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 $\Phi(\beta X)$
  • 1
    $\begingroup$ Who is the "author"? Do you have a reference? Is it possible that this particular author defines $\Phi(z)=\Pr(X \ge z)$? $\endgroup$ – whuber Apr 27 '16 at 15:18
  • 1
    $\begingroup$ Please see link to slide in problem. I don't think the author defines the cdf that way. I don't think that is a cdf because big X is the random variable. $\endgroup$ – user3022875 Apr 27 '16 at 17:04
  • 1
    $\begingroup$ Thank you for clearing that up (+1). You might not want to sweat the details, though--that doesn't seem to be the point of these slides. (They have nice conceptual illustrations but the mathematical exposition isn't the clearest.) If you are truly concerned about them, you would have been brought to a skidding halt at slide 38 which depicts a probability density (labeled "Logit") function that becomes negative for values greater than $3$ in size! $\endgroup$ – whuber Apr 27 '16 at 17:25
  • 1
    $\begingroup$ en.wikipedia.org/wiki/… isn't perfect but it may help you out here. You'll see statements of the form $\Pr[Y = 1 | X] = \Pr[\epsilon < \beta \, X]$, as you were expecting. $\endgroup$ – Adrian Apr 27 '16 at 17:30
  • 2
    $\begingroup$ Typos happen. God creates them to enhance our learning experience. :-) $\endgroup$ – whuber Apr 27 '16 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.