Example 4.1.11.
Suppose we observe a cell and measure the time \(T\) until a toxin molecule leaves the cell. Then \(T \sim \Exp(\lambda)\) for some unknown rate parameter \(\lambda\text{:}\)
\begin{gather*}
f(t) = \lambda e^{-\lambda t}.
\end{gather*}
Suppose we observe a molecule leave the cell at \(t = 0.3\) minutes. Then:
\begin{align*}
L(\lambda) \amp = \lambda e^{-0.3\lambda} \\
L'(\lambda) \amp = e^{-0.3\lambda} + \lambda e^{-0.3t} (-0.3) \\
\amp = e^{-0.3\lambda}\left[ 1 - 0.3\lambda\right]
\end{align*}
The critical point is \(\lambda = \frac{1}{0.3} \approx 3.33\text{.}\) Since \(\L' \gt 0\) on \((0, 3.33)\) and \(\L' \lt 0\) on \((3.33, \infty)\text{,}\) there is a global maximum at \(\lambda \approx 3.33\text{,}\) which is therefore the MLE.
The specific time \(0.3\) minutes doesn’t particularly matter in this calculation. Whatever the time \(t\text{,}\) essentially the same calculation will result in a MLE of \(\lambda = 1/t\text{.}\) But what if we collect multiple pieces of data?
| Molecule | Waiting Time | Rate Estimation |
| 1 | \(0.3\) | \(3.33\) |
| 2 | \(0.8\) | \(1.25\) |
| 3 | \(0.5\) | \(2.00\) |
| 4 | \(0.6\) | \(1.67\) |
| 5 | \(0.9\) | \(1.11\) |
How should take all of this data into account in our maximum likelihood estimation? We might consider taking the average of all of the separate rate estimations, which would give \(1.87\text{.}\) Is this the most likely? We need some mathematical justification.
To account for multiple, independent data points, we should multiply the probability densities for each in the creation of our likelihood function:
\begin{align*}
\L(\lambda) \amp = \left(\lambda e^{-0.3\lambda}\right)\left(\lambda e^{-0.8\lambda}\right)\left(\lambda e^{-0.5\lambda}\right)\left(\lambda e^{-0.6\lambda}\right)\left(\lambda e^{-0.9\lambda}\right) \\
\amp = \lambda^5 e^{-0.3\lambda - 0.8\lambda - 0.5\lambda - 0.6\lambda - 0.9\lambda } \\
\amp = \lambda^5 e^{-3.1\lambda}
\end{align*}
Now we can find the maximum:
\begin{align*}
\L'(\lambda) \amp = 5 \lambda^4 e^{-3.1\lambda} + \lambda^5 e^{-3.1\lambda} (-3.1) \\
\amp = 5 \lambda^4 e^{-3.1\lambda} + \lambda^5 e^{-3.1\lambda} (-3.1) \\
\amp = \lambda^4 e^{-3.1\lambda} \left[5 - 3.1\lambda \right]
\end{align*}
The only critical point is \(5/3.1 \approx 1.61\text{.}\) Since \(\L' \gt 0\) on \((0, 1.61)\) and \(\L' \lt 0\) on \((1.61, \infty)\text{,}\) there is a global maximum at \(\lambda = 1.61\text{,}\) which is therefore the MLE.
It may seem less clear how to generalize this calculation for other tables of data. Observe that the value \(3.1\) is the sum of the five times in the table, so \(3.1/5\) is the average time. The MLE turned out to be the reciprocal of the average time (just as the MLE with only one data point was the reciprocal of that one time). Notice that this does not match the guess we made previously of averaging the individual rate estimations for each data point.