Question
A ferry terminal measures each passenger's waiting time \(T\) in minutes. Assume independent waiting times from an exponential distribution with unknown rate \(\lambda>0\) per minute. The old display rounds each wait down to the nearest whole minute: it records \(Y=\lfloor T\rfloor\), not the exact value of \(T\).
Six displayed waiting times are 0, 0, 1, 2, 2, 3. Derive the likelihood for \(\lambda\) from the recorded integers and find its maximum-likelihood estimate. Compare it with the estimate obtained by incorrectly treating the displayed numbers as exact exponential observations. Which way does that shortcut bias this particular estimated rate, and why are the recorded zeros legitimate?
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Hint 1
A display reading k does not mean the wait equals k; it means \(k\le T<k+1\). Express this interval probability using the exponential survival function.
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Solution
For an exponential wait \(T\) with rate \(\lambda\), the survival function is \(\Pr(T\ge t)=e^{-\lambda t}\) for \(t\ge0\). A displayed integer \(Y=k\) describes the whole interval \(k\le T<k+1\), so
\[\Pr(Y=k)=e^{-\lambda k}-e^{-\lambda(k+1)}=e^{-\lambda k}(1-e^{-\lambda}),\quad k=0,1,2,\ldots.\]
These probabilities form a geometric distribution on the nonnegative integers and sum to one: \((1-e^{-\lambda})\sum_{k=0}^{\infty}e^{-\lambda k}=1\). In particular, \(\Pr(Y=0)=1-e^{-\lambda}>0\): a zero display represents a positive-length interval, not a zero-duration point event.
For \(n\) independent readings with total \(S=\sum_i y_i\), the correct likelihood is \(L(\lambda)=(1-e^{-\lambda})^n e^{-\lambda S}\). Hence
\[\ell(\lambda)=n\log(1-e^{-\lambda})-\lambda S,\qquad \ell'(\lambda)=\frac{n}{e^{\lambda}-1}-S.\]
Here \(n=6\) and \(S=0+0+1+2+2+3=8\). Setting the score to zero gives \(e^{\widehat\lambda}=1+n/S=7/4\), so
\[\boxed{\widehat\lambda=\log(7/4)\approx0.5596\ \mathrm{minute}^{-1}}.\]
The log-likelihood is strictly concave for \(\lambda>0\) because \(\ell''(\lambda)=-ne^\lambda/(e^\lambda-1)^2<0\); with \(S>0\), this stationary point is the unique interior global maximum. If instead the integers were wrongly treated as exact continuous exponential observations, the usual rate estimate would be \(\widetilde\lambda=n/\sum_i y_i=6/8=0.75\) per minute. The displayed integers systematically omit fractional waiting time, making their sample mean too small and this rate estimate too large. Algebraically, \(\log(1+x)<x\) for \(x>0\), so the correct \(\log(1+n/S)\) is below the naive \(n/S\) whenever \(S>0\). If every displayed wait had been zero (\(S=0\)), the rounded-data likelihood would instead increase toward its supremum as \(\lambda\to\infty\), with no finite MLE; that edge case does not arise here.
Assumptions
True waits are independent, exponentially distributed with a common positive rate \(\lambda\) per minute.
The device floors each wait to an integer minute rather than rounding to nearest or censoring at a threshold.
All six displayed waits are observed, including both zeros, and the rate is constant across them.
Final Answer
Correct MLE \(\hat\lambda=\ln(1+\frac68)=\ln(\frac74)\approx0.5596\ \mathrm{min}^{-1}\). Naive \(\hat\lambda=\frac68=0.75\ \mathrm{min}^{-1}\), which overestimates the rate. A displayed zero means a true wait in \([0,1)\) minutes.