# Plim of ols estimator

Suppose you are given 2 models.

Model of 1 has $$y$$ as the dependent variable and $$x$$ as the independent:

$$y = \beta_0 + \beta_1 x + \epsilon$$

Model 2 has $$w$$ as the dependent variable and $$x$$ as the independent variable

$$w = \lambda_0 + \lambda_1 x + u,$$

such that both models have their own disturbance term.

Then you are given a fitted model in which $$y$$ is dependent on $$w$$.

$$y = \phi_0 + b w + e$$

Now the question is to find plim of $$\hat{b}$$.

I tried making $$x$$ the subject from model 1, plugged the $$x$$ in model 2 and then plugged $$w$$ (model 2) in fitted. Now when I took $$Cov(xy)$$ after all this substitution my fitted model $$y$$ doesn't have and $$x$$. I can't further solve for $$plim$$.

• Please write explicitly the three model equations, so that the relations are clear. For example, we cannot tell whether a constant term(s) exist, or from which model does $\hat b$ is estimated Apr 29 '19 at 18:13
• @Mehreen I tried to write down some of the equations I think you are talking about. However, pls. chek to see if I am guessing right! Dec 19 '20 at 22:48

Here is what I think should be the answer based on your problem description. Otherwise this is hopefully still helpful for you to solve the problem.

I star by considering the model equations

$$(1) \ \ \ y = \beta_0 + \beta_1 x + e_y$$

$$(2) \ \ \ w = \lambda_0 + \lambda_1 x + e_w,$$

and then I manipulate equation (1) to get

$$(1b) \ \ \ y = \beta_0 + \frac{\beta_1}{\lambda_1} ( \lambda_1 x) + e_y,$$

from equation (2) I then find

$$(2b) \ \ \ w -\lambda_0 - e_w= \lambda_1 x,$$

which I insert in (1b) to get

$$y = \beta_0 - \frac{\beta_1}{\lambda_1}\lambda_0 + \frac{\beta_1}{\lambda_1} w - \frac{\beta_1}{\lambda_1} e_w+ e_y,$$

I then define $$b_0 :=\beta_0 - \frac{\beta_1}{\lambda_1}\lambda_0$$ and $$b_1:=\frac{\beta_1}{\lambda_1}$$ and $$u := - \frac{\beta_1}{\lambda_1} e_w+ e_y$$ to get the estimation model

$$y = b_0 + b_1 w + u$$

where I know the probability limit of $$\hat b_1$$ is

$$var(w)^{-1}cov(wy) = b_1 + var(w)^{-1}cov(wu),$$

where I note that $$cov(wu) = cov(w,-(\beta_1/\lambda_1) e_w+ e_y) = -\frac{\beta_1}{\lambda_1} var(e_w)$$.

I assume that $$cov(x,e_w) = cov(x,e_y) = cov(e_w,e_y) = 0$$.

Here is a small test code in R

N <- 1000
beta <- 1
lambda <- 2
x <- rnorm(N)
e_y <- rnorm(N)
e_w <- rnorm(N)
y <- 1 + beta*x + e_y
w <- 1 + lambda*x + e_w

model <- lm(y~w)

W <- cbind(rep(1,N),w)
bias <- -(beta/lambda)*solve(t(W)%*%W)%*%t(W)%*%e_w
coef(model) - bias[1,1]
1 - (beta/lambda)*1
coef(model) - bias[2,1]
beta/lambda


If your question is how to express the plim ($$b$$, say) of the slope estimator from the regression of $$y$$ on $$w$$ in terms of the plim's ($$a$$ and $$c$$, say) of the two slope estimators for your two models (one $$y$$ on $$x$$, and the other $$w$$ on $$x$$), my answer is no that's not possible. You did not give us enough information.

If $$y$$, $$x$$ and $$w$$ have unit variance (for simplification), what you want is $$b=cor(y,w)$$ expressed in terms of $$a=cor(y,x)$$ and $$c=cor(w,x)$$. But that's not possible. You cannot find $$b$$ from $$a$$ and $$c$$. $$Cor(y,w)$$ is not a function of $$cor(y,x)$$ and $$cor(w,x)$$. It is like we cannot find the joint distribution of $$(A,B)$$ from the individual marginal distributions of $$A$$ and $$B$$.

Here are two examples, with the same $$a$$ and $$c$$, but different $$b$$. (i) $$x$$ and $$y$$ are mutually independent ($$a=0$$), $$x$$ and $$w$$ are mutually independent ($$c=0$$), and $$y$$ and $$w$$ are mutually independent ($$b=0$$). (ii) $$x$$ and $$y$$ are mutually independent ($$a=0$$), $$x$$ and $$w$$ are mutually independent ($$c=0$$), and $$y=w$$ ($$b=1$$). You cannot find $$b$$ from $$a$$ and $$c$$.