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Waqas
Waqas
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I hope I understood your question. The equation for linear regression (without the intercept) can be written as follows: \begin{equation} y_i=\beta_1x_{i,1}+\beta_2x_{i,2}+…+\beta_{p}x_{i,p}+\epsilon_i. \end{equation} For $(a)$ using the transformation $x_i=x_i+(x_i * x_{i+1})$ and we get the following: \begin{equation} y_i=\beta_1x_{i,1}+\beta_1x^2_{i,1}+\beta_2x_{i,2}+\beta_2x_{i,2}+\beta_2x_{i,2}x_{i,3}+...+\beta_{p}x^2_{i,p}+\epsilon_i \end{equation} For $(b)$ using the transformation $x_i=x_i^2$ and we get the following: \begin{equation} y_i=\beta_1x^2_{i,1}+\beta_2x^2_{i,2}+…+\beta_{p}x^2_{i,p}+\epsilon_i. \end{equation} where $i=1,2,...,t$ and $t$ is some finite integer.

Note: $(a)$ and $(b)$ are still linear.

I hope I understood your question. The equation for linear regression (without the intercept) can be written as follows: \begin{equation} y_i=\beta_1x_{i,1}+\beta_2x_{i,2}+…+\beta_{p}x_{i,p}+\epsilon_i. \end{equation} For $(a)$ using the transformation $x_i=x_i+(x_i * x_{i+1})$ and we get the following: \begin{equation} y_i=\beta_1x_{i,1}+\beta_1x^2_{i,1}+\beta_2x_{i,2}+\beta_2x_{i,2}+\beta_2x_{i,2}x_{i,3}+...+\beta_{p}x^2_{i,p}+\epsilon_i \end{equation} For $(b)$ using the transformation $x_i=x_i^2$ and we get the following: \begin{equation} y_i=\beta_1x^2_{i,1}+\beta_2x^2_{i,2}+…+\beta_{p}x^2_{i,p}+\epsilon_i. \end{equation} where $i=1,2,...,t$ and $t$ is some finite integer.

Note: $(a)$ and $(b)$ are still linear.

I hope I understood your question. The equation for linear regression (without the intercept) can be written as follows: \begin{equation} y_i=\beta_1x_{i,1}+\beta_2x_{i,2}+…+\beta_{p}x_{i,p}+\epsilon_i. \end{equation} For $(a)$ using the transformation $x_i=x_i+(x_i * x_{i+1})$ and we get the following: \begin{equation} y_i=\beta_1x_{i,1}+\beta_1x^2_{i,1}+\beta_2x_{i,2}+\beta_2x_{i,2}+\beta_2x_{i,2}x_{i,3}+...+\beta_{p}x^2_{i,p}+\epsilon_i \end{equation} For $(b)$ using the transformation $x_i=x_i^2$ and we get the following: \begin{equation} y_i=\beta_1x^2_{i,1}+\beta_2x^2_{i,2}+…+\beta_{p}x^2_{i,p}+\epsilon_i. \end{equation} where $i=1,2,...,t$ and $t$ is some finite integer.

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Waqas
Waqas
  • 426
  • 4
  • 16

I hope I understood your question. The equation for linear regression (without the intercept) can be written as follows: \begin{equation} y_i=\beta_1x_{i,1}+\beta_2x_{i,2}+…+\beta_{p}x_{i,p}+\epsilon_i. \end{equation} For $(a)$ using the transformation $x_i=x_i+(x_i * x_{i+1})$ and we get the following: \begin{equation} y_i=\beta_1x_{i,1}+\beta_1x^2_{i,1}+\beta_2x_{i,2}+\beta_2x_{i,2}+\beta_2x_{i,2}x_{i,3}+...+\beta_{p}x^2_{i,p}+\epsilon_i \end{equation} For $(b)$ using the transformation $x_i=x_i^2$ and we get the following: \begin{equation} y_i=\beta_1x^2_{i,1}+\beta_2x^2_{i,2}+…+\beta_{p}x^2_{i,p}+\epsilon_i. \end{equation} wheterwhere $i=1,2,...,t$ and $t$ is some finite integer.

Note: $(a)$ and $(b)$ are still linear.

I hope I understood your question. The equation for linear regression (without the intercept) can be written as follows: \begin{equation} y_i=\beta_1x_{i,1}+\beta_2x_{i,2}+…+\beta_{p}x_{i,p}+\epsilon_i. \end{equation} For $(a)$ using the transformation $x_i=x_i+(x_i * x_{i+1})$ and we get the following: \begin{equation} y_i=\beta_1x_{i,1}+\beta_1x^2_{i,1}+\beta_2x_{i,2}+\beta_2x_{i,2}+\beta_2x_{i,2}x_{i,3}+...+\beta_{p}x^2_{i,p}+\epsilon_i \end{equation} For $(b)$ using the transformation $x_i=x_i^2$ and we get the following: \begin{equation} y_i=\beta_1x^2_{i,1}+\beta_2x^2_{i,2}+…+\beta_{p}x^2_{i,p}+\epsilon_i. \end{equation} wheter $i=1,2,...,t$ and $t$ is some finite integer.

Note: $(a)$ and $(b)$ are still linear.

I hope I understood your question. The equation for linear regression (without the intercept) can be written as follows: \begin{equation} y_i=\beta_1x_{i,1}+\beta_2x_{i,2}+…+\beta_{p}x_{i,p}+\epsilon_i. \end{equation} For $(a)$ using the transformation $x_i=x_i+(x_i * x_{i+1})$ and we get the following: \begin{equation} y_i=\beta_1x_{i,1}+\beta_1x^2_{i,1}+\beta_2x_{i,2}+\beta_2x_{i,2}+\beta_2x_{i,2}x_{i,3}+...+\beta_{p}x^2_{i,p}+\epsilon_i \end{equation} For $(b)$ using the transformation $x_i=x_i^2$ and we get the following: \begin{equation} y_i=\beta_1x^2_{i,1}+\beta_2x^2_{i,2}+…+\beta_{p}x^2_{i,p}+\epsilon_i. \end{equation} where $i=1,2,...,t$ and $t$ is some finite integer.

Note: $(a)$ and $(b)$ are still linear.

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Waqas
Waqas
  • 426
  • 4
  • 16

I hope I understood your question. The equation for linear regression (without the intercept) can be written as follows: \begin{equation} y_i=\beta_1x_{i,1}+\beta_2x_{i,2}+…+\beta_{p}x_{i,p}+\epsilon_i. \end{equation} For $(a)$ useusing the transformation $x_i=x_i+(x_i * x_{i+1})$ and we get the following: \begin{equation} y_i=\beta_1x_{i,1}+\beta_1x^2_{i,1}+\beta_2x_{i,2}+\beta_2x_{i,2}+\beta_2x_{i,2}x_{i,3}+...+\beta_{p}x^2_{i,p}+\epsilon_i \end{equation} For $(b)$ useusing the transformation $x_i=x_i^2$ and we get the following: \begin{equation} y_i=\beta_1x^2_{i,1}+\beta_2x^2_{i,2}+…+\beta_{p}x^2_{i,p}+\epsilon_i. \end{equation} wheter $i=1,2,...,t$ and $t$ is some finite integer.

Note: $(a)$ and $(b)$ are still linear.

I hope I understood your question. The equation for linear regression (without the intercept) can be written as follows: \begin{equation} y_i=\beta_1x_{i,1}+\beta_2x_{i,2}+…+\beta_{p}x_{i,p}+\epsilon_i. \end{equation} For $(a)$ use the transformation $x_i=x_i+(x_i * x_{i+1})$ \begin{equation} y_i=\beta_1x_{i,1}+\beta_1x^2_{i,1}+\beta_2x_{i,2}+\beta_2x_{i,2}+\beta_2x_{i,2}x_{i,3}+...+\beta_{p}x^2_{i,p}+\epsilon_i \end{equation} For $(b)$ use the transformation $x_i=x_i^2$: \begin{equation} y_i=\beta_1x^2_{i,1}+\beta_2x^2_{i,2}+…+\beta_{p}x^2_{i,p}+\epsilon_i. \end{equation} wheter $i=1,2,...,t$ and $t$ is some finite integer.

Note: $(a)$ and $(b)$ are still linear.

I hope I understood your question. The equation for linear regression (without the intercept) can be written as follows: \begin{equation} y_i=\beta_1x_{i,1}+\beta_2x_{i,2}+…+\beta_{p}x_{i,p}+\epsilon_i. \end{equation} For $(a)$ using the transformation $x_i=x_i+(x_i * x_{i+1})$ and we get the following: \begin{equation} y_i=\beta_1x_{i,1}+\beta_1x^2_{i,1}+\beta_2x_{i,2}+\beta_2x_{i,2}+\beta_2x_{i,2}x_{i,3}+...+\beta_{p}x^2_{i,p}+\epsilon_i \end{equation} For $(b)$ using the transformation $x_i=x_i^2$ and we get the following: \begin{equation} y_i=\beta_1x^2_{i,1}+\beta_2x^2_{i,2}+…+\beta_{p}x^2_{i,p}+\epsilon_i. \end{equation} wheter $i=1,2,...,t$ and $t$ is some finite integer.

Note: $(a)$ and $(b)$ are still linear.

added 52 characters in body
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Waqas
Waqas
  • 426
  • 4
  • 16
Source Link
Waqas
Waqas
  • 426
  • 4
  • 16