# Exponential distribution MLE with lifetime and frequency table

$$\begin{array}{c|c} \hline \text{Lifetime (months)} & \text{Observed frequency} \\ \hline 0-2 & 50 \\ \hline 2-4 & 35 \\ \hline 4-6 & 25 \\ \hline 6-8 & 15 \\ \hline 8-10 & 5 \\ \hline >10 & 10 \\ \hline\end{array}$$

The above is the lifetime and frequency table of $$140$$ light bulbs. I would like to calculate the MLE of $$\lambda$$.

I plan to use this formula:

$$\lambda = \frac{n}{x_1+x_2 + \ldots + x_n}$$

, where $$x_1$$ means the lifetime of first bulb but I only know the lifetime of the first bulb is between $$0$$ to $$2$$ months so how can I solve this question? Thank you in advance.

The probability (not density) that a lightbulb fails between time $$a$$ and time $$b$$ is $$e^{-\lambda a} - e^{-\lambda b} = e^{-\lambda a} \left(1 - e^{-\lambda (b-a)}\right)$$ The probability it fails after time $$c$$ is $$e^{-\lambda c}$$. But this will not lead to a simple expression for the maximum likelihood estimator, so you could resort to numerical methods; it seems to me that using this approach on your data, the calculation would produce $$\hat{\lambda}$$ of about $$0.25039$$
The simple alternative is an approximation and then take your simple expression. You might for example say that the representative figure for the first interval of $$0-2$$ is the midpoint of $$1$$, and similarly for the second interval of $$2-4$$ is $$3$$, etc. (For an exponential distribution, there is a good reason to think these might be slightly too high, but let's not worry about that now)
You would still need to choose a representative figure for the top range of above $$10$$ which theoretically heads off to infinity: personally I would have thought $$12$$ might be reasonable here, though others may have their own approach. Then your expression would give $$\hat{\lambda} = \frac{50+35+ 25 + 15 +5 + 10 }{50\times 1+35 \times 3+ 25 \times 5 + 15 \times 7+5 \times 9 + 10 \times 12} =\frac{140}{550} \approx 0.2545$$, not too far from my earlier calculation in the context of the ranges (and apparent rounding to $$5$$) in the original data