Logistic regression with latent variable problem I'm having problem solving this question:

Any help would be really valuable.
 A: $\newcommand{\x}{\mathbf{x}}\newcommand{\b}{\boldsymbol{\beta}'\x}$Hints:
Note that $\Phi$ (standard normal CDF) is the CDF of $\varepsilon$.
For part (a), try using the fact that that $\Phi(-u)=1-\Phi(u)$ and $P(-\varepsilon \le u) = P(\varepsilon \ge - u)$.
For (b), note that by definition of $Y^*$, $P(Y=1\mid \x) = P(-\varepsilon < \b)$. Use the result of (a) to arrive at the answer. 
A: In part (a) you need to show the symmetry property of the normal distribution. One way would be:
for $u\geq0$:
\begin{align}
\mathbb{P}(\epsilon\leq u) &= \int_{-\infty}^{u}\frac{1}{\sqrt{2\pi}}\exp(-\tfrac{1}{2}x^2)dx\\
 &= \int_{0}^{u}\frac{1}{\sqrt{2\pi}}\exp(-\tfrac{1}{2}x^2)dx + \int_{-\infty}^{0}\frac{1}{\sqrt{2\pi}}\exp(-\tfrac{1}{2}x^2)dx\\
&= \int_{-u}^{0}\frac{1}{\sqrt{2\pi}}\exp(-\tfrac{1}{2}(-x)^2)dx + \int_{0}^{\infty}\frac{1}{\sqrt{2\pi}}\exp(-\tfrac{1}{2}(-x)^2)dx\\
&= \int_{-u}^{\infty}\frac{1}{\sqrt{2\pi}}\exp(-\tfrac{1}{2}x^2)dx = \mathbb{P}(\epsilon \geq -u) = \mathbb{P}(-\epsilon \leq u).
\end{align}
for $u < 0$:
\begin{align}
\mathbb{P}(\epsilon\leq u) &= \int_{-\infty}^{u}\frac{1}{\sqrt{2\pi}}\exp(-\tfrac{1}{2}x^2)dx = \int_{-u}^{\infty}\frac{1}{\sqrt{2\pi}}\exp(-\tfrac{1}{2}(-x)^2)dx\\
&= \mathbb{P}(\epsilon \geq -u) = \mathbb{P}(-\epsilon \leq u). 
\end{align}
concluding for $u \in \mathbb{R},\mathbb{P}(\epsilon\leq u) = \mathbb{P}(-\epsilon \leq u)$
Part (b):
\begin{align}
\mathbb{P}(Y=1|x) &= \mathbb{P}(Y^{*}>0|x) = \mathbb{P}(-\epsilon <\bf{\beta}'x|x) \\
&=\mathbb{P}(\epsilon < \bf{\beta}'x|x) = \mathbb{P}(\epsilon \leq \bf{\beta}'x|x).
\end{align}
since $\epsilon$ is a continuous random variable $\mathbb{P}(\epsilon = \bf{\beta'x}|\bf{x}) = 0 $. Sorry, this is a regular 0 (not a vector of zeros). 
