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I'm having problem solving this question: enter image description here

Any help would be really valuable.

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    $\begingroup$ Add self study tag. $\endgroup$ Commented Jun 1, 2019 at 22:06

2 Answers 2

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

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  • $\begingroup$ I still have problem understanding how to reach 𝑃(πœ€ ≀ 𝑒) in part a, can you clarify? In the part b, why is 𝑃(πœ€ ≀ πœ·β€²π±)=Ξ¦(πœ·β€²π±) $\endgroup$ Commented Jun 1, 2019 at 16:45
  • $\begingroup$ $\newcommand{\x}{\mathbf{x}}\newcommand{\b}{\boldsymbol{\beta}'\x}$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 $P(\varepsilon < \b\mid \x)=\Phi(\b)$.) $\endgroup$ Commented Jun 1, 2019 at 18:55
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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).

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