User Tools


Mathematics 340 - Fall 2015

Homework

Exercises 3.12: 3, 41

Also:

  1. Suppose $X$ has a uniform distribution on the integers $-3, -2, -1, 0, 1, 2, 3$. Find the probability mass function for $Z = X^2$.
  2. Suppose $X$ and $Y$ are independent random variables, both with uniform distributions on the integers $-2, -1, 0, 1, 2$. Find the probability mass function for $T = X + Y$.
  3. Suppose $X$ and $Y$ are independent random variables, both with uniform distributions on the integers $-2, -1, 0, 1, 2$. Find the probability mass function for $S = X - Y$.
  4. Suppose $X$ has a uniform distribution on the integers $-2, -1, 0, 1, 2$. Find the probability mass function for $Z = X^2 - X$.
  5. Suppose $X$ and $Y$ are independent random variables, both with uniform distributions on the integers $1, 2, \ldots, 10$. Find the probability mass functions for $W = \max(X, Y)$ and $U = \min(X, Y)$.
  6. For $W$ and $U$ in the previous problem, what is $P(U < W)$? What is $P(U \le W)$? Are $U$ and $W$ independent?

For Problem Set due 5 October: Problems 2 and 5 from the above list.

Answers:

  1. $p_Z(z) = \begin{cases}\frac{1}{7},& \text{if } z = 0, \\ \frac{2}{7},& \text{if } z = 1, \\ \frac{2}{7},& \text{if } z = 4, \\ \frac{2}{7},& \text{if } z = 9, \\ 0,& \text{otherwise}.\end{cases}$
  2. $p_T(t) = \begin{cases}\frac{1}{25},& \text{if } t = -4, 4, \\ \frac{2}{25},& \text{if } t = -3, 3, \\ \frac{3}{25},& \text{if } t = -2, 2, \\ \frac{4}{25},& \text{if } t = -1, 1, \\ \frac{1}{5},& \text{if } t = 0, \\ 0,& \text{otherwise}.\end{cases}$
  3. $p_S(s) = \begin{cases}\frac{1}{25},& \text{if } s = -4, 4, \\ \frac{2}{25},& \text{if } s = -3, 3, \\ \frac{3}{25},& \text{if } s = -2, 2, \\ \frac{4}{25},& \text{if } s = -1, 1, \\ \frac{1}{5},& \text{if } s = 0, \\ 0,& \text{otherwise}.\end{cases}$
  4. $p_Z(z) = \begin{cases}\frac{2}{5},& \text{if } z = 0, \\ \frac{2}{5},& \text{if } z = 2, \\ \frac{1}{5},& \text{if } x = 6, \\ 0,& \text{otherwise}.\end{cases}$
  5. $p_W(w) = \begin{cases}\frac{1}{100},& \text{if } w = 1, \\ \frac{3}{100},& \text{if } w = 2, \\ \frac{1}{20},& \text{if } w = 3, \\ \frac{7}{100},& \text{if } w = 4, \\ \frac{9}{100},& \text{if } w = 5,\\ \frac{11}{100},& \text{if } w = 6, \\ \frac{13}{100},& \text{if } w = 7, \\ \frac{3}{20},& \text{if } w = 8, \\ \frac{17}{100},& \text{if } w = 9, \\ \frac{19}{100},& \text{if } w = 10, \\ 0,& \text{otherwise}.\end{cases}$ $p_U(u) = \begin{cases}\frac{19}{100},& \text{if } w = 1, \\ \frac{17}{100},& \text{if } w = 2, \\ \frac{3}{20},& \text{if } w = 3, \\ \frac{13}{100},& \text{if } w = 4, \\ \frac{11}{100},& \text{if } w = 5,\\ \frac{9}{100},& \text{if } w = 6, \\ \frac{7}{100},& \text{if } w = 7, \\ \frac{1}{20},& \text{if } w = 8, \\ \frac{3}{100},& \text{if } w = 9, \\ \frac{1}{100},& \text{if } w = 10, \\ 0,& \text{otherwise}.\end{cases}$
  6. $P(U < W) = \frac{9}{10}$, $P(U \le W) = 1$; $U$ and $W$ are not independent: for example, $P(U = 10, W = 1) = 0 \ne p_U(10)p_W(1)$