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 $plim$.
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 &lt;- 1000
beta &lt;- 1
lambda &lt;- 2
x &lt;- rnorm(N)
e_y &lt;- rnorm(N)
e_w &lt;- rnorm(N)
y &lt;- 1 + beta*x + e_y
w &lt;- 1 + lambda*x + e_w

model &lt;- lm(y~w)

W &lt;- cbind(rep(1,N),w)
bias &lt;- -(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