The random variable [math]X[/math] has survival function:

[[math]] S_X(x) = \frac{\theta^4}{(\theta^2 + x^2)^2} [[/math]]

Two values of [math]X[/math] are observed to be 2 and 4. One other value exceeds 4.

Calculate the maximum likelihood estimate of [math]\theta[/math].

• Less than 4.0
• At least 4.0, but less than 4.5
• At least 4.5, but less than 5.0
• At least 5.0, but less than 5.5
• At least 5.5

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

Key: E

[[math]] \begin{aligned} f(x) &= − S^{'}( x) = \frac{4x\theta^4}{(\theta^2 + x^2)^3} \\ L(\theta) &= f (2) f (4) S (4) = \frac{4(2)\theta^4}{(\theta^2 + 2^2)^3} \frac{4(4)\theta^4}{(\theta^2 + 4^2)^3} \frac{\theta^4}{(\theta^2 + 4^2)^2} = \frac{128\theta^{12}}{(\theta^2+4)^3(\theta^2 + 16)^5} \\ l (\theta ) &= \ln128 + 12 \ln(\theta) − 3\ln(\theta^ 2 + 4) − 5\ln(\theta^ 2 + 16) \\ l^{'}(\theta) &= \frac{12}{\theta} - \frac{6\theta}{\theta^2 + 4} - \frac{10\theta}{\theta^2 + 4} - \frac{10\theta}{\theta^2 + 16} = 0 \\ &12(\theta^4 + 20\theta^ 2 + 64) − 6(\theta^4 + 16 \theta^ 2 ) − 10(\theta^ 4 + 4 \theta^ 2 ) = 0 \\ 0 &= −4\theta^ 4 + 104 \theta^ 2 + 768 = \theta^ 4 − 26 \theta^2 − 192 \\ \theta^2 &= \frac{26 \pm \sqrt{26^2 + 4(192)}}{2} = 32 \\ \theta &= 5.657 \end{aligned} [[/math]]

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