Section 2.5

Cumulative Distribution Functions

Question 2.5.1

Consider rolling one fair six-sided die, so that \(\Omega = \{1,2,3,4,5,6\}\), and \(\mathbb{P}(\{\omega\}) = 1/6\) 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.

  1. \(F(x) = x\) for all \(x \in \mathbb{R}^1\)

  2. \(F(x) = \begin{cases} 0 & x < 0 \\ x & 0 \leq x < 1 \\ 1 & x \geq 1 \end{cases}\)

  3. \(F(x) = \begin{cases} 0 & x < -1 \\ x^2/9 & -1 \leq x < 3 \\ 1 & x \geq 3 \end{cases}\)

No, since \(F(x) > 1\) for \(x > 1\).

Yes.

No, since \(F(-1) > F(0)\) so \(F\) is not non-decreasing.

Question 2.5.3

Suppose \(F_X(x) = x^2\) for \(0 < x < 1\), and 0 otherwise. Compute each of the following.

  1. \(\mathbb{P}(X < 1/3)\)

  2. \(\mathbb{P}(1/4 < X < 1/2)\)

  3. \(\mathbb{P}(X < -1)\)

  4. \(\mathbb{P}(X = 3/7)\)

Note for all solutions a - d: Since \(X\) is continuous, \(\mathbb{P}(X < x) = \mathbb{P}(X \le x) = F_X(x)\)

\[ \mathbb{P}(X < 1/3) = F(1/3) = (1/3)^2 = 1/9 \]

\[ \begin{aligned} \mathbb{P}(1/4 < X < 1/2) &= \mathbb{P}(X < 1/2) - \mathbb{P}(X \leq 1/4) \\ &= F(1/2) - F(1/4) \\ &= (1/2)^2 - (1/4)^2 \\ &= 3/16 \end{aligned} \]

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 CDF \(F_X\).

The density of \(X\) is \(f(x) = 3e^{-3x}I_{[0, \infty)}(x)\). Since \(f_X(x)\) is only 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 \(x\).

  1. Compute the cumulative distribution function \(F_X(x) = \mathbb{P}(X \leq x)\) for all \(x \in \mathbb{R}\).

  2. What is \(\mathbb{P}(X \leq 2.5)\)?

  3. What is \(\mathbb{P}(1 < X \leq 3)\)?

  4. 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 \(\mathbb{P}(a < X \leq b) = F_X(b) - F_X(a)\): \[ \begin{aligned} \mathbb{P}(1 < X \leq 3) &= F_X(3) - F_X(1) \\ &= 0.8 - 0.1 \\ &= 0.7 \end{aligned} \]

\(\mathbb{P}(X=2) = \mathbb{P}(X \le 2) - \mathbb{P}(X <2) = \ F_X(2) - F_X(1) = 0.4 - 0.1 = 0.3\).

Question 2.5.6

In any given 15 minute block of time, the R4 shows up with the “SORRY, BUS FULL” sign 5 times. Assume the arrival times of full buses are independent. Let \(X\) be the time in minutes between full buses.

  1. Write out the CDF of \(X\). Be sure to include the support.

\(X\) is exponentially distributed, with rate parameter \(5/15 = 1/3\).

\(X \sim {\mathrm{Exp}}(1/3)\)

The CDF is \(F_X(x) = \left(1 - e^{-\frac{1}{3}x}\right)I_{[0, \infty)}(x)\)

  1. Using the CDF, find the median.

The median \(m\) satisfies \(F_X(m) = 1/2\). Since \(1/2 > 0\), we need \(m > 0\), so we use the branch \(1 - e^{-m/3}\): \[ \begin{aligned} 1 - e^{-m/3} &= \tfrac{1}{2} \\ e^{-m/3} &= \tfrac{1}{2} \\ -\tfrac{m}{3} &= \ln\left(\tfrac{1}{2}\right) = -\ln 2 \\ m &= 3\ln 2 \approx 2.079 \end{aligned} \] So the median time between full buses is \(3\ln 2 \approx 2.08\) minutes.

Question 2.5.7

Consider the scores of a midterm. Suppose that there are three types of students, modeled as follows:

  1. Poorly prepared students who did not study much (30%): \(X_1 \sim \mathcal{N}(\mu = 50, \sigma = 10)\).
  2. Well prepared students who studied a lot (60%): \(X_2 \sim \mathcal{N}(\mu = 80, \sigma = 8)\).
  3. Students who didn’t take the exam at all (10%): \(X_3 = 0\) with probability 1.

Find the CDF of the overall score distribution \(X\). Write your answer in terms of the standard normal CDF \(\Phi(\cdot)\).

Hint: If \(Y \sim \mathcal{N}(\mu, \sigma)\), then \(F_Y(y) = \Phi\left( \frac{y - \mu}{\sigma} \right)\).

We have that

\[\begin{aligned} F_X(x) &= 0.3 F_{X_1}(x) + 0.6 F_{X_2}(x) + 0.1 F_{X_3}(x) \\ &= 0.3 \Phi\left( \frac{x - 50}{10} \right) + 0.6 \Phi\left( \frac{x - 80}{8} \right) + 0.1 I_{[0,\infty)}(x). \end{aligned}\]

The indicator appears only for X3 making it a point mass. You can think of the CDF this way: What fraction of students who skipped the exam have a score less than 0? None. What fraction of students have a score less than or equal to zero? All. What fraction of students have a score less than or equal to 1? All. What fraction of students have a score less than or equal to 1000000? All. This is why the indicator is from 0 to infinity.

The normal components don’t need an indicator because the normal CDF already encodes all the boundary behavior automatically.