1. [math]X[/math] and [math]Y[/math] are Poisson distributed.
2. The first moment of [math]X[/math] is less than the first moment of [math]Y[/math] by 8.
3. The second moment of [math]X[/math] is 60% of the second moment of [math]Y[/math].

Calculate the variance of [math]Y[/math].

• 4
• 12
• 16
• 27
• 35

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

Solution: E

Let a be the mean and variance of [math]X[/math] and b be the mean and variance of [math]Y[/math]. The two facts are [math]a = b– 8 [/math] and [math]a + a_2 = 0.6(b + b_2)[/math]. Substituting the first equation into the second gives

[[math]] \begin{align*} b − 8 + (b − 8) &= 0.6b + 0.6b^2 \\ b − 8 + b 2 − 16b + 64 &= 0.6b + 0.6b^2 \\ 0.4b^2 − 15.6b + 56 &= 0 \\ b &= \frac{15.6 ± 15.62 − 4(0.4)(56)}{2(0.4)} \\ &= \frac{15.6 ± 12.4}{0.8}. \end{align*} [[/math]]

At [math]b = 4[/math], [math]a[/math] is negative, so the answer is 35.

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