Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
A20 – Probabilitas dan Statistika |
| Periode Ujian |
: |
Maret 2016 |
| Nomor Soal |
: |
28 |
SOAL
Misalkan X adalah suatu distribusi Poisson dengan parameter m. Jika m ialah suatu nilai eksperimen dari suatu peubah acak yang berdistribusi Gamma \((\alpha = 2,\beta = 1)\) , hitung Note : Cari suatu ekspresi yang menyatakan peluang gabungan dari X dan m. Kemudian cari bentuk integral m untuk menghitung distribusi marginal dari X.
- 11/16
- 7/16
- 2/9
- 2/7
- 13/16
PEMBAHASAN
| Diketahui |
X|M ~ Poisson (m)
M ~ Gamma \((\alpha = 2,\beta = 1)\) |
| Step 1 |
\({f_X}(x) = \int\limits_0^\infty {{P_{X|M}}(x|m)\,\,\,{f_M}(m)\,dm} \)
\({f_X}(x) = \int\limits_0^\infty {\frac{{{e^{ – m}}{m^x}}}{{x!}}\,\,\,\frac{{{m^{\alpha – 1}}{e^{ – \frac{m}{\beta }}}}}{{{\beta ^\alpha }\Gamma (\alpha )}}\,dm} \)
\({f_X}(x) = \frac{1}{{x!{\beta ^\alpha }\Gamma (\alpha )}}\int\limits_0^\infty {{e^{ – m\left( {1 + \frac{1}{\beta }} \right)}}{m^{x + \alpha – 1}}\,dm} \)
\(Subsitusi\,:\,t = m(1 + \frac{1}{\beta })\,\,;\,\,\frac{t}{{(1 + \frac{1}{\beta })}} = m\,\,;\,\,dm = \frac{{dt}}{{(1 + \frac{1}{\beta })}}\)
\({f_X}(x) = \frac{1}{{x!{\beta ^\alpha }\Gamma (\alpha )}}\int\limits_0^\infty {{e^{ – t}}{{\left( {\frac{t}{{1 + \frac{1}{\beta }}}} \right)}^{x + \alpha – 1}}\,\left( {\frac{{dt}}{{(1 + \frac{1}{\beta })}}} \right)} \)
\({f_X}(x) = \frac{1}{{x!{\beta ^\alpha }\Gamma (\alpha )}}{\left( {\frac{1}{{1 + \frac{1}{\beta }}}} \right)^{x + \alpha }}\int\limits_0^\infty {{e^{ – t}}{t^{x + \alpha – 1}}\,dt} \)
\({f_X}(x) = \frac{1}{{x!{\beta ^\alpha }\Gamma (\alpha )}}{\left( {\frac{1}{{1 + \frac{1}{\beta }}}} \right)^{x + \alpha }}\Gamma (x + \alpha )\) |
| Step 2 |
\((\alpha = 2,\beta = 1)\)
\({f_X}(x) = \frac{1}{{x!{1^2}\Gamma (2)}}{\left( {\frac{1}{{1 + \frac{1}{1}}}} \right)^{x + 2}}\Gamma (x + 2)\)
\({f_X}(x) = \frac{1}{{x!}}{\left( {\frac{1}{2}} \right)^{x + 2}}\Gamma (x + 2)\)
\({f_X}(x) = \frac{1}{{x!}}{\left( {\frac{1}{2}} \right)^{x + 2}}(x + 1)!\)
\({f_X}(x) = \frac{{x + 1}}{{{2^{x + 2}}}}\) |
| Step 3 |
\({f_X}(0) = \frac{{0 + 1}}{{{2^{0 + 2}}}} = \frac{1}{4}\)
\({f_X}(1) = \frac{{1 + 1}}{{{2^{1 + 2}}}} = \frac{2}{8}\)
\({f_X}(2) = \frac{{2 + 1}}{{{2^{2 + 2}}}} = \frac{3}{{16}}\) |
| Step 4 |
\({f_X}(X = 0,1,2) = \frac{1}{4} + \frac{2}{8} + \frac{3}{{16}}\)
\({f_X}(X = 0,1,2) = \frac{{11}}{{16}}\) |
| Jawaban |
a. 11/16 |
