2}}\)

Jika diberikan nilai \(\sigma _\varepsilon ^2 = 1,25\), hitunglah \({\gamma _0}\) dan \({\rho _1}\)

1. \({\gamma _0} = 1,267\) dan \({\rho _1} = 0,133\)
2. \({\gamma _0} = 1,131\) dan \({\rho _1} = 0,105\)
3. \({\gamma _0} = 1,267\) dan \({\rho _1} = 0,105\)
4. \({\gamma _0} = 1,131\) dan \({\rho _1} = 0,133\)
5. \({\gamma _0} = 1,131\) dan \({\rho _1} = 0,119\)

Diketahui	ARMA (3,2) dengan model \({y_t} = 0,5{y_{t – 1}} + 0,3{y_{t – 2}} + 0,2{y_{t – 3}} + 4,2 + {\varepsilon _t} – 0,4{\varepsilon _{t – 1}} + 0,1{\varepsilon _{t – 2}}\)   dengan \(\sigma _\varepsilon ^2 = 1,25\) Dengan bentuk umum  \({y_t} = {\phi _1}{y_{t – 1}} + {\phi _2}{y_{t – 2}} + {\phi _3}{y_{t – 3}} + \delta + {\varepsilon _t} + {\theta _1}{\varepsilon _{t – 1}} + {\theta _2}{\varepsilon _{t – 2}}\)
Rumus yang digunakan	Rumus Umum Varians dan Kovarians Model ARMA (p,q) \({y_t} = {\phi _1}{y_{t – 1}} + {\phi _2}{y_{t – 2}} + {\phi _3}{y_{t – 3}} + \delta + {\varepsilon _t} + {\theta _1}{\varepsilon _{t – 1}} + {\theta _2}{\varepsilon _{t – 2}}\) \(\rho \left( k \right) = \frac{{\gamma \left( k \right)}}{{\gamma \left( 0 \right)}}\) dan \(\gamma \left( i \right) = \sum\limits_{j = 1}^p {{\phi _j}} \gamma \left( {\left| {\tau – i} \right|} \right) + {\sigma ^2}\sum\limits_{k = i}^q {{\theta _j}{\psi _{j – i}}} \)   dengan   \({\psi _j} = \phi _1^{j – 1}\left( {{\phi _1} + {\theta _1}} \right)\) Sumber: http://www.maths.qmul.ac.uk/~bb/TimeSeries/TS_Chapter6_2_1.pdf
Proses pengerjaan	Dari rumus umum Varians dan Kovarians Model ARMA (3,2) diperoleh \(\gamma \left( 0 \right) = {\phi _1}\gamma \left( 1 \right) + {\phi _2}\gamma \left( 2 \right) + {\phi _3}\gamma \left( 3 \right) + \sigma _\varepsilon ^2 + {\theta _1}\left( {{\phi _1} + {\theta _1}} \right)\sigma _\varepsilon ^2 + {\theta _2}{\phi _1}\left( {{\phi _1} + {\theta _1}} \right)\sigma _\varepsilon ^2\) \(\gamma \left( 1 \right) = {\phi _1}\gamma \left( 0 \right) + {\phi _2}\gamma \left( 1 \right) + {\phi _3}\gamma \left( 2 \right) + {\theta _1}\sigma _\varepsilon ^2 + {\theta _2}\left( {{\phi _1} + {\theta _1}} \right)\sigma _\varepsilon ^2\) \(\gamma \left( 2 \right) = {\phi _1}\gamma \left( 1 \right) + {\phi _2}\gamma \left( 0 \right) + {\phi _3}\gamma \left( 1 \right) + {\theta _2}\sigma _\varepsilon ^2\) \(\gamma \left( 3 \right) = {\phi _1}\gamma \left( 2 \right) + {\phi _2}\gamma \left( 1 \right) + {\phi _3}\gamma \left( 0 \right)\) Kemudian lakukan proses substitusi diperoleh Karena nilai penyebut \(\left( {{\phi _1} + {\phi _2} + {\phi _3} – 1} \right)\left( {1 + {\phi _1} – {\phi _2} + {\phi _3}} \right)\left( {1 + {\phi _1}{\phi _3} + {\phi _2} – \phi _3^2} \right)\) \(= \left( {0,5 + 0,3 + 0,2 – 1} \right)\left( {1 + 0,5 – 0,3 + 0,2} \right)\left( {1 + 0,5 \cdot 0,2 + 0,3 – {{0,2}^2}} \right)\) \(= \left( 0 \right)\left( {1,4} \right)\left( {1,36} \right)\) \(= 0\) Maka tidak ada jawaban yang tepat.
Jawaban	Anulir