Find parameter of exponential distribution (first arrival at $x$ and second at $y$)

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I came across this interview question online:

Suppose you sit on the road side and observe cars driving by. Assume the distribution of cars driving by is according to an exponential distribution. Now you observe a first car after sitting for x mins, a second car after sitting for y min. Given those observations, can you estimate the parameter in the distribution?

I don't see how I am able to solve it without additional information. Thanks

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By definition there should be equal probability that the observations are on either side of the mean. The mean of an exponential distribution is $\frac{1}{\lambda}$. Therefore, $\frac {x+y}{2}= \hat {(\frac{1}{\lambda})}$ is an unbiased estimator, however the inverse of this is notably a biased estimator of $\lambda$ because $E(\hat \lambda - \lambda)=\frac{\lambda}{n-1}$. So $\hat {(\frac{1}{\lambda})}\approx \frac{1}{\hat \lambda}$ giving $\hat \lambda \approx \frac {2}{x+y}$; I don't know if an unbiased estimator exists but for such a small sample-size that might be as good as you can do...