# Maximum likelihood of Normal density under selection

Consider the density function given by

$$\left[\dfrac{\gamma_{\leq0} \mathbb{1}(t \leq 0) + \gamma_{>0} \mathbb{1}(t > 0)}{\gamma_{\leq0}\Phi\left(- \mu / \sigma\right) + \gamma_{>0}\Phi\left(\mu / \sigma\right)}\right] \phi\left(\dfrac{t - \mu}{\sigma}\right).$$

The parameters $$\gamma_{\leq 0}$$ and $$\gamma_{>0}$$ denote the respective probabilities of observing a value of $$t \in \mathbb{R}$$ if it is nonpositive and positive. In addition, $$\Phi(\cdot)$$ denotes the standard Normal CDF and $$\phi(\cdot)$$ the standard Normal pdf. The function $$\mathbb{1}(\cdot)$$ is the indicator function.

Suppose that the values of the parameters $$\gamma_{\leq 0}$$, $$\gamma_{>0}$$ and $$\sigma$$ are all known. Given a value of $$t$$, what is the value of $$\mu$$ that assigns the greatest density to $$t$$? I.e., what is the MLE of $$\mu$$?

If the density function were simply $$\phi\left(\dfrac{t - \mu}{\sigma}\right)$$, then the MLE of $$\mu$$ would be whichever value is equal to the observed $$t$$. But I'm not sure how to derive the MLE with the additional factor of $$\left[\dfrac{\gamma_{\leq0} \mathbb{1}(t \leq 0) + \gamma_{>0} \mathbb{1}(t > 0)}{\gamma_{\leq0}\Phi\left(- \mu / \sigma\right) + \gamma_{>0}\Phi\left(\mu / \sigma\right)}\right]$$. Thanks so much for any help.

To simplify the analysis, I will first deal with the special case where $$\sigma=1$$. Your likelihood function here can be written by removing the proportionality constants, which gives:

\begin{align} L_t(\mu) &= \frac{\exp(-\tfrac{1}{2}(t-\mu)^2)}{\gamma_{\leq0} \Phi(-\mu) + \gamma_{>0} \Phi(\mu)}. \\[6pt] \end{align}

Differentiating and applying the quotient rule gives:

\begin{align} \frac{dL_t}{d\mu}(\mu) &= \frac{\exp(-\tfrac{1}{2}(t-\mu)^2)}{[\gamma_{\leq0} \Phi(-\mu) + \gamma_{>0} \Phi(\mu)]^2} \begin{bmatrix} (t-\mu) [\gamma_{\leq0} \Phi(-\mu) + \gamma_{>0} \Phi(\mu)] \\ - \phi(\mu) [-\gamma_{\leq0} \Phi(-\mu) + \gamma_{>0} \Phi(\mu)] \end{bmatrix} \\[6pt] &= \frac{\exp(-\tfrac{1}{2}(t-\mu)^2)}{[\gamma_{\leq0} \Phi(-\mu) + \gamma_{>0} \Phi(\mu)]^2} \begin{bmatrix} \quad \ \ (t-\mu+\phi(\mu)) \gamma_{\leq0} \Phi(-\mu) \\ + (t-\mu-\phi(\mu)) \gamma_{>0} \Phi(\mu) \end{bmatrix} \\[6pt] &= L_t(\mu) \times \frac{(t-\mu+\phi(\mu)) \gamma_{\leq0} \Phi(-\mu) + (t-\mu-\phi(\mu)) \gamma_{>0} \Phi(\mu)}{\gamma_{\leq0} \Phi(-\mu) + \gamma_{>0} \Phi(\mu)]}, \\[6pt] \end{align}

which yields the score function:

$$s(\mu) = \frac{(t-\mu+\phi(\mu)) \gamma_{\leq0} \Phi(-\mu) + (t-\mu-\phi(\mu)) \gamma_{>0} \Phi(\mu)}{\gamma_{\leq0} \Phi(-\mu) + \gamma_{>0} \Phi(\mu)}.$$

(I will leave it as an exercise for you to check concavity or quasi-concavity of the log-likelihood function.) Critical points of the likelihood function occur when the score is zero, giving the critical point equation:

$$(t-\hat{\mu}+\phi(\hat{\mu})) \gamma_{\leq0} \Phi(-\hat{\mu}) + (t-\hat{\mu}-\phi(\hat{\mu})) \gamma_{>0} \Phi(\hat{\mu}) = 0.$$

You would need to solve this equation using numerical methods (e.g., Newton-Raphson) which would give you the critical points of the likelihood function, yielding the MLE for $$\mu$$.