Independent solution

How to solve this Exponential Distribution question

Answer in brief

This is an exponential memorylessness problem. Conditioning on survival to the stated age shifts the lifetime but leaves the residual variance equal to the square of the nine-year mean, so choice C is correct.

Setup

Setup

Let T denote the lifetime and condition on the information that T has reached age five. An exponential variable with mean nine has rate 1/9.

T∼Exp⁡(1/9),Var⁡(T)=92T\sim\operatorname{Exp}(1/9),\qquad \operatorname{Var}(T)=9^2

Model

Model

The memoryless property says that the additional lifetime beyond age five has the original exponential distribution.

T∣(T≥5)=d5+Y,Y∼Exp⁡(1/9)T\mid(T\ge 5)\overset{d}{=}5+Y,\qquad Y\sim\operatorname{Exp}(1/9)

Compute

Compute

Adding a constant changes the conditional mean but not the variance. The variance therefore comes entirely from the residual variable Y.

Var⁡(T∣T≥5)=Var⁡(5+Y)\operatorname{Var}(T\mid T\ge5)=\operatorname{Var}(5+Y)
Var⁡(5+Y)=Var⁡(Y)=1(1/9)2=81\operatorname{Var}(5+Y)=\operatorname{Var}(Y)=\frac{1}{(1/9)^2}=81

Answer

Answer

The conditional lifetime variance is 81 square years.

81(C)\boxed{81\quad\text{(C)}}