# Logistic regression with latent variable problem

I'm having problem solving this question:

Any help would be really valuable.

• Add self study tag. Commented Jun 1, 2019 at 22:06

$$\newcommand{\x}{\mathbf{x}}\newcommand{\b}{\boldsymbol{\beta}'\x}$$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.

• I still have problem understanding how to reach 𝑃(𝜀 ≤ 𝑢) in part a, can you clarify? In the part b, why is 𝑃(𝜀 ≤ 𝜷′𝐱)=Φ(𝜷′𝐱) Commented Jun 1, 2019 at 16:45
• $\newcommand{\x}{\mathbf{x}}\newcommand{\b}{\boldsymbol{\beta}'\x}$For a), use $P(\varepsilon \ge -u)=1- P(\varepsilon \le u)$. For b), it is because $P(\varepsilon \le a)=\Phi(a)$ for all $a$; let $a =\b$. (I have been thinking of $\x$ as a constant, but if it is a random variable, we would write $P(Y=1\mid \x)=P(-\varepsilon < \b \mid \x)$, and since we are conditioning on $\x$, it is effectively treated as a constant still. So $P(\varepsilon < \b\mid \x)=\Phi(\b)$.) Commented Jun 1, 2019 at 18:55

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).