Expected Wait for Two Independent Train Lines
You arrive at a train station at a random time. Two train lines, $A$ and $B$, serve the station. Trains on line $A$ arrive according to a Poisson process with rate $\lambda_1$ (so the inter-arrival time is $\text{Exp}(\lambda_1)$), and trains on line $B$ arrive independently with rate $\lambda_2$ (inter-arrival time $\text{Exp}(\lambda_2)$).
You board whichever train comes first, regardless of the line.
- Find the expected time you wait until the next train arrives.
- Evaluate your answer when $\lambda_1 = 1/10$ and $\lambda_2 = 3/20$.
Hints
- Since you arrive at a random time, the memoryless property of the exponential distribution tells you the residual wait on each line is still exponential with the original rate. Your wait is determined by whichever train comes first.
- The wait until the first arrival is $T = \min(T_A, T_B)$. Use the tail expectation formula $E[T] = \int_0^{\infty} P(T \geq t) \, dt$ and exploit independence to factor the joint survival function.
- The product $e^{-\lambda_1 t} \cdot e^{-\lambda_2 t} = e^{-(\lambda_1 + \lambda_2)t}$ is the survival function of an $\text{Exp}(\lambda_1 + \lambda_2)$ random variable, so the expected wait is $1/(\lambda_1 + \lambda_2)$.
Worked Solution
How to Think About It: You show up at the station and two independent Poisson clocks are ticking -- one for each line. You leave on whichever fires first. Before writing anything, think about rates: line $A$ sends a train every 10 minutes on average, line $B$ every $20/3 \approx 6.7$ minutes. Together, trains arrive at a combined rate. By the memoryless property of the exponential distribution, it does not matter when you arrive -- the residual wait on each line is still exponential with the same rate. So your wait is just the minimum of two independent exponentials.
Quick Estimate: Line $A$ alone gives an average wait of 10 minutes; line $B$ alone gives about 6.7 minutes. Having both lines can only help, so the wait is less than 6.7 minutes. A natural guess: the rates add, so the expected wait is $1/(\lambda_1 + \lambda_2) = 1/(1/10 + 3/20) = 1/(5/20) = 4$ minutes. That passes the sniff test -- it is well below either individual mean.
Approach: Use the survival function (tail integral) formula for the expectation of a non-negative random variable, combined with the independence of the two lines.
Formal Solution:
Let $T_A \sim \text{Exp}(\lambda_1)$ and $T_B \sim \text{Exp}(\lambda_2)$ be the residual waiting times for each line (justified by memorylessness). Your actual wait is $T = \min(T_A, T_B)$.
Using the tail expectation formula:
$$E[T] = \int_0^{\infty} P(T \geq t) \, dt = \int_0^{\infty} P(T_A \geq t, \, T_B \geq t) \, dt$$
By independence:
$$= \int_0^{\infty} P(T_A \geq t) \cdot P(T_B \geq t) \, dt = \int_0^{\infty} e^{-\lambda_1 t} \cdot e^{-\lambda_2 t} \, dt = \int_0^{\infty} e^{-(\lambda_1 + \lambda_2) t} \, dt = \frac{1}{\lambda_1 + \lambda_2}$$
In fact, the minimum of independent exponentials is itself exponential: $T \sim \text{Exp}(\lambda_1 + \lambda_2)$. This follows because the survival functions multiply, and the product of exponential survival functions is again an exponential survival function.
Plugging in $\lambda_1 = 1/10$ and $\lambda_2 = 3/20$:
$$E[T] = \frac{1}{1/10 + 3/20} = \frac{1}{2/20 + 3/20} = \frac{1}{5/20} = \frac{20}{5} = 4 \text{ minutes}$$
Answer: The expected wait is $\dfrac{1}{\lambda_1 + \lambda_2}$. For the given rates, this equals $4$ minutes.
Intuition
The core principle here is that independent Poisson processes superpose: if you merge two independent Poisson streams with rates $\lambda_1$ and $\lambda_2$, you get a single Poisson stream with rate $\lambda_1 + \lambda_2$. Rates add because arrivals from different sources do not interfere with each other. This is exactly the same logic behind why competing exponential clocks -- the kind that appear everywhere in continuous-time Markov chains, queuing theory, and reliability engineering -- resolve to the minimum, which is itself exponential with the summed rate.
In practice, this shows up constantly. A market maker watching two independent liquidity pools sees order flow at the combined rate. A server with two independent failure modes fails at the sum of the individual failure rates. The pattern generalizes to $n$ independent exponential clocks: the minimum has rate $\sum \lambda_i$. The common mistake is to average the waiting times instead of adding the rates -- if you naively averaged 10 and 6.7 you would get about 8.3 minutes, which is too high. Always think in terms of rates when combining independent processes.