What is a two point support in this lemma?

Question

What is the terminology of two point support in this lemma?

Answer 1

The meaning is that the random variable $T$ takes on the two values $r$ and $\bar{r}$ with probabilities $p$ and $1-p$ respectively. Consequently, one of $T-r$ and $T-\bar{r}$ always has value $0$ and so $E[(T-r)(T-\bar{r})] = E[0] = 0$. This expression allows for an easy way of determining the stated relationship between $r$, $\bar{r}$ and the given values of $\mu$ and $\sigma^2$. It is also possible to grind out the same result from the definitions $$\begin{align} \mu &= pr + (1-p)\bar{r}\\ \sigma^2 &= p(r-\mu)^2 + (1-p)(\bar{r}-\mu)^2 \end{align}$$ eliminating $p$ in the process, but the calculations take longer and are more tedious.

Answer 2

It is probably the support of the distribution (http://en.wikipedia.org/wiki/Support_(measure_theory)).

That is why you see in the proof that the expected value of $(T-r)(T-\overline{r})$is zero.

Edit: If the distribution is supported at those two points then $T$ takes either of the values $r$ or $\overline{r}$ with probability one. In terms of integrals. If $\mu$ is the distribution $E((T-r)(T-\overline{r}))=\int (T-r)(T-\overline{r})d\mu(T)=(T-r)(T-\overline{r})|_{T=r}\mu(\{r\})+(T-r)(T-\overline{r})|_{T=\overline{r}}\mu(\{\overline{r}\})$