Variance of the mean of a random variable with a normal distribution I am trying to understand the methods of a medical-statistics paper I am reading. This is a direct excerpt:

The model considered is a 2-group clinical trial (treatment v
  control), with true (unknown) treatment effect denoted by $\mu$. The
  outcome is continuous and normally distributed with variance assumed
  known and equal to 0.5 (this is for convenience, so that the variance
  of an estimate of treatment effect based on a pair of observations
  equals 1).
The trial will have 100 observations in each group, considered as 100
  pairs of differences (random variables $X_i$  with realisations $x_i$  :
  $i=1\dots100$). Each $X_i$  is distributed $\mathcal{N}(\mu,1)$, and their mean $\bar{X}$ -
  distributed $\mathcal{N}(\mu,0.01)$ with realisation $\bar{x}$ - will be used to estimate the
  treatment effect.

I have a few questions:


*

*Is "the outcome" (second sentence) referring to the outcome of each arm of the clinical trial or to a single treatment effect (i.e. difference between the two arms)?

*Why does the variance of two outcomes with a variance of 0.5 equal 1 when estimating the treatment effect based on a pair of observations?

*Each $X_i$ is distributed $\mathcal{N}(\mu,1)$, so why is $\bar{X} \sim \mathcal{N}(\mu,0.01)$?
Also, if you can think of a more appropriate title for this question so it might be indexed and better searched in the future, please suggest it.
 A: 
Is "the outcome" (second sentence) referring to the outcome of each arm of the clinical trial or to a single treatment effect (i.e. difference between the two arms)?

Hm, hard to tell. My guess, based on the second paragraph, it's probably the difference for a pair, between the treatment vs non-treatment

Why does the variance of two outcomes with a variance of 0.5 equal 1 when estimating the treatment effect based on a pair of observations?

It assumes that the observations are independent. I.e., the variance of a sum equal the sum of the variances.

Each $X_i$ is distributed $\mathcal{N}(\mu,1)$, so why is $\bar{X} \sim \mathcal{N}(\mu,0.01)$?

Remember variance is defined as
\begin{aligned}\operatorname {Var} (X)=\operatorname {E} \left[(X-\operatorname {E} [X])^{2}\right]\\ =\operatorname {E} \left[X^{2}\right]-\operatorname {E} [X]^{2}\end{aligned}
Here, you are comparing the variance of the random variable vs the variance of the sample mean. The variance of your sample is (where $\sigma^2$ is the variance:
$$n\left({\frac {\sigma ^{2}}{n^{2}}}\right)={\frac {\sigma ^{2}}{n}}$$
(precisely, this is for the population variance and mean, but the article probably simplifies the concept of sample mean)
