Losses come from a mixture of an exponential distribution with mean 100 with probability p and an exponential distribution with mean 10,000 with probability 1 − p.

Losses of 100 and 2000 are observed.

Determine the likelihood function of p.

• [math]\left (\frac{pe^{-1}}{100} \frac{(1-p)e^{-0.01}}{10,000}\right) \left( \frac{pe^{-20}}{100}\frac{(1-p)e^{-0.2}}{10,000}\right)[/math]
• [math]\left (\frac{pe^{-1}}{100} \frac{(1-p)e^{-0.01}}{10,000}\right) + \left( \frac{pe^{-20}}{100}\frac{(1-p)e^{-0.2}}{10,000}\right)[/math]
• [math]\left (\frac{pe^{-1}}{100}+ \frac{(1-p)e^{-0.01}}{10,000}\right) \left( \frac{pe^{-20}}{100} + \frac{(1-p)e^{-0.2}}{10,000}\right)[/math]
• [math]\left (\frac{pe^{-1}}{100} +\frac{(1-p)e^{-0.01}}{10,000}\right) + \left( \frac{pe^{-20}}{100}+\frac{(1-p)e^{-0.2}}{10,000}\right)[/math]
• [math]p\left (\frac{pe^{-1}}{100} +\frac{(1-p)e^{-0.01}}{10,000}\right) + (1-p)\left( \frac{pe^{-20}}{100}+\frac{(1-p)e^{-0.2}}{10,000}\right)[/math]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Key: C

[[math]]\begin{aligned} f(x) & =p \frac{1}{100} e^{-x / 100}+(1-p) \frac{1}{10,000} e^{-x / 10,000} \\ L(100,200) & =f(100) f(2000) \\ & =\left(\frac{p e^{-1}}{100}+\frac{(1-p) e^{-0.01}}{10,000}\right)\left(\frac{p e^{-20}}{100}+\frac{(1-p) e^{-0.2}}{10,000}\right) \end{aligned}[[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.