As I am working through my old university notes, due to excess of time lately, I've stumbled upon the following exercise:
Let's consider a medical treatment in which the same drug is used twice. We know that it has a probability $\alpha$ of curing given patient on each trial, but we don't know it's exact value. We use it independently on 100 test subjects. 40 of them got better after the first dose, and 20 after a second one. The rest of them were uncured.
Now we want to derive MLE estimator of said parameter $\alpha$.
I've tried to wrap my mind around it, and at first glance, it seemed that we are dealing with simple 0-1 distribution, but then I've got in trouble in incorporating the second dose onto it. As it seems to me now we are dealing with two dependant series of observations:
- first trial where each person is treated independently and we have 0-1 distribution as mentioned before
- second trial where the remainder of the uncured people from the first trial is being treated. Now even though we again treat them independently, we are dependant on the first trial. Also here it seems that even that in this trial we again have probability $\alpha$ of success for each person, the total probability of success is now given in terms of $(1-\alpha)\alpha$ as we once have failed. But here we don't have 0-1 distribution, rather something in terms of 00 and 01, where first 0 denote the first failure. But this would suggest that the first trial variable should also be similar, as it seems to me.
And here I've got stuck, as I struggle to define when with which or which at all distribution should I be dealing with and how to define it. Any help and explanation would be appriciated.
note what can happen:
Probability to be cured at first strike: $\alpha$
Probability to be cured at second strike: $(1-\alpha)\alpha$
Probability to be uncured: $(1-\alpha)^2$
Obviously $\alpha+(1-\alpha)\alpha+(1-\alpha)^2=1$
Your likelihood is the following
$L(\alpha) \propto \alpha^{40}((1-\alpha)\alpha)^{20}((1-\alpha)^2)^{40}$
$L(\alpha) \propto \alpha^{60}(1-\alpha)^{100}$
take the logarithm, derive with respect to $\alpha$, set =0, solve with respect to $\alpha$ and you will find
$\hat{\alpha}_{ML}=\frac{3}{8}$