I'm reading a paper in epidemiology.


While I'm reading this article, there is a formula about multiple linear regression.(in this paper, (2) is that.) The author write it without much explanation. I'm trying to calculate it, However, it is very hard to do it.

This formula starts with these assumption.

\begin{equation} U = \alpha_U+\beta_UE+\epsilon_U\\ M = \alpha_M+\beta_MU+\epsilon_M\\ D = \alpha_D+\beta_DU+\epsilon_D\\ \end{equation}

so, we can get D equation with implementation of U

\begin{equation} D = c_1+\beta_D\beta_UE+\epsilon_1\\ \end{equation}

and after adjusting M, the author write that we can get this formula.

\begin{equation} D = c_2+\frac{\beta_D\beta_U}{1+\beta_M^2}E+\frac{\beta_D\beta_M}{1+\beta_M^2}M+\epsilon_2\\ \end{equation}

I'm trying to calculate this formula. As far as I know, the coefficient of multiple regression can be obtained with this formula.

when the multiple linear regression formula is

\begin{equation} \hat{y}=\hat{\beta_0}+\hat{\beta_1} x_1 + \hat{\beta_2} x_2\\ \text{then}\; \hat{{\beta_1}}=\frac{\Sigma \hat{r_{i1}} y_i}{\Sigma r_{i1}^2}\;\;\;(1) \end{equation}

the r is OLS residuals from a simple regression of x_1 on y_1.

Let's suppose that we want to calculate coefficient of E. We can get regression of E on M from assumption.

\begin{equation} E=\frac{1}{\beta_M \beta_U}M+c_3+\epsilon_3\\ \text{then, the r is}\; E-\frac{1}{\beta_M \beta_U}M+c_3\;\;\;(2) \end{equation}

then, we put (2) in (1) and the y is from assumption.

\begin{equation} y_i = D_i = c_1+\beta_D\beta_UE_i+\epsilon_1\\ \end{equation}

and beta of E from multiple linear regression formula is

\begin{equation} \frac{\Sigma (c_1+\beta_D\beta_UE_i+\epsilon_1)(E_i-\frac{1}{\beta_M \beta_U}M_i+c_{3i})}{\Sigma(E_i-\frac{1}{\beta_M \beta_U}M_i+c_{3i})^2} \end{equation}

as you can see, it is almost impossible to calculate it. Though some term can be zero, due to zero sample mean of r, it is still very difficult to calculate it. I tried to make it explicit, but I could not find the answer.

Is there a better way to get coefficient of E?


1 Answer 1


I'm not familiar with the subject of the paper, so I don't have a motivation for the conversion, but from a pure mathematical perspective, the desired formulation can be reached via some algebraical tricks: $$\begin{align}D&=c_1+\beta_D\beta_UE+\epsilon_1=c_1+\epsilon_1+\beta_D\beta_U\frac{1+\beta_M^2}{1+\beta_M^2}E\\&=c_1+\epsilon_1+\frac{\beta_D\beta_U}{1+\beta_M^2}E+\frac{\beta_D\beta_M}{1+\beta_M^2}\beta_M\beta_UE\\&=c_1+\epsilon_1+\frac{\beta_D\beta_U}{1+\beta_M^2}E+\frac{\beta_D\beta_M}{1+\beta_M^2}(M-\alpha_M-\epsilon_M-\beta_M\alpha_U-\beta_M\epsilon_U)\\&=c_2+\epsilon_2+\frac{\beta_D\beta_U}{1+\beta_M^2}E+\frac{\beta_D\beta_M}{1+\beta_M^2}M\end{align}$$

  • $\begingroup$ I really appreciate your help. Though it seems that it is not usual way of calculating coefficient, it helped me a lot. $\endgroup$
    – ESKim
    Commented Aug 4, 2019 at 16:05

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