Logistic regression versus mean and variance, which estimate is more efficient and why?

Question

Let

$$\begin{array}{} Y_i & \sim& Bernoulli(0.5) \\ X_i|Y_i &\sim& N(\mu_{Y_i},\sigma^2) \end{array}$$

In this case we can consider independent pairs of observations $X_i,Y_i$ following the model of a logistic regression where the $Y_i$ has a conditional Bernoulli distribution with

$$P(Y_i=1|X_i=x) = \frac{1}{1+e^{-(a+bx)}}$$

where $a = \frac{\mu_1^2-\mu_0^2}{2\sigma^2}$ and $b=\frac{\mu_0-\mu_1}{\sigma^2}$

So, we can estimate these parameters $a$ and $b$ in two different ways

• Perform logistic regression.
• Estimate $\mu_i$ and $\sigma$ directly with the assumed normal distribution, then compute to the parameters $a$ and $b$.

The two methods will not give the same result. Which method is most efficient (has the expected squared error)?

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To clarify my question a bit further. It is also about understanding.

Here is a simulation that shows that the estimate of mean and variance results in smaller error. And it could answer the question.

• But is it general?
• And why wouldn't logistic regression perform the same?
• What sort of information does the knowledge about the distribution of $X_i$ add and could this be incorporated into the logistic regression (e.g. by adding weights) to make it perform better?

set.seed(1)

sim = function(n, mu) {

  Y = rbinom(n,1,0.5)
  X = rnorm(n,Y*mu,1)

  mod = glm(Y ~ X, family = binomial)
  glm_est = mod$coefficients

  m = c(mean(X[Y==0]), mean(X[Y==1]))
  V = (sum((X-m[Y+1])^2)/(n-2))

  mm_est = c(-diff(m^2)/2/V,
             diff(m)/V)

  return(c(glm_est, mm_est))
}

mu = 1
z = replicate(10^3, sim(200, mu))

true_a = -mu^2/2
true_b = mu

### glm_errors
mean((z[1,]-true_a)^2) # 0.03592216
mean((z[2,]-true_b)^2) # 0.03245813

### gaussian_mm_errors
mean((z[3,]-true_a)^2) # 0.01323784
mean((z[4,]-true_b)^2) # 0.03075781