the expected amount of time that the 3:00 appointment spends at the doctor’s office.

Question

A doctor has scheduled two appointments, one at 2:00 P.M. and the other at 3:00 P.M. The amounts of time that appointments last are independent exponential random variables with mean 60 minutes. Assuming that both patients are on time, find the expected amount of time that the 3:00 appointment spends at the doctor’s office.

I tried to find my own solution by using conditional expectation where :

$T_1 (\text{time appointment 2:00 P.M leaves before 3:00 P.M})$

$T_2 (\text{time appointment 2:00 P.M leaves after 3:00 P.M})$

$\mathbb{E}(T_1)=\mathbb{E}(T_2)=60.$

My solution is:

$$\mathbb{E}(\text{T spent time})=\mathbb{E}(T{|}T_1\le 60)\mathbb{P}(T_1\le 60)+\mathbb{E}(T{|}T_2\ge 60)\mathbb{P}(T_2\ge 60)$$ $$=\mathbb{E}(T_1)\int _0^{60}\:\frac{1}{60}e^{-\frac{1}{60}t_1}dt_1+\mathbb{E}(T_2)\int _{60}^{\infty }\frac{1}{60}e^{-\frac{1}{60}t_2}dt_2$$

$$=60(1-e^{-1})+60(e^{-1})=60.$$

I see this is wrong answer , and I'm not sure if

$$\mathbb{E}(T{|}T_1\le 60)=\mathbb{E}(T_1)$$ $$\mathbb{E}(T{|}T_2\ge 60)=\mathbb{E}(T_2)$$

Answer 1

Let's measure time in hours. The random treatment times $X$ and $Y$ of the two patients have probability density functions $$f_X(t)=f_Y(t)=e^{-t}\qquad(t\geq0)\ .$$ The time $T$ that the second patient spends at the doctor's office is equal to a possible overtime of the first patient plus his own treatment time. It follows that $T=(X-1)^+ +Y$, whereby the so-called positive part $\>a^+$ of a real number $a$ is defined as $0$ if $a<0$ and as $a$ if $a\geq0$. It follows that $$\eqalign{E(T)&=\int_0^\infty (t-1)^+\>f_X(t)\>dt+\int_0^\infty t\>f_Y(t)\>dt \cr &=\int_1^\infty(t-1)\>e^{-t}+\int_0^\infty t\> e^{-t}\>dt=1+{1\over e}\ .\cr}$$

Answer 2

Hint:

Let $T$ be the total amount of time 3:00 appointment spends in the doctor's office and $Y$ be the amount of time 3:00 apppointment gets serviced by the doctor and $X$ be the amount of time 2:00 appointment gets serviced in the doctor's office.

What is asked is $E(T)$

There are three cases as you have mentioned. The first case is $X> 60$ minutes further split into two as Y waits for the doctor during X's exceeding the 60 minutes, then the second case is Y gets serviced after that and the third case is Y gets serviced while $X\le 60$ minutes.

Thus $P(Y) = P((\text{Y waiting while X exceeds 60})/X>60).P(X>60) + P(Y>X/X>60).P(X>60) + P(Y/X<60).P(X<60)$

$P(X>60) = e^{-1}$ and $P(X<60) = 1-e^{-1}$

$P(Y \text{waiting while X exceeds 60}/X>60) = P(\text{X's remaining service time after 60 minutes}) = P(\text{X service time})$.

Remark: Application of memoryless property as the mean rate of remaining service is the same rate as service.

Thus $E(T) = E(X).P(X>60) + E(Y).P(X>60) + E(Y).P(X<60) = 2E(Y)(e^{-1})+ E(Y)(1-e^{-1})$

$E(X) = E(Y)$.

$$= 120e^{-1} + 60(1-e^{-1}) = 60+60e^{-1}$$