Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Probabilitas dan Statistika |
| Periode Ujian |
: |
November 2018 |
| Nomor Soal |
: |
20 |
SOAL
Misal \({X_1},{X_2},{X_3}\) adalah sebuah sampel acak yang saling bebas (mutually independent) dari suatu distribusi diskret dengan fungsi peluang sebagai berikut :
\(p(x) = \left\{ \begin{array}{l} \frac{1}{3},\,x = 0\\ \frac{2}{3},\,x = 1\\ 0,\,\,\,lainnya \end{array} \right.\)
Tentukanlah fungsi pembangkit momen, M(t) dari \(Y = {X_1},{X_2},{X_3}\)
- \(\frac{{19}}{{27}} + \frac{8}{{27}}{e^t}\)
- \(1 + 2{e^t}\)
- \({\left( {\frac{1}{3} + \frac{2}{3}{e^t}} \right)^3}\)
- \(\frac{1}{{27}} + \frac{8}{{27}}{e^{3t}}\)
- \(\frac{1}{3} + \frac{2}{3}{e^{3t}}\)
| Step 1 |
\(Y = {X_1},{X_2},{X_3}\)
\(P(Y = 1) = P({X_1},{X_2},{X_3} = 1)\)
\(P(Y = 1) = P({X_1} = 1)P({X_2} = 1)P({X_3} = 1)\)
\(P(Y = 1) = \frac{2}{3}\frac{2}{3}\frac{2}{3}\)
\(P(Y = 1) = \frac{8}{{27}}\) |
| Step 2 |
\(P(Y = 0) = 1 – P(Y = 1)\)
\(P(Y = 0) = 1 – \frac{8}{{27}}\)
\(P(Y = 0) = \frac{{19}}{{27}}\) |
| Maka |
\(M(t) = E[{e^{tY}}]\)
\(M(t) = \sum {{e^{tY}}} P(Y = y)\)
\(M(t) = {e^0}\frac{{19}}{{27}} + {e^t}\frac{8}{{27}}\)
\(M(t) = \frac{{19}}{{27}} + \frac{8}{{27}}{e^t}\) |
| Jawaban |
a. \(\frac{{19}}{{27}} + \frac{8}{{27}}{e^t}\) |
