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

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.
