exercise:Ef20ccd421

Jan 15'24

Exercise

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
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300
400
500

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ABy Admin

Jan 15'24

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]]