Tell me more ×
Cross Validated is a question and answer site for statisticians, data analysts, data miners and data visualization experts. It's 100% free, no registration required.

I have the following regression equation, where $Y^*$ is wages:

$$Y^* = \beta_0 + \beta_1 X_1 + \beta_2 X_1^2 + \beta_3 X_1^3 + \beta_4 X_1^4$$

I have been given the Beta values as, $(1.6 , 0.2 , -0.3 , 0.04 , -0.002)$, respectively.

I create a new variable $W$,

$$W=\begin{cases} 0, & \text{if }Y^* \leq 30 \\ Y^*-30, & \text{if }Y^* \gt 30 \end{cases}$$

So I reckon this is a case of left censoring?

The question is how can I compute the marginal effect at $X_1=10$ in this setting ?

I found a formula which said that the $\text{marg effect} = \beta_j*\Phi(\frac{X\beta}{ \sigma})$, but I have no clue how to find $\sigma$..?

Hope you can help :)

share|improve this question

3 Answers

Two points:

  1. The value of $\sigma$ should be a part of the output of the regression model. It is the estimated standard deviation of the error term. There is no way to get it from the $\beta$'s only.
  2. The formula for the marginal effect that you are quoting would apply only if you had 4 different predictors, where you could conceptually fix the values of all the other predictors and modify only the predictor of interest ($X_1$). In your model that is impossible: if $X_1$ changes, so does $X_1^2$, etc.
share|improve this answer
I don't get to estimate the model, so i can't find sigma - so how do i find the partial effects og X_1 in such a model ? – Mark Moritzen May 13 '12 at 9:30
As far as I can tell, there is no way to do it without $\sigma$. – Aniko May 14 '12 at 16:00

I don't see any way of solving this in general without $\sigma$. However, if you are willing to assume that $x$ and $y^*$ are jointly normal, a consistent method of moments estimator for \begin{equation}\Pr[y^*>30]=\Phi(\frac{X\beta}{\sigma})\end{equation} is $\frac{n_{U}}{N}$ where $n_{U}$ is the number of uncensored observations and $N$ is the total number of observations. This makes your marginal effect of x on $y^*$ \begin{equation}(\beta_{1}+2\beta_{2}10+3\beta_{3}10^2+3\beta_{3}10^3)\frac{n_{U}}{N}.\end{equation}

This obviously requires additional information.

share|improve this answer

As you have written it, this is not a stochastic model.

It should have a stochastic error term,

$Y_i^* = \beta_0 + \beta_1 X_{1i} + \beta_2 X_{1i}^2 + \beta_3X_{1i}^2 + \beta_4 X_{1i}^4 +\varepsilon_i$

where typically a parametric assumption is made for the error term, something like $\varepsilon_i \sim N(0, \sigma^2)$ which is where the $\sigma$ comes from in the canonical Tobit model.

Lastly, your outcome variable is generated by the censoring rule

$W_i = \left(Y_i^*- 30\right)\mathbb{1}\left[Y_i^* > 30\right]$

For this model of the outcome, the marginal effects of the $j$-th regressor is given by the forumla that you have presented:

$\beta_j\Phi\left(\frac{\mathbf{X}_i'\boldsymbol{\beta}}{\sigma}\right)$

where the argument of the standard normal CDF is the entire linear index $\mathbf{X}_i'\boldsymbol{\beta}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.