The CDF of a normal variable is
$P(X \leq x)$, where $X$ is a random variable. This also written as $\Phi (x)$ so if $\Phi (\cdot)$ is the normal CDF, then $\Phi (0)$ is $P(X<0) = 50 \% $ because the normal is symmetric.
Now I am looking at the derivation of a Probit model and have a question. The derivation states:
A Probit model is defined as $Y = \Phi(X\beta +\epsilon) $ where $\epsilon \sim N(0,1)$ so the normal CDF is around the left hand side. In this case let Y only take on the values 0 or 1.
A probit can be written in terms of latent a variable $Y^*$ as
$Y^* = X \beta+\epsilon, \epsilon \sim N(0,\sigma)$
Now I am looking at slide 44 here: http://www.columbia.edu/~so33/SusDev/Lecture_9.pdf
In probits we don't observe this latent variable $Y^*$ but it leads to our observed outcome Y like this:
$y_i = 0$ if $y^* \le 0$
$y_i = 1$ if $y^* > 0$
so if $y^*$ is greater than some boundary in this case 0 we see the value 1.
so the probability $y_i = 1$ given the independent variable $x_i$ is
$Pr(y_i = 1 | x_i) = Pr(y^*_i >0 | x_i) = Pr( \beta X_i + \epsilon_i > 0)= Pr(\epsilon_i > -\beta X)$
above we have what looks like a CDF but notice the $>$ instead of the $\le$. In this case $\epsilon$ is the random variable in the CDF
The author's next line is
= $\Phi( -\beta X)$
My question is how does $Pr(\epsilon_i > -\beta X) = \Phi( -\beta X)$
$Pr(\epsilon_i > -\beta X)$ doesn't look like a CDF.
wouldn't you need to get $Pr(\epsilon_i \le ?? )$ to have a CDF and use $\Phi (??)$#
@Adrian referencing your wiki on logisit regression
$Pr(\epsilon_i > -\beta X )$ leads to $Pr(\epsilon_i < \beta X )$ from multiplying by -1
In my case
I have $Pr(\epsilon_i > -\beta X)$ and if multiplied by -1 = $Pr(\epsilon_i < \beta X)$ i.e. $\Phi(beta X)$ which is opposite what slide 44 shows
Any Thoughts?after thought now convinced slide 44 is typo after wiki on probit shows $\Phi(\beta X)$