4
$\begingroup$

Smith is waiting for his two friends Lee and Yang to visit his house. The time until Lee arrives is Exp($\lambda_1$) and the time until Yang arrives is Exp($\lambda_2$). After arrival, Lee stays an amount of time that is Exp($\mu_1$), whereas Yang stays an amount of time that is Exp($\mu_2$).

All four random variables are independent.

What is the expected time of the first departure?

Hint: Let X be the time of the first departure, and write X=F+Y where F is the time of the first arrival and Y is the additional amount of time until the first departure. Compute E(Y) by conditioning on who arrived first.

Attempt:

Denote S=Lee's arrive time; U=Lee's stay time; T=Yang's arrive time; V=Yang's stay time.

$E\{first arrival\}=E[X]=E[\min(S,T)]=\frac 1{\lambda_1+\lambda_2}$

After that, everything refreshes, then the additional waiting time is again:

$E[A]=E[\min(U,V)]=\frac 1{\mu_1+\mu_2}$.

Therefore, total time of expected first departure = $\frac 1{\mu_1+\mu_2}+\frac 1{\lambda_1+\lambda_2}$.

Which is wrong

$\endgroup$
  • $\begingroup$ It might be less confusing if you don't use the same letters to denote different things. What is A? Lee's arrival time or the additional time till first departure? $\endgroup$ – Graham Kemp Mar 4 '16 at 0:48
  • $\begingroup$ Everything doesn't refresh after first arrival. Suppose Lee arrives first: On that condition: the additional time until first departure is then the minimum time until: Lee's departure, or Yang's arrival and departure. $\endgroup$ – Graham Kemp Mar 4 '16 at 0:56
2
$\begingroup$

Everything doesn't refresh after first arrival.   The first person could depart before the other person even arrives.

Suppose Lee arrives first: On that condition: the additional time until first departure is then the minimum time until: Lee's departure, or Yang's arrival and departure.

$\begin{align}\mathsf E(X) = & ~\mathsf E[(S+U)\wedge (T+V)] \\ = & ~ \mathsf E(S\wedge T) + \mathsf P(S<T)~\mathsf E [U\wedge (T+V)]+\mathsf P(S>T)~\mathsf E[(S+U)\wedge V] \\ = & ~\mathsf E(S\wedge T) + \mathsf P(S<T)~(\mathsf E(U\wedge T)+\mathsf P(U>T)~\mathsf E(U\wedge V))+\mathsf P(S>T)~(\mathsf E(S\wedge V)+\mathsf P(S<V)~\mathsf E(U\wedge V)) \\ = & ~\frac{1}{\lambda_1+\lambda_2} + \frac{\lambda_1}{\lambda_1+\lambda_2}~(\frac{1}{\mu_1+\lambda_2}+\frac{\lambda_2}{\mu_1+\lambda_2}~\frac{1}{\mu_1+\mu_2})+\frac{\lambda_2}{\lambda_1+\lambda_2}~(\frac{1}{\lambda_1+\mu_2}+\frac{\lambda_1}{\lambda_1+\mu_2}~\frac{1}{\mu_1+\mu_2}) \end{align}$

$\endgroup$
  • $\begingroup$ I kept trying to find a simpler form for the solution but I suppose there isn't one :( $\endgroup$ – Math1000 Mar 4 '16 at 2:48
0
$\begingroup$

We have $$ \mathbb E[A] = \frac1{\lambda_1+\lambda_2} $$ and \begin{align} \mathbb E[F] &= \mathbb E[F\mid \text{Lee arrives first }]\mathbb P(\text{Lee arrives first})\\ &+ \mathbb E[F\mid \text{Yang arrives first }]\mathbb P(\text{Yang arrives first})\\ &= \left(\frac1{\mu_1+\lambda_2} + \left(\frac{\lambda_2}{\mu_1+\lambda_2}\right)\left(\frac1{\mu_1+\mu_2} \right) \right) \left( \frac{\lambda_1}{\lambda_1+\lambda_2}\right)\\ &+\left(\frac1{\mu_2+\lambda_1} + \left(\frac{\lambda_1}{\mu_2+\lambda_1}\right)\left(\frac1{\mu_1+\mu_2} \right) \right) \left( \frac{\lambda_2}{\lambda_1+\lambda_2}\right). \end{align} It follows that \begin{align} \mathbb E[X] &= \mathbb E[A] + \mathbb E[F]\\ &= \left(\frac1{\lambda_1+\lambda_2}\right)\left(1 + \frac{\lambda _1 \left(\lambda _2+\mu _1+\mu _2\right)}{\left(\mu _1+\mu _2\right) \left(\lambda _2+\mu _1\right)} + \frac{\lambda _2 \left(\lambda _1+\mu _1+\mu _2\right)}{\left(\mu _1+\mu _2\right) \left(\lambda _1+\mu _2\right)} \right). \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.