Probability Density Function in a Tobit Model

Question

I have been taught the following derivation of the tobit model from a latent underlying variable model:

Suppose:

• $y*=$ latent variable (e.g. utility)
• $y=max[0,y*]$

And $$y* = \mathbf X\beta + u,\qquad u|\mathbf X \sim Normal(0,\sigma^2) $$

Where the intercept has been absorbed into $\mathbf X_i$ for notational simplicity.

According to lectures and the textbook (Wooldridge, 2012, p. 597), the density of $y$ given $\mathbf X$ is:

$$ P(y=0| \mathbf X )\quad = \quad P(y*<0 \ | \ \mathbf X) $$ $$ \qquad \qquad \qquad \qquad \quad = \quad P( \mathbf X \beta \ + \ \epsilon<0 \ | \ \mathbf X) $$ $$\qquad \qquad \qquad \quad \ = \quad P( \epsilon<- \mathbf X \beta \ | \ \mathbf X)$$

$$ \qquad \qquad \qquad \quad \ = \quad P \left( \frac{\epsilon}{\sigma} < - \frac{\mathbf X \beta}{\sigma} \ | \ \mathbf X \right) $$

$$\qquad \qquad \qquad \quad \quad = \quad \phi \left(- \frac{\mathbf X \beta}{\sigma} \right), \ \frac{\epsilon}{\sigma} \sim N(0,1) $$

I understand the derivation up to here. According to the lecture, it then follows that the density of y given x is:

$$ \frac{1}{\sqrt{2\pi \sigma^2}} e^{-(y-\mathbf X \beta)^2/2\sigma^2) }$$

How is this implied by the preceding line? In my mind, if the standard normal function is given by:

$$ \phi(v) = \frac{1}{\sqrt{2\pi}} e^\frac{-v^2}{2} $$

where $v = \frac{\epsilon}{\sigma} = \frac{y- \mathbf X \beta}{\sigma} $

wouldn't this make the density: $$ \phi \left(\frac{y- \mathbf X\beta}{\sigma} \right) = \frac{1}{\sqrt{2\pi}} e^\frac{-(y- \mathbf X \beta)^2}{2 \sigma^2} $$

I apologize for the shoddy formatting, first time trying MathJax.

Thank you!

Answer 1

I think your notations are a little bit confusing. I think $\phi \left(- \frac{\mathbf x \beta}{\sigma} \right)$, should be$ ...P(y*<0 \ | \ \mathbf x)... =\Phi \left(- \frac{\mathbf x \beta}{\sigma} \right)$ which is CDF.

Above CDF is for $y=0$, note since $y=max[0,y*]$, $y$ only can be $\ge 0$

For $y>0$ then $y=y*$. The pdf of $y$ can be get from $y*$

Since $y* = \mathbf x\beta + u,\qquad u|\mathbf x \sim Normal(0,\sigma^2)$. (Note the book does not use captial $X$)

Therefore $y=y*\sim N(x\beta,\sigma^2),y>0$ i.e $(x\beta+u)\sim N(x\beta,\sigma^2)$

The pdf of $y*$ is

$$\frac{1}{\sqrt{2\pi \sigma^2}} e^{-(y-\mathbf X \beta)^2/2\sigma^2) }$$

which is also pdf of $y$ for $y>0$

Also your last formula for change of variable is not correct, you missed Jacobian $\frac{1}{\sigma}$, I think.