Suppose we observe a cell, measuring the time T until a toxin molecule leaves the cell.
Then T \sim \Exp(\lambda) for some \lambda, with pdf
f(t) =\lambda e^{-\lambda t}, \lambda \gt 0
If we see a toxin molecule leave at 0.3 min, what's the MLE for \lambda?
\L(\lambda) = \lambda e^{-0.3 \lambda} \quad \text{(density, not probability)}
We want to maximize \L(\lambda) over \lambda \in (0, \infty), an open interval.
So we'll use the "Open Interval Method".
\L'(\lambda) \amp = e^{-0.3\lambda} + \lambda e^{-0.3\lambda}(-0.3)
\amp = e^{-0.3\lambda}\left( 1 - 0.3 e^{-0.3\lambda}\right)
Then \L'(\lambda) = 0 when \lambda = \frac{1}{0.3}\approx 3.33.
Checking \lambda values to the left and the right:
\L'(1) \amp = (+)(+) = (+)
\L'(10) \amp = (+)(-) = (-)
So \L'(\lambda) \gt 0 (and therefore \L(\lambda) is increasing) on (0, 1/0.3), and \L'(\lambda) \lt 0 (and therefore \L(\lambda) is decreasing) on (1/0.3, \infty).
Now we can conclude that \widehat{\lambda} = 1/0.3 \approx 3.33 is the location of a global (and not just local) maximum value.
More generally, if the observed time is t, then the MLE will be \widehat{\lambda} = \frac{1}{t}.
What if we had more data points? For example, suppose two toxin molecules leave the cell at t_1 = 0.3 min and t_2 = 0.5 min?
Waiting Times
| Molecule |
Time |
Rate Estimate |
| 1 |
0.3 |
1/0.3 \approx 3.33 |
| 2 |
0.5 |
1/0.5 = 2 |
How do we combine these data points? We could take the average of the rate estimates:
\frac{3.33 + 2}{2} \approx 2.67
Alternatively, we could average the times first, then create a new rate estimate from the average time:
\frac{0.3 + 0.5}{2} \amp = 0.4
\frac{1}{0.4} \amp = 2.5
Both of these make some sense, but let's do a careful computation to be certain which way is correct (if either of them is!).
\L(\lambda) \amp = \left( \lambda e^{-0.3 \lambda} \right)\left( \lambda e^{-0.5 \lambda} \right)
\amp = \lambda^2 e^{-0.3 \lambda - 0.5 \lambda}
\amp = \lambda^2 e^{- 0.8 \lambda}
Using the Open Interval Method:
\L'(\lambda) \amp = 2\lambda e^{-0.8 \lambda} + \lambda^2 e^{-0.8 \lambda} (-0.8)
\amp = \lambda e^{-0.8 \lambda} \left[ 2 - 0.8\lambda \right]
So \L'(\lambda) = 0 when \lambda = \frac{2}{0.8} = 2.5.
Testing points to the left and right:
\L'(1) \amp = (+)(+)(+) = (+)
\L'(5) \amp = (+)(+)(-) = (-)
Now \L'(\lambda) \gt 0 (\L(\lambda) is increasing) on (0, 2.5), and \L'(\lambda) \lt 0 (\L(\lambda) is decreasing) on (2.5, \infty).
Therefore, the MLE is \widehat{\lambda} = \frac{2}{0.8} = 2.5.
Tracing the values 2 and 0.8 throughout the calculation, we can see that the value 2 will generally match the number of waiting times collected, and the value 0.8 will be the sum of the waiting times.
So, generally, with collected waiting times of t_1, \dotsc, t_n, the MLE will be:
\widehat{\lambda} = \frac{n}{t_1 + \dotsb + t_n} = \frac{1}{\text{avg time}}.