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Here is the background of the question. I am stuck on part b. Here is what I have so far. I would appreciate any help. Thank you!

Let $M = \min(X_1, X_2)$ where $X_1$ represents the exponential process of service time for clerk 1 and $X_2$ is for clerk 2. Then, we have 2 cases: clerk 2 is slower initially and when clerk 1 is slower initially. We then get $P(X_1>X_2\mid M=X_1)P(M=X_1) + P(X_2>X_1\mid M=X_2)P(M=X_2)$. I know how to find $P(M=X_1)$ and $P(M=X_2)$ ($\frac{\lambda_1}{\lambda_1+\lambda_2}$ for $P(M=X_1)$), but I am unsure how to use the memoryless property of exponential distributions to solve the first conditional probability.

StubbornAtom
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Emily
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1 Answers1

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Recall that exponential distributions are memory-less. Therefore, at the time when the first of Jake and Naomi get done with their service, Paul would go to the server that becomes empty. Say he goes to server 1 at time $t=s$. Then, due to memoryless property, $\mathbb{P}[X_2>t+s|X_2>s] = \mathbb{P}[X_2>t]$. Thus, we are comparing the distributions for $X_1$ and $X_2$ afresh.

Concretely, let $\lambda_1 = \frac{1}{3}, \lambda_2 = \frac{1}{5}$. Then $\lambda_{min(1,2)} = \lambda_1 + \lambda_2 = \frac{8}{15}$ (See linked question)

Thus the expected time at which Paul enters the service is $T_{entry} = \frac{1}{\lambda_{min(1,2)}} = \frac{15}{8}$

Now for part A there are two cases as you have pointed out. $\mathbb{P}[M=X_1] = \frac{\lambda_1}{\lambda_1+\lambda_2}=\frac{5}{8}$ and $\mathbb{P}[M=X_2]=\frac{3}{8}$

If $M=X_1$, then time taken for Paul to finish service in expectation is $\mathbb{E}[T_{total|M=X_1}] = T_{entry} + \frac{1}{\lambda_1} = \frac{15}{8} + 3 = \frac{39}{8}$

Similarly, $\mathbb{E}[T_{total|M=X_2}] = T_{entry} + \frac{1}{\lambda_2} = \frac{15}{8} + 5 = \frac{55}{8}$

Finally we have by iterated expectation, $\mathbb{E}[T_{total}] = \mathbb{E}_M\big[\mathbb{E}_{T|M}[T_{total}|M=X_i]\big] = \frac{5}{8}\cdot\frac{39}{8}+\frac{3}{8}\cdot\frac{55}{8}=\frac{45}{8}$

For part B, we again have two cases. If $M=X_1$ in the first iteration when Paul enters the system, then $M=X_2$ in the second iteration for Paul to finish last. Similarly, if $M=X_2$ in the first iteration, then $M=X_1$ in the second iteration. The total probability of this happening is

$\mathbb{P}(\text{Paul finishes last}) = \frac{\lambda_1}{\lambda_1+\lambda_2}\cdot \frac{\lambda_2}{\lambda_1+\lambda_2}+\frac{\lambda_2}{\lambda_1+\lambda_2}\cdot \frac{\lambda_1}{\lambda_1+\lambda_2} = \frac{2\lambda_1\lambda_2}{(\lambda_1+\lambda_2)^2} = \frac{\frac{2}{15}}{(\frac{8}{15})^2} = \frac{15}{32}$

Rahul Madhavan
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