11 Probability Mass and Probability Density Functions
11.1 Probability Mass Function
A probability mass function (pmf) gives the probability that a discrete random variable is exactly equal to some value. \[p_X(x) = P\{X = x\}\] Example: A committee of 4 is to be selected from a group consisting of 5 men and 5 women. Let \(X\) be the random variable that represents the number of women in the committee. Create the probability mass distribution.
Solution: For \(x \in \{0,1,2,3,4\}\) we have \(p_X(x) = \frac{{5 \choose x}{5\choose 4-x}}{10 \choose 4}\). The probability mass distribution can be described by the table
| \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \(p_X(x)\) | \(\frac{5}{210}\) | \(\frac{50}{210}\) | \(\frac{100}{210}\) | \(\frac{50}{210}\) | \(\frac{5}{210}\) |
Example: Consider the experiment of rolling a fair die twice. Let \(X(i, j) = \max\{i, j\}\). Find the equation of \(p_X(x)\).
Solution: The pmf of \(X\) is \[p_X(x)=\begin{cases}\frac{2x-1}{36} & \text{if } x\in \{1,2,3,4,5,6\}\\0 & \text{otherwise}\end{cases} = \frac{2x-1}{36}\times I_{\{1,2,3,4,5,6\}}\] where the indicator function is defined as \(I_A = \begin{cases}1 & \text{if } x \in A \\ 0 & \text{otherwise}\end{cases}\). ### Properties If the range of a discrete random variable \(X\) with range \(\mathfrak R(X) = \{x : p(x) > 0\} = \{x_i\}_{i\ge1}\) where \(x_i < x_{i+1}\); then \(p_X(x_1) = F(x_1)\) and * \(p_X(x_i) = F(x_i) - F(x_{i-1})\) * \(\sum_{x : } p_X(x_i) = 1\)
11.2 Probability Density Function
We call the function \(f_X\) the probability density function (pdf) of an (absolutely) continuous random variable \(X\) if for any (measurable) \(A\): \[P\{X\in A\} = \int_A f_X\] with \(m(\cdot)\) being the Lebesgue measure.
11.2.1 Properties
- \(P\{a < X \le b\} = \int_a^b f_X(x)dx\)
- \(P\{X = a\} = 0\)
- \(F(a) = \int_{-\infty}^a f_X(x) dx\)
- \(\int_\mathbb R f_X dm = 1\)
- \(f_X(x) = F'(x)\)
Example: If the probability density of \(X\) is given by \[f_X(x) = 6x(1-x) \hskip 6em 0 < x < 1\] find the probability density of \(Y = X^3\).
Solution: We have \[\begin{align}F_Y(y) &= P\{Y \le y\} = P\{X^3 \le y\} = P\{X \le y^{1/3}\} \\ &=\int_0^{y^{1/3}} 6x(1-x) dx = 3y^{2/3} - 2y \end{align}\] Hence, \(f_Y(y) = F'_Y(y) = 2(y^{1/3} - 1)\)
Example: Suppose that the function \(f(x)\) defined below is the density function of some random variable \(X\). \[f(x) = \begin{cases}e^{-x} & x \ge 0 \\ 0 & x < 0\end{cases}\] Calculate \(P(-1\le X \le 1)\).
Solution: \[\begin{align}P(-1 \le x \le 1) &= \int_{-1}^1 f(x) dx \\&= \int_{0}^1 e^{-x}dx \\ &= 1-e^{-1} \end{align}\]