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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 $plim$.

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    $\begingroup$ 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 $\endgroup$ – Alecos Papadopoulos Apr 29 '19 at 18:13
  • $\begingroup$ @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! $\endgroup$ – Jesper Hybel Dec 19 '20 at 22:48
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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)[1] - bias[1,1]
1 - (beta/lambda)*1
coef(model)[2] - bias[2,1]
beta/lambda
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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 $c$.

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