# How to isolate two distinct effects of a unique independent variable in multiple regression analysis?

Salut!

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

## Problem

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

$$$$Y_t = \beta_a*A_t + \varepsilon_t \;\;\;\;\;\;\; (1)$$$$

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 :

$$$$Y_t = \beta_a*A_t + \beta_b*B_t + \varepsilon_t \;\;\;\;\;\;\; (2)$$$$

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)?

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

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 :)

• 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? – Nick Koprowicz Jan 4 at 0:47
• 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. – Mxml Jan 4 at 21:01

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.