Scientists are searching for a vaccine for a disease. You are given:

(i) 100,000 lives age [math]x[/math] are exposed to the disease

(ii) Future lifetimes are independent, except that the vaccine, if available, will be given to all at the end of year 1

(iii) The probability that the vaccine will be available is 0.2

(iv) For each life during year [math]1, q_{x}=0.02[/math]

(v) For each life during year 2, [math]q_{x+1}=0.01[/math] if the vaccine has been given, and [math]q_{x+1}=0.02[/math] if it has not been given

Calculate the standard deviation of the number of survivors at the end of year 2.

• 100
• 200
• 300
• 400
• 500

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Answer: D

This is a mixed distribution for the population, since the vaccine will apply to all once available.

Available?

[math](A)[/math]	[math]\operatorname{Pr}(A)[/math]	[math]{ }_{2} p \mid A[/math]	[math]E(S \mid A)[/math]	[math]\operatorname{Var}(S \mid A)[/math]	[math]E\left(S^{2} \mid A\right)[/math]
Yes	0.2	0.9702	97,020	2,891	[math]9,412,883,291[/math]
No	0.8	0.9604	96,040	3,803	[math]9,223,685,403[/math]
			[math]E(S)[/math]		[math]E\left(S^{2}\right)[/math]
			96,236		[math]9,261,524,981[/math]
		[math]\operatorname{Var}(S)[/math]	157,285
		[math]S D(S)[/math]	397

As an example, the formulas for the "No" row are

[math]\operatorname{Pr}(\mathrm{No})=1-0.2=0.8[/math]

[math]{ }_{2} p[/math] given [math]\mathrm{No}=(0.98[/math] during year 1[math])(0.98[/math] during year 2[math])=0.9604[/math]

[math]E(S \mid \mathrm{No}), \operatorname{Var}(S \mid[/math] No [math])[/math] and [math]E\left(S^{2} \mid\right.[/math] No [math])[/math] are just binomial, [math]n=100,000 ; \mathrm{p}([/math] success [math])=0.9604[/math]

[math]E(S), E\left(S^{2}\right)[/math] are weighted averages,

[math]\operatorname{Var}(S)=E\left(S^{2}\right)-E(S)^{2}[/math]

Or, by the conditional variance formula:

[[math]] \begin{aligned} \operatorname{Var}(S) & =\operatorname{Var}[E(S \mid A)]+E[\operatorname{Var}(S \mid A)] \\ & =0.2(0.8)(97,020-96,040)^{2}+0.2(2,891)+0.8(3,803) \\ & =153,664+3,621=157,285 \\ \operatorname{StdDev}(S) & =397 \end{aligned} [[/math]]

Copyright 2024 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.