How to estimate coefficients of logistic model Consider model $logit(p)=a+bx$. I would like to get an analytic formula of $a$ and $b$ like in linear regression. In linear regression, we can get a formula of estimates of $a$ and $b$.
I tried using MLE. But it is too complicated for me.
I am doing research about the statistical genetics. Here $p$ is penetrance and $x$ is genotype. In linear regression, we can use the coefficient $b$ to represent the covariance matrix of $x$ and trait $y$ (both of them should be standardized). I am thinking whether we can use the $b$ in logistic model to recover the information about the covariance matrix of $x$ and $y$ (binary trait) in meta-analysis (they do not provide individual level data). But some cohort will prove the odds ratio and coefficient t-statistic. I do not know how to use it to recover the covariance matrix of $x$ and $y$. 
But some summary statistics will be provided like estimated odds ratio, Coefficient t-statistic.
 A: I strongly encourage you to work through the case where $x$ can take two values (e.g. 0 = dominant, 1 = recessive as in your example),  as it a useful exercise for understanding the logistic model and also the MLE approach.
The model in question is supposed to be 
$$
Y_i \sim \mathrm{Bernoulli}(p_i),
$$
where $p_i = \mathrm{logistic}(a + bx_i)$, $i = 1,\ldots,n$.  Suppose the observed values are $y_1, \ldots, y_n$.
Homework:


*

*What is the probability of observing this particular string of $y_i$
(assuming they are all independent).  Hint: split according to
whether $y_i = 0$ or $1$ and similarly for $x_i$.

*What is the log of this?

*What happens to the log likelihood if $a$ gets large or small (negative)?  Same question for $b$.

*Conclude that you can maximise the log likelihood by differentiating in $a$ and $b$ then finding where the derivative is
zero (separately for $a$ and $b$).


OK, for the final answer:
Write $\#_{l,m} = \#\{i : y_i = l, x_i = m\}$, so $\#_{0,1}$ is the number of observations with $y_i = 0$ and $x_i = 1$ etc.
You'll see the problem can be parameterised more easily in terms of 
$$
p = \frac{e^{a}}{1 + e^{a}}
$$
and
$$
r = \frac{e^{a+b}}{1 + e^{a+b}}.
$$
Assuming I didn't make a mistake, we find
$$
r = \frac{\#_{1,1}}{\#_{0,1} + \#_{1,1}},
$$
i.e. the fraction of cases with $y_i = 1$ when the independent variable has $x_i = 1$.
Then,
$$
p = \frac{\#_{1,0}}{\#_{0,0} + \#_{1,0}}
$$
is the fraction when $x_i = 0$.
So, $a + b = \ln(r/(1-r))$ is the log odds for $y_i = 1$ when $x_i = 1$.
Finally, $a = \ln(p/(1-p))$ and $b = \ln(r(1-p)/p(1-r))$ is the log odds ratio.
