# Finding a consistent estimator for area under simple regression line

I am trying to solve the following problem:

Take the following simple linear regression model, where $$x_i \in \mathbb R$$:

$$y_i=\beta_0 + x_i \beta_1 + \epsilon_i$$

Given that:

• $$\mathbb E[\epsilon_i]=0$$
• $$\mathbb E[\epsilon_i|x_i]=0$$
• $$\beta_0 >0$$
• $$\beta_1 <0$$

Let $$\theta_0$$ represent the area under the regression line. Propose a consistent estimator of $$\theta_0$$.

I have began by finding the integral of $$y_i$$ with respect to $$x_i$$. That is,

$$\theta_0= \int \beta_0 + x_i \beta_1 + \epsilon_i$$ $$dx_i=\beta_0x_i + {x_i^2\over{2}}{\beta_1} + \epsilon_ix_i$$

I am considering proposing an MLE and proceeding to find the derivative of the log-likelihood of this expression. However, since this expression is rather complicated and I foresee the MLE computation turning incredibly thorny, I suspect I may be approaching this incorrectly. Any thoughts?

I guess you mean the area $$S(\beta_0, \beta_1)$$ between $$0$$ and the intersection of the regression line with the $$x$$ axis. Then, you can calculate it whether with the triangle formula, or integral. Namely, the intersection point is $$-\beta_0 / \beta_1$$, thus the area is $$S = \beta_0 \beta_0 / (2|\beta_1|) = \frac{\beta_0^2}{2|\beta_1|},$$ or using integration $$\int_0 ^ {-\beta_0 / \beta_1} E[Y|X=x]dx = - \beta_0 \frac{\beta_0}{\beta_1} + \frac{\beta_1 \beta_0 ^ 2}{2\beta_1^2} = - \frac{\beta_0^2}{2\beta_1}.$$ Use the invariance of the MLE, and just replace the $$\beta_0$$ and $$\beta_1$$ with their MLE, and you get the MLE of $$S$$, i.e., $$S_{MLE} = S(\hat\beta_0, \hat\beta_1)$$.