<p>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.</p>
<p>I star by considering the model equations</p>
<p><span class="math-container">$$(1) \ \ \ y = \beta_0 + \beta_1 x + e_y $$</span></p>
<p><span class="math-container">$$(2) \ \ \ w = \lambda_0 + \lambda_1 x + e_w, $$</span></p>
<p>and then I manipulate equation (1) to get</p>
<p><span class="math-container">$$(1b) \ \ \ y = \beta_0 + \frac{\beta_1}{\lambda_1} ( \lambda_1 x) + e_y, $$</span></p>
<p>from equation (2) I then find</p>
<p><span class="math-container">$$(2b) \ \ \ w -\lambda_0 - e_w=  \lambda_1 x,$$</span></p>
<p>which I insert in (1b) to get</p>
<p><span class="math-container">$$y = \beta_0 - \frac{\beta_1}{\lambda_1}\lambda_0 + \frac{\beta_1}{\lambda_1} w - \frac{\beta_1}{\lambda_1} e_w+ e_y,$$</span></p>
<p>I then define <span class="math-container">$b_0 :=\beta_0 - \frac{\beta_1}{\lambda_1}\lambda_0$</span> and <span class="math-container">$b_1:=\frac{\beta_1}{\lambda_1}$</span>  and <span class="math-container">$u :=  - \frac{\beta_1}{\lambda_1} e_w+ e_y$</span> to get the estimation model</p>
<p><span class="math-container">$$y = b_0 + b_1 w + u$$</span></p>
<p>where I know the probability limit of <span class="math-container">$\hat b_1$</span> is</p>
<p><span class="math-container">$$var(w)^{-1}cov(wy) = b_1 + var(w)^{-1}cov(wu),$$</span></p>
<p>where I note that <span class="math-container">$cov(wu) = cov(w,-(\beta_1/\lambda_1) e_w+ e_y) = -\frac{\beta_1}{\lambda_1} var(e_w)$</span>.</p>
<p>I assume that <span class="math-container">$cov(x,e_w) = cov(x,e_y) = cov(e_w,e_y) = 0$</span>.</p>
<p>Here is a small test code in R</p>
<pre><code>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
</code></pre>

CC BY-SA 4.0
2020-12-20T00:19:21.490Z
2020-12-20T00:19:21.490Z
Jesper Hybel
<p>If your question is how to express the plim (<span class="math-container">$b$</span>, say) of the slope estimator from the regression of <span class="math-container">$y$</span> on <span class="math-container">$w$</span> in terms of the plim's (<span class="math-container">$a$</span> and <span class="math-container">$c$</span>, say) of the two slope estimators for your two models (one <span class="math-container">$y$</span> on <span class="math-container">$x$</span>, and the other <span class="math-container">$w$</span> on <span class="math-container">$x$</span>), my answer is no that's not possible. You did not give us enough information.</p>

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

<p>Here are two examples, with the same <span class="math-container">$a$</span> and <span class="math-container">$c$</span>, but different <span class="math-container">$b$</span>. (i) <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are mutually independent (<span class="math-container">$a=0$</span>), <span class="math-container">$x$</span> and <span class="math-container">$w$</span> are mutually independent (<span class="math-container">$c=0$</span>), and <span class="math-container">$y$</span> and <span class="math-container">$w$</span> are mutually independent (<span class="math-container">$b=0$</span>). (ii) <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are mutually independent (<span class="math-container">$a=0$</span>), <span class="math-container">$x$</span> and <span class="math-container">$w$</span> are mutually independent (<span class="math-container">$c=0$</span>), and <span class="math-container">$y=w$</span> (<span class="math-container">$b=1$</span>). You cannot find <span class="math-container">$b$</span> from <span class="math-container">$a$</span> and <span class="math-container">$c$</span>.</p>

CC BY-SA 4.0
2018-11-30T13:13:09.827Z
2018-11-30T13:13:09.827Z
chan1142
<p>Suppose you are given 2 models.</p>
<p>Model of 1 has <span class="math-container">$y$</span> as the dependent variable and <span class="math-container">$x$</span> as the independent:</p>
<p><span class="math-container">$$y = \beta_0  + \beta_1 x + \epsilon$$</span></p>
<p>Model 2 has <span class="math-container">$w$</span> as the dependent variable and <span class="math-container">$x$</span> as the independent variable</p>
<p><span class="math-container">$$w = \lambda_0 + \lambda_1 x + u,$$</span></p>
<p>such that both models have their own disturbance term.</p>
<p>Then you are given a fitted model in which <span class="math-container">$y$</span> is dependent on <span class="math-container">$w$</span>.</p>
<p><span class="math-container">$$y = \phi_0 + b w + e$$</span></p>
<p>Now the question is to find plim of <span class="math-container">$\hat{b}$</span>.</p>
<p>I tried making <span class="math-container">$x$</span> the subject from model 1, plugged the <span class="math-container">$x$</span> in model 2 and then plugged <span class="math-container">$w$</span> (model 2) in fitted. Now when I took <span class="math-container">$Cov(xy)$</span> after all this substitution my fitted model <span class="math-container">$y$</span> doesn't have and <span class="math-container">$x$</span>. I can't further solve for <span class="math-container">$plim$</span>.</p>

CC BY-SA 4.0
2019-04-29T18:13:51.510Z
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
Alecos Papadopoulos
CC BY-SA 4.0
2020-12-19T22:48:16.150Z
@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!
Jesper Hybel
<econometrics>
Plim of ols estimator
Mehreen Rauf
# Plim of ols estimator

(Asked by: username_0 on 2018-11-26)

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 $plim$.

**Tags:** econometrics
> **Comments:**
> **username_1**: 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  
> **username_2**: @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!  

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## 2 answers

### Answer by username_2 on 2020-12-20. Score: 3

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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### Answer by username_3 on 2018-11-30. Score: 0

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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