# regression coefficient on sum of regressors

Say I have the true linear model with normal errors:

$y = \beta_1 X_1 + \beta_2 X_2 + \epsilon$

However, I only observe $Z = X_1 + X_2$, so I estimate instead:

$y = \delta (X_1 + X_2) + e$,

can I express $\delta$ in terms of $\beta_1$ and $\beta_2$ (and perhaps $X_1$ and $X_2$)?

Observably when I generate data and estimate the results $\delta$ seems to be the average of the coefficients weighted by the variance of the regressors, but I can't seem to derive a solution.

## Incorporating hints:

Thanks for the hints, which are helpful. Here is an attempt -- although I feel I haven't gotten to the bottom of it yet:

$y = a(X_1 + X_2) + b(X_1 - X_2) + \epsilon$ (thanks @whuber!)

$y = aX_1 + aX_2 + bX_1 - bX_2 + \epsilon$,

$y = (a + b)X_1 + (a - b)X_2 + \epsilon$, therefore

$\beta_1 = (a+b)$ and $\beta_2 = (a - b)$.

Solving for $a$ and $b$ we get:

$a = (\beta_1 + \beta_2)/2$ and $b = (\beta_1 - \beta_2)/2$

Effectively I am estimating $a$ while omitting $b$, so

$y = a(X_1 + X_2) + u,$

$u = b(X_1 - X_2) + \epsilon$, therefore my estimate of $\delta$ above will be the estimate of $a$ with omitted variable bias. Calling $X_1 + X_2 = Z$, my estimate of $\hat{\delta}$ is therefore:

$\hat{\delta} = (Z'Z)^{-1}(Z'[Za + (X_1 - X_2)b + \epsilon])$

Ignoring $\epsilon$ as it disappears in expectation, I get

$\hat{\delta} = a + (Z'Z)^{-1}(Z'(X_1 - X_2))b$

$\hat{\delta} = \frac{\beta_1 + \beta_2}{2} + (Z'Z)^{-1}Z'[X_1 - X_2] \frac{\beta_1 - \beta_2}{2}$

Does this look on the right track? Can we reduce the trailing second expression?

I was thinking that since the expression $(Z'Z)^{-1}Z'[X_1 - X_2]$ looks like a regression of $X_1 - X_2$ on $Z$, I could perhaps re-write it as $Var(X_1 + X_2)^{-1}Cov(X_1 + X_2, X_1 - X_2) = Var(X_1+X_2)^{-1}(Var(X_1) - Var(X_2))$?

• Hint: Comparing your model to the original one as written in the form $$y = \alpha(X_1+X_2) + \beta(X_1-X_2)+\epsilon$$ is routine and your model is the case $\beta=0.$
– whuber
Apr 19, 2018 at 13:24
• Another hint: you can't do it only with $\beta_1$ and $\beta_2$, you need information about the variance of these estimates Apr 19, 2018 at 14:26
• Thanks for the hints @whuber and Richard Border, I made some edits to reflect progress. The expression I derived still looks a little complicated -- is there a simpler way of thinking about it?
– gfgm
Apr 19, 2018 at 15:35
• Yes, there is a simpler way: this multiple regression can be accomplished in two stages of simple regression, as described at stats.stackexchange.com/questions/46185/….
– whuber
Apr 19, 2018 at 15:44
• Thanks @whuber the link was helpful, I think I found the easier way your were indicating. Is the citation you are referencing in the linked post the book Data Analysis and Regression?
– gfgm
Apr 20, 2018 at 8:58

With a great deal of prodding (full credit to @whuber) I seem to have solved it:

We re-write $$y = \beta_1X_1 + \beta_2X_2 + \epsilon$$ as

$$y = a(X_1 + X_2) + b(X_1 - X_2) + \epsilon$$

$$y = aX_1 + aX_2 + bX_1 - bX_2 + \epsilon$$,

$$y = (a + b)X_1 + (a - b)X_2 + \epsilon$$, therefore

$$\beta_1 = (a+b)$$ and $$\beta_2 = (a - b)$$.

Solving for $$a$$ and $$b$$ we get:

$$a = (\beta_1 + \beta_2)/2$$ and $$b = (\beta_1 - \beta_2)/2$$

What I estimate is equivalent to

$$y = \delta(X_1 + X_2) + u,$$

$$u = b(X_1 - X_2) + \epsilon$$. We know that the $$k$$th coefficient of a multiple regression can be recovered by first partialling out the regressors other than $$k$$. Thus, if I let $$z = X_1 + X_2$$ and $$q = X_1 - X_2$$, the regression coefficient $$b$$ in:

$$y = az + bq + \epsilon$$ can be obtained with the steps

(1) $$y = \delta z + u$$,

(2) $$q = \lambda z + e$$

(3) $$u = be + \epsilon$$. Substituting expressions back in,

$$y = \delta z + be + \epsilon$$ from (1).

$$y = \delta z + b(q - \lambda z) + \epsilon$$ from (2). Therefore:

$$y = (\delta - b \lambda)z + bq + \epsilon$$. Now we know that $$(\delta - b\lambda ) = a = (\beta_1 + \beta_2)/2$$, so we solve for $$\delta$$ which yields:

$$\delta = (\beta_1 + \beta_2)/2 + \lambda (\beta_1 - \beta_2)/2$$

Here is some R code to check the solution, or to play around with:

rep.func <- function(N  = 100, b1 = 2, b2 = 10, s1 = 1, s2 = 5) {
x1 <- rnorm(N, 0, s1)
x2 <- rnorm(N, 0, s2)
eps <- rnorm(N)
y <- x1*b1 + x2*b2 + eps
z <- x1 + x2
q <- x1 - x2
lambda <- lm(q ~ z - 1)$$coefficients est.delta <- lm(y ~ z - 1)$$coefficients
est.betas <- lm(y ~ x1 + x2 - 1)$coefficients derived.delta <- sum(est.betas)/2 + lambda * (est.betas[1] - est.betas[2])/2 c("est.delta" = est.delta, "derived.delta" = derived.delta) } rep.func() #> est.delta.z derived.delta.z #> 9.77236 9.77236 library(ggplot2) res <- t(replicate(1000, rep.func())) all.equal(res[,1], res[,2]) #> [1] TRUE ggplot(as.data.frame(res), aes(est.delta.z, derived.delta.z)) + geom_point() + geom_smooth(method='lm') + theme_minimal()  ### Update for case with P predictors, September 2020 Returning to this much later to generalize to case with P predictors at request. The underlying principle is the same: we are going to 1. re-parameterize the true data generating process so that it contains a term for the quantity we are interested in, 2. determine the value of the re-parameterized coefficients in terms of the original coefficients, 3. evaluate the effect of ommitted variable bias on the estimation of the re-parameterized coefficient With $$P$$ predictors the true data generating process is: $$y = \sum_{p=1}^{P} \beta_p x_p + \epsilon$$. This true data generating process can be re-parameterized in a way that is mathematically equivalent as $$y = a (\sum_{p=1}^{P} x_p) + \sum_{p=2}^{P} b_p (x_1 - x_p) + \epsilon \\ y = (a + \sum_{p=2}^{P} b_p) x_1 + \sum_{p=2}^{P}(a - b_p)x_p + \epsilon$$ Now we can solve for these new parameters (a and b's) in terms of the old parameters ($$\beta$$'s). $$a + \sum_{p=2}^{P} b_p = \beta_1 \\ a - b_2 = \beta_2 \\ ... \\ a - b_P = \beta_P$$ therefore $$a = \frac{\sum_{p=1}^{P} \beta_P}{P} = \bar{\beta} \\ b_p = \bar{\beta} - \beta_p$$. Now lets simplify our life a little by setting $$z = \sum_{p=1}^{P}x_p \\ Q = x_1 - x_p \\$$ so we can write $$y = \bar{\beta}z + QA' + \epsilon$$ where $$Q$$ is an $$N \times (P-1)$$ matrix of our column differences and $$A$$ is a $$(P-1)$$-length row-vector collecting coefficients. We now move to problem 3, namely that we are not actually estimating the model above we are estimating $$y = \delta z + u$$ We just repeat the process above for when P = 2 (the application of Frisch-Waughl-Lovell) but try not to lose track of dimensions: First we partial $$z$$ out of each column of $$Q$$ with some abuse of notation $$q_p = \lambda_p z + e_p$$ so $$\Lambda$$ would be a vector of length $$(P-1)$$, and $$e$$ would be an $$N$$ by $$(P-1)$$ matrix. Then we can re-write $$u$$ with $$z$$ removed: $$u = eA' + \epsilon$$ Plug-in terms $$y = \delta z + eA' + \epsilon \\ y = \delta z + QA' - (\Lambda A') z + \epsilon \\ y = (\delta - \Lambda A') z + QA' + \epsilon \\ \delta = \bar{\beta} + \Lambda A'$$ In words, this seems to say that the regressor in the short regression of $$y$$ on the sum of predictors, will be equal to the average of the regression coefficients in the original model, plus a 'bias' term equal in magnitude to the inner product of the ommitted coefficients in the reparameterization and the regression coefficients from a regression of the ommitted variables on the included variable. Since its always risky to get your math from randos on the internet here's a simulation to support the derivation: ## Load mvtnorm because more interesting # when predictors are correlated library(mvtnorm) library(ggplot2) # create correlation matrix set.seed(42) rep_func <- function(N = 1000, P =10) { # how the predictors are correlated sig <- matrix(round(runif(P^2),2), nrow = P) diag(sig) <- 1:P sig[lower.tri(sig)] <- t(sig)[lower.tri(sig)] # Draw covariates X <- matrix(rmvnorm(N, mean = 1:P, sigma = sig), nrow = N) # draw true parameters betas <- rnorm(P) # draw true errors eps <- rnorm(N) # gen y y <- X %*% betas + eps # predictor sums Z <- rowSums(X) # create some helpers x1 <- X[,1] Q <- x1 - X[,2:P] # delta through regression of y on sum of predictors delta <- lm(y ~ Z - 1)$coef

# true coefficients from reparameterization
agg.reg <- lm(y ~ Z + Q - 1)$coef a <- agg.reg[1] A <- agg.reg[2:P] # lambdas lambda <- lm(Q ~ Z - 1)$coef

# compare deltas
delta_est <- sum(A * lambda) + mean(betas)

c("lm_delta" = delta[[1]], "derived_delta" = delta_est)
}

rep_func()
#>      lm_delta derived_delta
#>     0.3196803     0.3229088

res <- as.data.frame(t(replicate(1000, rep_func())))
#> Warning in rmvnorm(N, mean = 1:P, sigma = sig): sigma is numerically not
#> positive semidefinite

ggplot(res, aes(lm_delta - derived_delta)) + geom_density() +
geom_vline(xintercept = 0) +
hrbrthemes::theme_ipsum() +
xlim(c(-1*sd(res$$lm_delta), 1*sd(res$$lm_delta))) +
ggtitle("LM estimated delta - derived delta",
subtitle = "X-axis +/- 1 standard error of estimate")


Created on 2020-09-20 by the reprex package (v0.3.0)

• Thank you so much for this great explanation of the k predictor case! Sep 20, 2020 at 15:36
• Sorry for another follow-up question. Is it also possible to solve this problem without regressing y on Z and Q as you do with agg.reg <- lm(y ~ Z + Q - 1)$coef? I liked in the original solution, that there is only a regression of y on the the two predictors to get the estimates for beta, but the rest of the solution is simply up to Z and Q (and thus$X_1$and$X_2$). Is there any way to write this like you did in your initial post without involving y again? I'm talking about your solution$\hat{\delta} = \frac{\beta_1 + \beta_2}{2} + (Z'Z)^{-1}Z'[X_1 - X_2] \frac{\beta_1 - \beta_2}{2}$Sep 20, 2020 at 19:20 • Yeah, each element of$A$is$\bar{\beta} - \beta_p$so you could re-write the solution as$\delta = \bar{\beta}+\Lambda (\bar{\beta} - \beta_p)\$ where the final expression in parens should be read as a (P-1) length vector
– gfgm
Sep 20, 2020 at 20:32
• Thank you so much, this worked perfect! Sep 20, 2020 at 21:23