Independent solution

How to solve this Exponential Distribution question

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)}}