Insects of a certain species were exposed to cold temperature and how long the insects survived was recorded. The survival times of 9 of the 10 insects, in hours, are given below.
0.8, 0.6, 3.7, 5.3, 0.9, 3.4, 0.4, 6.8, 5.3
The experiment was stopped after 7 hours with one insect surviving longer than 7 hours.
(i) If survival times are assumed to follow the distribution f(x) = λ exp(−λx), x > 0, λ > 0 find the maximum likelihood estimate of λ.
(ii) The standard error of λ is calculated to be 0.09. Suppose we actually want an estimate of the mean survival time, µ, and its standard error. Write down the estimate of the mean survival time and use the delta method to find an approximate standard error of µ.
(iii) Explain what the EM-algorithm is and how it can be used with censored data.
(iv) Derive the E-step and M-step for this example
I've done (i), (ii), and (iii):
(i) $\hat{\lambda} = 5/19$
(ii) se$(\hat{\mu}) = 1.2996$
(iii) Maximisation algorithm used for finding MLEs. Iteratively finds the expectation and maximisation of the incomplete data.
But I cannot figure out how to get the E-Step at all!
I started trying to find E[loglik($\lambda$)] but I just ended up in a muddle.
Any help with understanding how to find the E-Step for this data would be greatly appreciated, thanks!
EDIT:
I feel like I'm getting somewhere:
The missing data is the time for insect 10, $x_{10}$. We assume this is known and set $x_{10} = s$.
$g(y|\lambda) = p(Y=s) \cdot \prod_{i=1}^{9} p(Y=y_i)$
$= \lambda exp(-s\lambda) \cdot \prod_{i=1}^{9} \lambda exp(-\lambda y_i)$
$= \lambda^{10} exp(-s\lambda) \cdot \prod_{i=1}^{9} exp(-\lambda y_i)$$log(g(y|\lambda) = 10log(\lambda)-s\lambda- \sum_{i=1}^9 \lambda y_i $
$= 10log(\lambda) - \lambda(s+27.2)$
I'm slightly confused from here though. Do I simply put $E(s) = P(Y=s) = 1/\lambda$ and hence $s^{(m)} = 1/\lambda^{(m-1)}$ ?
EDIT 2:
From this link: http://www.mrc-bsu.cam.ac.uk/wp-content/uploads/EM_slides.pdf
$E[Y|Y>s] = s + 1/\lambda$ hence $s^{(m)} = s^{(m-1)} - 1/\lambda^{(m)}$ and I think we might have our answer!