correlation coefficient in linear regression My interest is to develop a relation of the correlation coefficient when the data (both the dependent and independent variables) have measurement errors.
Intro
The measured values are related to the true / actual values by:
\begin{align}
\newcommand{\Var}{{\rm Var}}\newcommand{\cov}{{\rm Cov}}
x_i &= x_{t,i} + \varepsilon_{x,i}  \\
y_i &= y_{t,i} + \varepsilon_{y,i} 
\end{align}
where $\varepsilon_{x,i}$ and $\varepsilon_{y,i}$ are the random measurement errors on $x_i$ and $y_i$, respectively. $x_t$ and $y_t$ are the true values, $\varepsilon_{x,i} \sim N(0, \sigma_x^2)$ and $\varepsilon_{y,i} \sim N(0, \sigma_y^2)$. Finally, $\sigma_x$ and $\sigma_y$ are known.
When the data are measured without error, the correlation coefficient is:
$$ r = \frac{\cov(x_t,y_t)}{\sqrt{\Var(x_t)\Var(y_t)}} $$
Case 1: Measurement errors are the same for each data point
I was able to derive the formula for the correlation coefficient in case of measurement errors. In this case $\sigma_x^2$, $\sigma_y^2$ and $\sigma_{x,y}$ are the same for each data point. Using the properties of variance and covariance, the correlation coefficient is
\begin{align}
r &= \frac{\cov(x_t,y_t)}{\sqrt{\Var(x_t)\Var(y_t)}} \tag{1}  \\[10pt] 
  &= \frac{\cov(x,y) - \sigma_{x,y}}{\sqrt{(\Var(x) - \sigma_x^2) (\Var(y) - \sigma_y^2)} } 
\end{align}
where $\sigma_{x,y}$ is the covariance.
Case 2: Measurement errors are NOT the same for each data point


*

*assuming $\sigma_{x,y} = 0$, then the numerator in equation $(1)$ simplifies to:
$\cov(x_t,y_t) = \cov(x, y)$

*In this case $\sigma_{x,i}$ are not identical. Hence I cannot use the property of the variance which says that:
\begin{align}
\Var(x) &= \Var(x_t + \varepsilon_x)  \\
        &= \Var(x_t) + \Var(\varepsilon_x)  \\
        &= \Var(x_t) + \sigma_x^2 
\end{align}
So how can I derive a formula for the correlation coefficient in case the measurement errors are not identical?
 A: My understanding of your question is that you want to compute an estimate of 
$$V=\frac{1}{n}\sum_i(X_i+\epsilon_i-\bar{X}-\bar{\epsilon})^2$$
where $\bar{u}$ is the empirical mean of any vector $u=(u_1,\dots,u_n)$, $X_i$ are iid with variance $var(X)$, and $\epsilon_i/\sigma_i$ are iid with variance 1 and mean zero. 
My answer $V$ splits into 3 different temrs: $V=V_1+V_2+V_3$, 
with 
$$V_1=\frac{1}{n}\sum_i(X_i-\bar{X})^2$$
$$V_2=\frac{1}{n}\sum_i(\epsilon_i-\bar{\epsilon})^2$$
$$V_3=\frac{2}{n}\sum_i(X_i-\bar{X})(\epsilon_i-\bar{\epsilon})$$
From your assumption $V_3$ tends to be null and $V_1$ tends to equal $var(X)$ (with my notation, my $X$ is your "$x_t$"), hence what you want is the limit of $V_2$. You observe that in the case when $\sigma_1=\sigma_2=\dots=\sigma_n=\sigma$ then $V_2$ tends to $\sigma^2$. 
You really have to understand your question is about the limit (which wasn't really clear from you formulation because you mix theoretical and empirical quantities) for $V_2$. Under suitable conditions that might involve further moment convergence, I'm pretty sure $V_2$ converges to it expectation which means:
$$V_2 \rightarrow \lim_n\frac{1}{n}\sum_i\sigma_i^2$$  
