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Essentials of Mathematical Probability and Statistics

John Travis

Section 5.3 Probability Functions

In the formulas below, we will presume that we have a random variable 5.2.1 \(X\) which maps the sample space S onto some range of real numbers \(R\text{.}\) From this set, we then can define a probability function \(f(x)\) which acts on the numerical values in \(R\) and returns another real number. We attempt to do so to obtain (for discrete values) P(sample space value s)\(= f(X(s))\text{.}\) That is, the probability of a given outcome s is equal to the composition which takes s to a numerical value x which is then plugged into f to get the same final values.
For example, consider a random variable which assigns a 1 when you roll a 1 on a six-sided die and 0 otherwise. Presuming each side is equally likely, \(f(1) = \frac{1}{6}\) and \(f(0) = \frac{5}{6}\text{.}\)

Definition 5.3.1. Probability "Mass" Function.

Given a discrete random variable 5.2.1 \(X\) on a space \(R\text{,}\) a probability mass function on \(X\) is given by a function \(f:R \rightarrow \mathbb{R}\) such that:
\begin{align*} & \forall x \in R , f(x) \gt 0\\ & \sum_{x \in R} f(x) = 1\\ & A \subset R \Rightarrow P(X \in A) = \sum_{x \in A}f(x) \end{align*}
For \(x \not\in R\text{,}\) you can use the convention \(f(x)=0\text{.}\)

Definition 5.3.2. Probability "Density" Function.

Given a continuous random variable \(X\) on a space \(R\text{,}\) a probability density function on \(X\) is given by a function \(f:R \rightarrow \mathbb{R}\) such that:
\begin{align*} & \forall x \in R , f(x) \gt 0\\ & \int_{R} f(x) dx = 1\\ & A \subset R \Rightarrow P(X \in A) = \int_{A} f(x) dx \end{align*}
For \(x \not\in R\text{,}\) you can use the convention \(f(x)=0\text{.}\)
For the purposes of this book, we will use the term "Probability Function" to refer to either of these options.
Consider \(f(x) = x/10\) over R = {1,2,3,4}. Then, f(x) is obviously positive for each of the values in R and certainly
\begin{equation*} \sum_{x \in R} f(x) = f(1) + f(2) + f(3) + f(4) = 1/10 + 2/10 + 3/10 + 4/10 = 1. \end{equation*}
Therefore, f(x) is a probability mass function over the space \(R\text{.}\)
Consider \(f(x) = x^2/c\) for some positive real number c and presume \(R\) = [-1,2]. Then f(x) is nonnegative (and only equals zero at one point). To make \(f(x)\) a probability density function 5.3.2, we must have
\begin{equation*} \int_{x \in R} f(x) = 1. \end{equation*}
In this instance you get
\begin{equation*} 1 = \int_{-1}^2 x^2/c = x^3/(3c) |_{-1}^2 = \frac{8}{3c} - \frac{-1}{3c} = \frac{3}{c} \end{equation*}
Therefore, \(f(x)\) is a probability density function over \(R\) provided \(c = 3\text{.}\)

Definition 5.3.5. Distribution Function.

Given a random variable \(X\) on a space \(R\text{,}\) a probability distribution function on \(X\) is given by a function
\begin{equation*} F:\mathbb{R} \rightarrow \mathbb{R} \ni \displaystyle F(x)=P(X \le x). \end{equation*}
Using \(f(x) = x/10\) over \(R\) = {1,2,3,4} again, note that \(F(x)\) will only change at these four domain values. We get
Table 5.3.7. Discrete Distribution Function Example
X F(x)
\(x \lt 1\) 0
\(1 \le x \lt 2\) 1/10
\(2 \le x \lt 3\) 3/10
\(3 \le x \lt 4\) 6/10
\(4 \le x \) 1
Consider \(f(x) = x^2/3\) over \(R\) = [-1,2]. Then, for \(-1 \le x \le 2\text{,}\)
\begin{equation*} F(x) = \int_{-1}^x u^2/3 du = x^3/9 + 1/9. \end{equation*}
Notice, \(F(-1) = 0\) since nothing has yet been accumulated over values smaller than -1 and \(F(2) = 1\) since by that time everything has been accumulated. In summary:
Table 5.3.9. Continuous Distribution Function Example
X F(x)
\(x \lt -1\) 0
\(-1 \le x \lt 2\) \(x^3/9 + 1/9\)
\(2 \le x\) 1
Let a = inf(\(R\)). Then, for \(x \lt a,\)
\begin{equation*} F(x) = P(X \le x) \le P(X \lt a) = 0 \end{equation*}
since none of the x-values in this range are in \(R\text{.}\)
Let b = sup(\(R\)). Then, for \(x \ge b,\)
\begin{equation*} F(x) = P(X \le x) = P(X \le b) + P( b \lt X \le x) = P(X \le b) = 1 \end{equation*}
since all of the x-values in this range are in R and therefore will either sum over or integrate over all of \(R\text{.}\)
Case 1: \(R\) discrete
\begin{align*} \forall x_1,x_2 \in \mathbb{Z} \ni x_1 \lt x_2\\ F(x_2) & = \sum_{x \le x_2} f(x) \\ & = \sum_{x \le x_1} f(x) + \sum_{x_1 \lt x \le x_2} f(x)\\ & \ge \sum_{x \le x_1} f(x) = F(x_1) \end{align*}
Case 2: \(R\) continuous
\begin{align*} \forall x_1,x_2 \in \mathbb{R} \ni x_1 \lt x_2\\ F(x_2) & = \int_{-\infty}^{x_2} f(x) dx \\ & = \int_{-\infty}^{x_1} f(x) dx + \int_{x_1}^{x_2} f(x) dx\\ & \ge \int_{-\infty}^{x_1} f(x) dx\\ & = F(x_1) \end{align*}
Assume \(x \in R\) for some discrete \(R\text{.}\) Then,
\begin{equation*} F(x) - F(x-1) = \sum_{u \le x} f(u) - \sum_{u \lt x} f(u) = f(x) \end{equation*}
For a and b as noted, consider
\begin{align*} F(b) - F(a) & = \int_{-\infty}^b f(x) dx - \int_{-\infty}^a f(x) dx\\ & = \int_a^b f(x) dx \\ & = P(a \lt x \le b) \end{align*}
We will assume that \(F(x)\) is a continuous function. With that assumption, note
\begin{equation*} P(a-\epsilon \lt x \le a) = \int_{a-\epsilon}^a f(x) dx = F(a) - F(a-\epsilon) \end{equation*}
Take the limit as \(\epsilon \rightarrow 0^+\) to get the result.
Assume \(X\) is continuous and \(f\) and \(F\) as above. Notice, by the definition of \(f\text{,}\) \(\lim_{x \rightarrow \pm \infty} f(x) = 0\) since otherwise the integral over the entire space could not be finite.
Now, let \(A(x)\) be any antiderivative of \(f(x)\text{.}\) Then, by the Fundamental Theorem of Calculus,
\begin{align*} F(x) & = \int_{-\infty}^x f(u) du\\ & = A(x) - \lim_{u \rightarrow -\infty} A(u) \end{align*}
Hence, \(F'(x) = A'(x) - \lim_{u \rightarrow -\infty} A'(u) = f(x)\) as desired.

Definition 5.3.17. Percentiles for Random Variables.

For \(0 \lt p \lt 1\text{,}\) the \(100p^{th}\) percentile is the largest random variable value c that satisfies
\begin{equation*} F(c) = p. \end{equation*}
For continuous random variables over an interval \(R = [a,b]\text{,}\) you will solve for c in the equation
\begin{equation*} \int_a^c f(x) dx. \end{equation*}
For discrete random variables, it is unlikely that a particular percentile will land exactly on one of the elements of \(R\) but you will want to take the smallest value in \(R\) so that \(F(c) \ge p\text{.}\)
The 50th percentile (as before) is also known as the median.
For our earlier example with \(f(x) = x^2/3\) on R = [-1,2], the 50th percentile (i.e. the median) is found by starting with p = 0.5 and then solving
\begin{equation*} F(c) = 0.5 \end{equation*}
or
\begin{equation*} c^3/9 + 1/9 = 1/2 \end{equation*}
or
\begin{equation*} c^3 + 1 = 9/2. \end{equation*}
After solving for c, you find
\begin{equation*} \text{median} = \sqrt[3]{7/2} \approx 1.518. \end{equation*}
TBA, using one of the table examples from above.
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