problem 91

(a): If $F = G$ , Then, $X_{j}$ is equally likely to be in any of the $m + n$ positions in the ordered list. $E (R) = m \sum j = 1 E (R_{j}) = m \sum j = 1 \frac{(m + n) (m + n + 1)}{2} \frac{1}{m + n} = m \frac{m + n + 1}{2} .$
(b): $R_{j} = (\sum_{k = 1}^{n} I_{Y_{k}} + \sum_{k \neq j} I_{X_{k}} + 1)$ where $I_{Y_{k}}$ are the indicator random variables for $X_{j}$ being larger than $Y_{k}$ and $I_{X_{k}}$ are the indicator random variables for $X_{j}$ being larger than $X_{k}$ . Note that $E (I_{Y_{k}}) = p$ for all k since the Ys are iid, and $E (I_{X_{k}}) = 1 / 2$ - $X_{j}$ and $X_{k}$ are iid and never equal, so they are equally likely to be bigger or smaller than the other. Then $E (R_{j}) = np + (m - 1) / 2 + 1$ .Thus, $E (R) = m (np + (m - 1) ∕ 2 + 1)$ .