Pembahasan Ujian PAI: A50 - No. 22 - November 2014

Pembahasan Soal Ujian Profesi Aktuaris

Institusi	:	Persatuan Aktuaris Indonesia (PAI)
Mata Ujian	:	Metoda Statistika
Periode Ujian	:	November 2014
Nomor Soal	:	22

SOAL

Sekumpulan dari \(n\) orang diamati sampai semuanya meninggal, dengan kematian dikelompokkan dalam interval yang tetap. Jika \(Var\left[ {\hat S\left( t \right)} \right] = 0,0009\) , \(Var\left[ {\hat S\left( r \right)} \right] = 0,0016\) , dan covarians-nya adalah \(Cov\left[ {\hat S\left( t \right),\hat S\left( r \right)} \right] = 0,0008\) . Tentukan \(E\left[ {\hat S\left( t \right)} \right]\)

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Tidak ada jawaban yang benar

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Diketahui	Sekumpulan dari \(n\) orang diamati sampai semuanya meninggal, dengan kematian dikelompokkan dalam interval yang tetap. \(Var\left[ {\hat S\left( t \right)} \right] = 0,0009\) \(Var\left[ {\hat S\left( r \right)} \right] = 0,0016\) covarians-nya adalah \(Cov\left[ {\hat S\left( t \right),\hat S\left( r \right)} \right] = 0,0008\)
Rumus yang digunakan	\(E\left[ {\hat S\left( t \right)} \right] = S\left( t \right)\) \(Var\left[ {\hat S\left( t \right)} \right] = \frac{{S\left( t \right)F\left( t \right)}}{n} = \frac{{S\left( t \right)\left( {1 – S\left( t \right)} \right)}}{n}\) \(Cov\left[ {\hat S\left( t \right),\hat S\left( r \right)} \right] = \frac{{F\left( t \right)S\left( r \right)}}{n}\)
Proses pengerjaan	\(Cov\left[ {\hat S\left( t \right),\hat S\left( r \right)} \right] = \frac{{F\left( t \right)S\left( r \right)}}{n} \Leftrightarrow 0.0008n = F\left( t \right)S\left( r \right)\) \(Var\left[ {\hat S\left( t \right)} \right] = \frac{{S\left( t \right)F\left( t \right)}}{n} \Leftrightarrow 0.0009n = S\left( t \right)F\left( t \right)\) \(Var\left[ {\hat S\left( r \right)} \right] = \frac{{S\left( r \right)F\left( r \right)}}{n} \Leftrightarrow 0.0016n = S\left( r \right)F\left( r \right)\) \(\frac{{F\left( t \right)S\left( r \right)}}{{0.0008}} = \frac{{S\left( r \right)F\left( r \right)}}{{0.0016}} \Leftrightarrow \frac{{F\left( t \right)}}{{F\left( r \right)}} = \frac{8}{{16}}\) \(\frac{{F\left( t \right)S\left( r \right)}}{{0.0008}} = \frac{{S\left( t \right)F\left( t \right)}}{{0.0009}} \Leftrightarrow \frac{{S\left( t \right)}}{{S\left( r \right)}} = \frac{9}{8}\) \(Cov\left[ {\hat S\left( t \right),\hat S\left( r \right)} \right] = \frac{{F\left( t \right)S\left( r \right)}}{n}\) \(0.0008n = \left[ {\frac{{0.0009n}}{{S\left( t \right)}}} \right] \cdot \left[ {\frac{{0.0016n}}{{F\left( r \right)}}} \right]\) \(S\left( t \right)F\left( r \right) = \frac{9}{8}\left( {0.0016n} \right)\) \(S\left( t \right)F\left( r \right) = 0.0018n\) \(\left[ {1 – F\left( t \right)} \right]\left[ {1 – S\left( r \right)} \right] = 0.0018n\) \(1 – F\left( t \right) – S\left( r \right) + F\left( t \right)S\left( r \right) = 0.0018n\) \(S\left( t \right) – \frac{8}{9}S\left( t \right) + 0.0008n = 0.0018n\) \(S\left( t \right) = 9\left( {0.0018n – 0.0008n} \right)\) \(S\left( t \right) = 0.009n\) \(0.0009n = S\left( t \right)F\left( t \right)\) \(0.0009n = 0.009n\left( {1 – 0.009n} \right)\) \(n = \frac{{0.009n – \left( {0.009n} \right)\left( {0.009n} \right)}}{{0.0009}}\) \(n = 10n – 0.09{n^2}\) \(0.09{n^2} = 9n\) \(n = \frac{9}{{0.09}} = 100\) \(E\left[ {\hat S\left( t \right)} \right] = S\left( t \right) = 0.009n = 0.009\left( {100} \right) = 0.9\)
Jawaban	D. 0,9