Show that the variance of the longitudinal estimate is $ \dfrac{2 \sigma^2(1-\rho)} { n} $ rather than $\dfrac{\sigma^2(\rho)}{n}$ 
patients are randomızed to elther treatment, their pressures are measured at baseline, treatment is administered for two weeks and $\mathrm{BP}$ is then measured a second time. The treatment effect is estimated by the average change in $\mathrm{BP}$ for group A minus the average change for group $\mathrm{B}$.


Suppose that BP measurements from distinct individuals are independent with variance $\sigma^2$ and that the correlation between repeated observations on a single individual is $\rho$. Then the variance of the longitudinal estimate is $$ 2 \sigma^2(1-\rho) / n .$$

I cannot understand why the variance of the longitudinal estimate is not $\dfrac{\sigma^2(\rho)}{n}$ and I do not understand the origin of the scalar multiplication by two.
https://doi.org/10.1002/sim.4780111406
 A: If $Var(X)=\sigma_X^2$ and $Var(Y)=\sigma_Y^2$ and $Cor(X,Y)=\rho$,
and assuming without loss of generality $E[X]=E[Y]=0$, so $E[X^2]=E[Y^2]=\sigma^2$,
then $Cov(X,Y)=E[XY]=\rho \sigma_X^{\,} \sigma_Y^{\,}$,
and thus $Var(X-Y)= E[(X-Y)^2]=E[X^2]-2E[XY]+E[Y^2]= \sigma_X^2 - 2\rho \sigma_X^{\,} \sigma_Y^{\,} +  \sigma_Y^2$.
If $\sigma_X^2=\sigma_Y^2 = \sigma^2$ we can simplify this to $Var(X-Y)=2(1-\rho)\sigma^2$.
In this question, that is the variance of the change in blood pressure of an individual.  The division by $n$ is to see the variance in the average change across $n$ individuals.
A: The variance of the longitudinal estimator of the treatment effect should be
$$
\begin{align}
&\mathbb{V}\left(\frac{1}{n}\sum_{i=1}^n\left(X_{A,t_2,i}-X_{A,t_1,i}\right)-\frac{1}{n}\sum_{j=1}^n\left(X_{B,t_2,j}-X_{B,t_1,j}\right)\right) \\
&=\frac{1}{n^2}\left[\mathbb{V}\left(\sum_{i=1}^n\left(X_{A,t_2,i}-X_{A,t_1,i}\right)\right)+\mathbb{V}\left(\sum_{j=1}^n\left(X_{B,t_2,j}-X_{B,t_1,j}\right)\right)\right] \\
&= 
\frac{1}{n^2} \cdot 2 \cdot  \left[ \mathbb{V}\left(\sum_{i=1}^nX_{A,t_2,i}\right) + \mathbb{V}\left(\sum_{i=1}^nX_{A,t_1,i}\right) - 2\cdot \operatorname{Cov}\left(\sum_{i=1}^nX_{A,t_2,i}, \sum_{i=1}^nX_{A,t_1,i}\right) \right] \\
&= 
\frac{2}{n^2} \left(n \cdot \sigma^2 + n \cdot \sigma^2 - 2 \cdot n \cdot \rho \cdot \sigma^2 \right) = \frac{2}{n}  \left(2 \cdot \sigma^2 - 2 \cdot \rho \cdot  \sigma^2 \right) \\
&= 
\frac{4}{n} \sigma^2 \left(1-\rho\right).
\end{align}
$$
