Question 2.5.1
Consider rolling one fair six-sided die, so that \(\Omega = \{1,2,3,4,5,6\}\), and \(\mathbb{P}(s) = 1/6\) for all \(\omega \in \Omega\). Let \(X\) be the number showing on the die, so that \(X(\omega) = \omega\) for \(\omega \in \Omega\). Let \(Y = X^2\). Compute the cumulative distribution function \(F_Y(y) = \mathbb{P}(Y \leq y)\), for all \(y \in \mathbb{R}^1\).
\[
F_Y(y) =
\begin{cases}
0 & y < 1 \\
1/6 & 1 \leq y < 4 \\
2/6 & 4 \leq y < 9 \\
3/6 & 9 \leq y < 16 \\
4/6 & 16 \leq y < 25 \\
5/6 & 25 \leq y < 36 \\
1 & 36 \leq y
\end{cases}
\]
Question 2.5.2
For each of the following functions \(F\), determine whether or not \(F\) is a valid cumulative distribution function.
\(F(x) = x\) for all \(x \in \mathbb{R}^1\)
\(F(x) = \begin{cases} 0 & x < 0 \\ x & 0 \leq x < 1 \\ 1 & x \geq 1 \end{cases}\)
\(F(x) = \begin{cases} 0 & x < -1 \\ x^2/9 & -1 \leq x < 3 \\ 1 & x \geq 3 \end{cases}\)
No, since \(F(x) > 1\) for \(x > 1\).
No, since \(F(-1) > F(0)\) so \(F\) is not non-decreasing.
Question 2.5.3
Suppose \(F_X(x) = x^2\) for \(0 < x < 1\). Compute each of the following.
\(\mathbb{P}(X < 1/3)\)
\(\mathbb{P}(1/4 < X < 1/2)\)
\(\mathbb{P}(X < -1)\)
\(\mathbb{P}(X = 3/7)\)
\[
\mathbb{P}(X < 1/3) = F(1/3) = (1/3)^2 = 1/9
\]
\[
\begin{aligned}
\mathbb{P}(1/4 < X < 1/2) &= \mathbb{P}(X < 1/2) - \mathbb{P}(X \leq 1/4) \\
&= F(1/2) - F(1/4) \\
&= (1/2)^2 - (1/4)^2 \\
&= 3/16
\end{aligned}
\]
Since \(0 \leq \mathbb{P}(X < -1) \leq \mathbb{P}(X \leq 0) = F(0) = 0\), we have \(\mathbb{P}(X < -1) = 0\).
\[
\mathbb{P}(X = 3/7) = \mathbb{P}(X \leq 3/7) - \mathbb{P}(X < 3/7) = F(3/7) - F(3/7) = 0
\]
Question 2.5.4
Let \(X \sim {\mathrm{Exp}}(3)\). Compute the function \(F_X\).
The density of \(X\) is \(f(x) = 3e^{-3x}\) (\(x \geq 0\)). Since \(X\) is defined on \([0, \infty)\), \(F(x) = \mathbb{P}(X \leq x) = 0\) for all \(x < 0\). For \(x \geq 0\): \[
F(x) = \int_{-\infty}^{x} f(t)\, dt = \int_0^x 3e^{-3t}\, dt = -e^{-3t}\Big|_0^x = -e^{-3x} + 1 = 1 - e^{-3x}
\]
Therefore the CDF is:
\[
F(x) = \begin{cases} 0 & x < 0\\
1 - e^{-3x} & x \ge 0
\end{cases}
\]
Question 2.5.5
A vending machine dispenses either 1, 2, 3, or 4 snacks per malfunction event, according to the probability mass function \[
p_X(1) = 0.1, \qquad p_X(2) = 0.3, \qquad p_X(3) = 0.4, \qquad p_X(4) = 0.2,
\] with \(p_X(x) = 0\) for all other values of \(x\).
Compute the cumulative distribution function \(F_X(x) = \mathbb{P}(X \leq x)\) for all \(x \in \mathbb{R}\).
What is \(\mathbb{P}(X \leq 2.5)\)?
What is \(\mathbb{P}(1 < X \leq 3)\)?
Confirm that \(\mathbb{P}(X = 2)\) = 0.3 using only the CDF.
Since \(X\) is discrete, we compute \(F_X(x)\) by summing \(p_X\) over all values less than or equal to \(x\): \[
F_X(x) =
\begin{cases}
0 & x < 1 \\
0.1 & 1 \leq x < 2 \\
0.4 & 2 \leq x < 3 \\
0.8 & 3 \leq x < 4 \\
1 & x \geq 4
\end{cases}
\]
Note the running totals: \(F_X(1) = 0.1\); \(F_X(2) = 0.1+0.3 = 0.4\); \(F_X(3) = 0.4+0.4 = 0.8\); \(F_X(4) = 0.8+0.2 = 1\).
Since \(F_X\) is constant on \([2, 3)\), and \(2.5\) falls in that interval: \[
\mathbb{P}(X \leq 2.5) = F_X(2.5) = 0.4
\]
Using the identity \(\mathbb{P}(a < X \leq b) = F_X(b) - F_X(a)\): \[
\begin{aligned}
\mathbb{P}(1 < X \leq 3) &= F_X(3) - F_X(1) \\
&= 0.8 - 0.1 \\
&= 0.7
\end{aligned}
\]
\(\mathbb{P}(X=2) = \mathbb{P}(X \le 2) - \mathbb{P}(X <2) \ F(2) - F(1) = 0.4 - 0.1 = 0.3\).