Pembahasan Ujian PAI: A20 - No. 18 - Juni 2016

Pembahasan Soal Ujian Profesi Aktuaris

Institusi	:	Persatuan Aktuaris Indonesia (PAI)
Mata Ujian	:	A20 – Probabilitas dan Statistika
Periode Ujian	:	Juni 2016
Nomor Soal	:	18

SOAL

Misalkan\(f({x_1},{x_2}) = 21x_1^2x_2^3\,,\,0 < {x_1} < {x_2} < 1\) , dan 0 untuk lainnya, adalah fungsi kepadatan peluang gabungan antara \({X_1}\) dan \({X_2}\) . Berapa rataan dari \({X_1}\) kondisional diberikan \({X_2} = {x_2}\,,\,0 < {x_2} < 1\,?\)

21/32
11/32
3/8
23/35
1/7

Kunci Jawaban & Pembahasan

Rumus	\(Rataan\,bersyarat\,{X_2} = {x_2}\,\) \(E[E[{X_1}\|{X_2} = {x_2}]] = E[{X_1}]\)
Step 1	\(f[{x_1}] = \int\limits_{{x_1}}^1 {21x_1^2x_2^3\,d} {x_2}\) \(f[{x_1}] = 21x_1^2\left( {\frac{{1 – x_1^4}}{4}} \right)\) \(f[{x_1}] = \frac{{21}}{4}\left( {x_1^2 – x_1^6} \right)\)
Step 2	\(E[{X_1}] = \int\limits_0^1 {{x_1}\left( {\frac{{21}}{4}\left( {x_1^2 – x_1^6} \right)} \right)} \,d{x_1}\) \(E[{X_1}] = \frac{{21}}{4}\int\limits_0^1 {\left( {x_1^3 – x_1^7} \right)} d{x_1}\) \(E[{X_1}] = \frac{{21}}{4}\left( {\frac{1}{4} – \frac{1}{8}} \right)\) \(E[{X_1}] = \frac{{21}}{4}\left( {\frac{1}{8}} \right)\) \(E[{X_1}] = \frac{{21}}{{32}}\)
Jawaban	a. 21/32

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