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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?

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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., $S_{MLE} = S(\hat\beta_0, \hat\beta_1)$.

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  • $\begingroup$ I came to the same conclusion, but I couldn't justify convergence in probability. Could you provide a reference on the MLE invariance that proves the last passage? $\endgroup$
    – Snoop
    Commented Nov 19, 2021 at 23:58
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    $\begingroup$ stats.stackexchange.com/questions/77573/… $\endgroup$
    – V. Vancak
    Commented Nov 20, 2021 at 0:01
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    $\begingroup$ Thanks! @V. Vancak $\endgroup$
    – Snoop
    Commented Nov 20, 2021 at 0:03

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