First of all thank you if you're reading this post and I wish you all the best for 2020!

Edit 1: All variables are continuous variables (no dummy or categorical)

Edit 2: An illustration of the problem has been added at the end of the post


My question can be illustrated with a simple regression as follow :

\begin{equation} Y_t = \beta_a*A_t + \varepsilon_t \;\;\;\;\;\;\; (1) \end{equation}

Assume that the effect of variable $A_t$ on $Y_t$ have two (and only two) specific channels, meaning that $A_t$ have two distinct effects (let's call them $e_1$ and $e_2$ for the sake of simplicity) on $Y_t$.

In equation (1), we can only conclude about the overall effect of $A$ on $Y$; where $\beta_a$ would represent both the sum of the effects $e_1 + e_2$, right?

Now assume we can characterize one of the channel using another variable, say $B_{t}$. This variable $B_t$ allows us to control for the effect $e_1$, but also capture other effects. Including it in Equation (1), we obtain :

\begin{equation} Y_t = \beta_a*A_t + \beta_b*B_t + \varepsilon_t \;\;\;\;\;\;\; (2) \end{equation}

When we estimate both Equation $(1)$ and $(2)$, we find two different $\hat{\beta_a}$. But, can we draw any conclusion from there about $e_1$ and $e_2$?

My intuition is that:

  • In Equation $(1)$: $\hat{\beta_a}$ accounts for $e_1 + e_2$
  • In Equation $(2)$: $\hat{\beta_a}$ accounts only for $e_2$ since we control for $e_1$ with the variable $B_t$

Is my intuition correct? If yes, is there a way I can estimate/measure $e_1$ from there ($\beta_b \neq e_1$)?

Potential solution:

Would integrating an interaction term of $A$ and $B$ be a way to capture $e_1$ (as follow)?

\begin{equation} Y_t = \beta_a*A_t + \beta_b*B_t + \beta_{ab}*A_tB_t + \varepsilon_t \;\;\;\;\;\;\; (3) \end{equation}

In which:

  • $\hat{\beta_a}$ : would account for $e_2$
  • $\hat{\beta_b}$ : would account for the general effect of $B_t$
  • $\hat{\beta_{ab}}$ : would account for $e_1$ (which would change based on the level of $B_t$)

Any help would be much appreciated. I hope the problem is clear enough, don't hesitate to ask if you need further information/specification. Thank you :)

Illustration of the problem

  • 3
    $\begingroup$ When you say that the variable has two specific channels, do you mean that it's a factor (or categorical) variable with two levels? And those levels are e1 and e2? $\endgroup$ Jan 4, 2020 at 0:47
  • $\begingroup$ Hi Nick, thank you for your answer. Actually A, B and Y are all continuous. By the two channels I mean that when A changes, it has two direct impacts on Y. As an exemple you could think of Y as the average price in a market, and A as the quantity of cheap goods produced in this market. If A increases, the average price on the market will drop because more cheap products might be sold (effect 1), but also because the market share of middle-range and expensive products might be reduced (effect 2). I don't know if the exemple is really good, but that's the idea. $\endgroup$
    – Mxml
    Jan 4, 2020 at 21:01

1 Answer 1


Multiple regression model always capture effect et ceteris paribus for each variable, meaning when every other variables are held constant. I think this property is enough to justify my first intuition, that is:

  • In Equation $(1)$: $\hat{\beta_a}$ accounts for $e_1 + e_2$
  • In Equation $(2)$: $\hat{\beta_a}$ accounts only for $e_2$; $\hat{\beta_b}$ accounts for the general effect of $B_t$

$e_1$ is not captured anymore in equation $(2)$, but rather goes is in the error term $\varepsilon_t$, am I right ?

Thus, adding an interaction term, see equation $(3)$, would rather leads :

  • $\hat{\beta_a}$: accounts for $e_2$
  • $\hat{\beta_b}$: accounts for the general effect of $B_t$
  • $\hat{\beta_{ab}}$: accounts for the interaction between $e_2$ and $B_t$

I can argue without the measure of $e_1$, but I'm still very curious if there is a way to measure it?

Since we know the value of ($e_1$ + $e_2$) and the value of $e_2$ only, I guess there is some sort of method, maybe I've learned it before but can't remember. Any help would be welcome please.

  • $\begingroup$ Consider incorporating this elaboration directly into your post. It doesn’t seem to answer the question. $\endgroup$
    – utobi
    Oct 30, 2022 at 7:39

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