Exponential Distribution MLE The lifetime of a type of component has an exponential distribution with rate λ per hour. Ten of these components were tested but the only recorded results were that 3 components had failed within 100 hours and 7 had survived that time.
I have been asked to 
(a) Find the maximum likelihood estimate of λ.
(b) Find the approximate standard error of this estimator.
Can you help me please?
 A: The probability of a component's surviving $100$ hours is $e^{-100\lambda}$.  So the probability of the observed outcome, given $\lambda$ is $(e^{-100\lambda})^7 (1-e^{-100\lambda})^3$.  So the likelihood function is
$$
L(\lambda) = (e^{-100\lambda})^7 (1-e^{-100\lambda})^3
$$
and its logarithm is
$$
\ell(\lambda) = -700\lambda + 3\log(1-e^{-100\lambda}).
$$
Differentiation yields
$$
-700 + \frac{3(100e^{-100\lambda})}{1-e^{-100\lambda}}.
$$
That is $0$ if its product with $1-e^{-100\lambda}$ is $0$, thus if
$$
-700(1-e^{-100\lambda}) + 300e^{-100\lambda}=0.
$$
So
$$
1000e^{-100\lambda} = 700.
$$
Can you get the answer from there?
A: Another way of looking at the problem is that we have observed $7$ occurrences of an event of unknown probability $p$ on $10$ independent trials of the experiment. The maximum-likelihood estimate $\hat{p}$ of the unknown quantity
$p$ is thus the observed relative frequency, that is,
$$\hat{p} = \frac{7}{10}.$$
In this case, $p$ is of the form $\exp(-1000\lambda)$ and thus the
maximum-likelihood estimate of $\lambda$ is the solution to the
equation 
$$\exp(-100\hat{\lambda}) = \frac{7}{10}.$$
