IAI Actuarial Core Principles · Actuarial Statistics · Hypothesis testing and goodness of fit
A random sample of n = 50 observations from an exponential distribution with mean 1/λ has sample mean 2.5. A likelihood ratio test of H0: λ = 0.5 against H1: λ unrestricted is used. What is −2 log Λ (to 2 decimal places)?
The statistic is about 2.69, found by doubling the difference between the maximised log-likelihood at λ=0.4 and that at λ=0.5.
- A0.46
- B0.93Correct
- C1.86
- D3.20
- 0.19
Explanation
MLE of λ is 1/2.5 = 0.4. Log-likelihood = n ln λ − λ Σx, with Σx = 125. At 0.4: 50 ln0.4 − 50 = −45.8145−50 = −95.8145. At 0.5: 50 ln0.5 − 62.5 = −34.6574−62.5 = −97.1574. Difference = 1.3429, times 2 = 2.69. Check: 2n[(λ0/λ̂ −1) − ln(λ0/λ̂)] = 100[0.25 − ln1.25] = 100[0.25−0.22314]=2.69.
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