The cumulative distribution function of a random variable \(X\) is the function \(F: \mathbb{R} \rightarrow [0, 1]\) defined by
\[F(t) = P(X \leq t).\]
The cumulative distribution function is refered to as the cdf for short.
Example: Let \(X\) be the discrete random variable \(P(X = 0) = \frac{1}{4}\) and \(P(X = 1) = \frac{3}{4}.\) Then the cdf of \(X\) is \(F(t) = 0\) for \(t < 0,\) \(F(t) = \frac{1}{4}\) for \(0 \leq t < 1,\) and \(F(t) = 1\) for \(t \geq 1.\)
Example: Let \(X\) have cdf \(F(t) = 1-e^{-t}\) for \(t > 0\) and 0 otherwise. Find \(P(1 < X < 3).\)
Since \(F(t)\) is a continuous function and not a jump function, \(X\) is a continuous random variable.
The event in question is \(\{1 < X < 3\}.\) We can rewrite this event as \(\{1 < X\} \cap \{X < 3\} = \{X < 3\} - \{X \leq 1\}.\)
\begin{align}
P(1 < X < 3) & = P(\{X < 3\} - \{X \leq 1\}) \\
& = P(X < 3) - P(X \leq 1) \\
& = P(X < 3) - P(X < 1) \\
& = (1 - e^{-3}) - (1 - e^{-1}) \\
& = e^{-1} - e^{-3}
\end{align}
The second line follows because \(\{X \leq 1\} \subset \{X < 3\}.\) The third line follows because \(P(X \leq 1) = P(X < 1).\)
Properties
Every cdf \(F\) must satisfy the following properties:
\(F\) is increasing
\(F(t) \in [0, 1]\) for all numbers \(t\)
\(\lim_{t \rightarrow -\infty}F(t) = 0\)
\(\lim_{t \rightarrow \infty}F(t) = 1\)
All of these properties follow direction from the definition that \(F(t) = P(X < t).\)
cdf and pdf
Proposition: If \(X\) is a continuous random variable with cdf \(F\) and pdf \(f,\) then both of the following are true for every value \(t\):
\[\int_{-\infty}^t f(s)ds = F(t)\]
\[F'(t) = f(t)\]
▼ Proof:
The first equality follows from the definitions of pdf and cdf. In particular, \(F(t) = P(X < t),\) and \(P(X < t) = \int_{-\infty}^t f(s)ds.\)
The second result follows from the first and the Fundamental Theorem of Calculus.
\begin{align}
F'(t) & = \left(\int_{-\infty}^t f(s)ds\right)' \\
& = f(t)
\end{align}
Check your understanding:
1. Which of these is a cumulative distribution function?
Unanswered
Solution: The answer cannot be A since \(F(-1) = -1 < 0\) and a cdf is always greater than or equal to \(0.\) The answer cannot be B because \(F(2) = 2 > 1\) and a cdf must always be between 0 and 1, inclusive. The answer cannot be C since \(F(-0.5) = -0.5.\)
The function in D is a cdf.
2. The pdf of \(X\) is \(f(x) = \frac{1}{2}\) for \(1 < x < 3.\) What is the cdf of \(X?\)
Unanswered
Solution: The cdf of a random variable is the integral of the pdf. Since the pdf is non-zero for \(1 < x < 3,\) \(F(x) = 0\) for \(x \leq 1\) and \(F(x) = 1\) for \(x \geq 3.\) So,
\[F(x) = \int f(x) dx = \int \frac{1}{2} dx = \frac{1}{2}x+C\]
To find \(C,\) notice that \(F(1) = 0\) since the pdf is \(0\) for \(x < 1.\) Plugging in \(x = 1\) we get \(\frac{1}{2}+C = 0,\) or \(C = -\frac{1}{2}.\) Therefore, \(F(x) = \frac{1}{2}x - \frac{1}{2}\) for \(1 < x < 3.\)
3. The random variable \(X\) has cdf \(F(x) = 0\) for \(x < 0,\) \(F(x) = x^2\) for \(0 < x < 1,\) and \(F(x) = 1\) for \(x > 1.\) What is \(P(X > 0.5)?\)
4. The random variable \(X\) has cdf \(F(x) = 0\) for \(x < 0,\) \(F(x) = \frac{\sqrt{x}}{2}\) for \(0 < x < 4,\) and \(F(x) = 1\) for \(x > 4.\) What is \(P(1 < X < 3)?\)