Answers: PMFs, PDFs, and CDFs
These are the answers to Questions: PMFs, PDFs, and CDFs.
Please attempt the questions before reading these answers!
Q1
1.1.
The given PMF is valid because:
Non-negativity: All \(P(X = x) \geq 0\)
Honesty: The sum of all probabilities equals 1: \[ \sum_{x=1}^{4} p(x) = \sum_{x=1}^{4} P(X = x) = \dfrac{1}{10} + \dfrac{1}{5} + \dfrac{1}{2} + \dfrac{1}{5} = 1 \]
\(P(X = 4) = \dfrac{1}{5}\).
1.2.
The given PMF is valid because:
Non-negativity: All \(P(X = x) \geq 0\)
Honesty: The sum of all probabilities equals 1: \[ \sum_{x=1}^{4} p(x) = \sum_{x=1}^{4} P(X = x) = 0.25 + 0.35 + 0.05 + 0.2 + 0.1 = 1 \]
\(P(X = 3 \textsf{ or } X = 4) = 0.05 + 0.2 = 0.25\)
1.3.
The completed PMF table for the biased coin toss is:
| \(x\) | Heads | Tails |
|---|---|---|
| \(P(X=x)\) | 0.3 | 0.7 |
This is a valid PMF because:
Non-negativity: Both \(P(X = x) \geq 0\)
Honesty: The sum of both probabilities equal 1: \[ \sum_{x}p(x) = \sum_{x}P(X = x) = 0.3 + 0.7 = 1 \] #### 1.4. {-}
This is not a valid PMF since it fails the honesty condition:
Honesty: The sum of the given probabilities does not equal 1: \[ \sum_{x=1}^{7} p(x) = \sum_{x=1}^{7} P(X = x) = 0.1 + 0.05 + 0.05 + 0.3 + 0.25 + 0.75 + 0.35 = 1.85 \neq 1 \]
1.5.
\(\displaystyle P(\textsf{Blue}) = \dfrac{3}{10} = 0.3\)
The PMF for the given scenario is:
| \(x\) | Red | Blue | Green |
|---|---|---|---|
| \(P(X=x)\) | 0.5 | 0.3 | 0.2 |
This is a valid PMF because:
Non-negativity: All \(P(X = x) \geq 0\)
Honesty: The sum of all three probabilities equals to 1: \[ \sum_{x}p(x) = \sum_{x}P(X = x) = 0.5 + 0.3 + 0.2 = 1 \]
1.6.
For the given PMF to be valid, you must have \(p = \dfrac{1}{10}\).
For \(p = \dfrac{1}{10}\), then \(P(X = 3) = \dfrac{3}{10}\).
Q2
2.1.
This is a valid PDF because:
Non-negativity: \(\displaystyle f(x) \geq 0\) for all values of \(x\).
Honesty: \(\displaystyle \int_{-\infty}^{\infty} f(x) \, \textrm{d}x = \int_{0}^{2} \dfrac{1}{2} \, \textrm{d}x = \left[\, \dfrac{x}{2} \,\right]_{0}^{2} = 1\)
\(\displaystyle P(1 \leq x \leq 2) = \int_{1}^{2} \dfrac{1}{2} \, \textrm{d}x = \left[\, \dfrac{x}{2} \,\right]_{1}^{2} = \dfrac{1}{2}\)
2.2.
This is a valid PDF because:
Non-negativity: \(\displaystyle f(x) \geq 0\) for all values of \(x\)
Honesty: \(\displaystyle \int_{-\infty}^{\infty} f(x) \, \textrm{d}x = \int_{0}^{1} \dfrac{x}{2} \, \textrm{d}x = \left[\, x^2 \,\right]_{0}^{1} = 1\)
\(\displaystyle P(0.5 \leq X \leq 1) = \int_{0.5}^{1} 2x \, \textrm{d}x = \left[ x^2 \right]_{0.5}^{1} = 1^2 - (0.5)^2 = 1 - 0.25 = 0.75\)
\(\displaystyle P(0.25 \leq X \leq 0.75) = \int_{0.25}^{0.75} 2x \, \textrm{d}x = \left[ x^2 \right]_{0.25}^{0.75} = (0.75)^2 - (0.25)^2 = 0.5625 - 0.0625 = 0.5\)
2.3.
This is a valid PDF because:
Non-negativity: \(\displaystyle f(x) \geq 0\) for all values of \(x\)
Honesty: \(\displaystyle \int_{-\infty}^{\infty} f(x) \, \textrm{d}x = \int_{3}^{7} \dfrac{1}{4} \, \textrm{d}x = \left[ \dfrac{x}{4} \right]_{3}^{7} = 1\)
\(\displaystyle P(3 \leq X \leq 6) = \int_{3}^{6} \dfrac{1}{4} \, \textrm{d}x = \left[ \dfrac{x}{4} \right]_{3}^{6} = \dfrac{6}{4} - \dfrac{3}{4} = \dfrac{3}{4}\)
2.4.
This is not a valid PDF since it does not meet the honesty condition:
Honesty: \(\displaystyle \int_{-\infty}^{\infty} f(x) \, \textrm{d}x = \int_{1}^{4} \dfrac{1}{9} \, \textrm{d}x + \int_{5}^{7} \dfrac{1}{4} \, \textrm{d}x \neq 1\)
Calculating the individual integrals:
\(\displaystyle \int_{1}^{4} \dfrac{1}{9} \, \textrm{d}x = \dfrac{1}{9} \left[ x \right]_{1}^{4} = \dfrac{1}{3}\)
\(\displaystyle \int_{5}^{7} \dfrac{1}{4} \, \textrm{d}x = \dfrac{1}{4} \left[ x \right]_{5}^{7} = \dfrac{1}{2}\)
And adding them together:
\(\displaystyle \int_{-\infty}^{\infty} f(x) \, \textrm{d}x = \dfrac{1}{3} + \dfrac{1}{2} = \dfrac{5}{6} \neq 1\)
2.5.
For the given PDF to be valid, you must have \(k = 3\).
\(\displaystyle P(0.2 \leq X \leq 0.3) = \int_{0.2}^{0.3} 3 x^2 \, \textrm{d}x = 3 \left[ \dfrac{x^3}{3} \right]_{0.2}^{0.3} = \left[ x^3 \right]_{0.2}^{0.3} = 0.019\)
2.6.
This is a valid PDF because:
Non-negativity: \(f(x) \geq 0\) for all values of \(x\)
Honesty: \(\displaystyle \int_{-\infty}^{\infty} f(x) \, \textrm{d}x = \int_{0}^{0.5} 4x \, \textrm{d}x + \int_{0.5}^{0.75} (4 - 4x) \, \textrm{d}x + \int_{0.75}^{1} 0.5 \, \textrm{d}x\)
Calculating the individual integrals:
\(\displaystyle \int_{0}^{0.5} 4x \, \textrm{d}x = \left[ 2x^2 \right]_{0}^{0.5} = 0.5\)
\(\displaystyle \int_{0.5}^{0.75} (4 - 4x) \, \textrm{d}x = \left[ 4x - 2x^2 \right]_{0.5}^{0.75} = 0.375\)
\(\displaystyle \int_{0.75}^{1} 0.5 \, \textrm{d}x = \left[ 0.5x \right]_{0.75}^{1} = 0.125\)
and adding them together gives \(0.5 + 0.375 + 0.125 = 1\).
Q3
3.1.
\(F(3) = P(X \leq 3) = 0.1 + 0.3 + 0.5 = 0.9\)
\(P(X > 2) = 1 - P(X \leq 2) = 1 - (0.1 + 0.3 + 0.5) = 1 - 0.9 = 0.1\)
3.2.
The CDF for values \(0.5\), \(1\), and \(2\):
\(\displaystyle F(0.5) = \int_0^{0.5} \dfrac{1}{2} \, \textrm{d}x = \left[ \dfrac{x}{2} \right]_0^{0.5} = \dfrac{0.5}{2} = 0.25\)
\(\displaystyle F(1) = \int_0^{1} \dfrac{1}{2} \, \textrm{d}x = \left[ \dfrac{x}{2} \right]_0^{1} = \dfrac{1}{2} = 0.5\)
\(\displaystyle F(2) = \int_0^{2} \dfrac{1}{2} \, \textrm{d}x = \left[ \dfrac{x}{2} \right]_0^{2} = \dfrac{2}{2} = 1\)
\(\displaystyle F(3) = 1\) (since the CDF for any \(x \geq 2\) is \(1\).)
3.3.
The CDF at points \(4\), \(5\), and \(6\):
\(\displaystyle F(4) = \int_3^4 \dfrac{1}{4} \, \textrm{d}x = \left[ \dfrac{x}{4} \right]_3^4 = \dfrac{4}{4} - \dfrac{3}{4} = \dfrac{1}{4}\)
\(\displaystyle F(5) = \int_3^5 \dfrac{1}{4} \, \textrm{d}x = \left[ \dfrac{x}{4} \right]_3^5 = \dfrac{5}{4} - \dfrac{3}{4} = \dfrac{2}{4} = \dfrac{1}{2}\)
\(\displaystyle F(6) = \int_3^6 \dfrac{1}{4} \, \textrm{d}x = \left[ \dfrac{x}{4} \right]_3^6 = \dfrac{6}{4} - \dfrac{3}{4} = \dfrac{3}{4}\)
\(P(X > 5) = 1 - F(5) = 1 - \dfrac{1}{2} = \dfrac{1}{2}\).
3.4.
This is not a valid CDF because the CDF should be non-decreasing as \(x\) increases.
Version history and licensing
v1.0: initial version created 12/24 by Sophie Chowgule as part of a University of St Andrews VIP project.