Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
A20 – Probabilitas dan Statistika |
| Periode Ujian |
: |
Maret 2016 |
| Nomor Soal |
: |
11 |
SOAL
Misalkan X adalah peubah acak Poisson dengan \(E[X] = \ln \,2\) . Hitung \(E[cos(\pi X)]\)
- 0
- ¼
- ½
- 1
- 2 ln2
PEMBAHASAN
| Diketahui |
\(X\, \sim Poisson\,(\lambda )\)
\(E[X] = \lambda = ln2\) |
| Kalkulasi |
\(P(X = x) = \frac{{{e^{ – \ln 2}}{{\left( {\ln 2} \right)}^x}}}{{x!}}\)
\(E[cos(\pi X)] = \sum\limits_{x = 0}^\infty {\cos (\pi x)} \frac{{{e^{ – \ln 2}}{{\left( {\ln 2} \right)}^x}}}{{x!}}\)
\(E[cos(\pi X)] = {e^{ln2}}^{^{ – 1}}\sum\limits_{x = 0}^\infty {\frac{{{{\left( {\ln 2} \right)}^x}}}{{x!}}} \left( {1 – 1 + 1 – 1 + …} \right)\)
\(E[cos(\pi X)] = {2^{ – 1}}\sum\limits_{x = 0}^\infty {\frac{{{{\left( {\ln 2} \right)}^x}{{\left( { – 1} \right)}^x}}}{{x!}}} \)
\(E[cos(\pi X)] = \frac{1}{2}\sum\limits_{x = 0}^\infty {\frac{{{{\left( { – \ln 2} \right)}^x}}}{{x!}}} \)
\(E[cos(\pi X)] = \frac{1}{2}{e^{ – \ln 2}}\)
\(E[cos(\pi X)] = \frac{1}{2}\left( {\frac{1}{2}} \right)\)
\(E[cos(\pi X)] = \frac{1}{4}\) |
| Jawaban |
b. ¼ |
