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  • Probability Distribution Function

Probability Mass Function

• The distribution function of a Discrete Random Variable

• Can be expressed as a table, a mathematical expression, or a graph.

• Properties: Theorem 26

  • For all $x$ in the support of $X$.
    • $0\le p(x)\le 1$
    • $\sum _{\mathrm{\forall }i}^{}p(x)=1$

• Example: Toss a fair coin

  • Let $X$ represent $\mathrm{\#}$ of heads observed from tossing a fair coin twice. (assuming tosses are independent^[Statistical Independence] )
  • $S=\{\underbrace{\{T,T\}},\{H,T\},\{T,H\},\{H,H\}\}$
    • $\frac{1}{4}$ chance of happening
    • $\frac{1}{4}=0H$
    • $\frac{1}{2}=1H$
    • $\frac{1}{4}=2H$
  • Table:
    • $\begin{array}{cccc}X=x& 0& 1& 2\\ P(X=x)& \frac{1}{4}& \frac{1}{2}& \frac{1}{4}\end{array}$
      • This satisfies the requirements above ^[Probability Mass Function#^fe5edf]
  • Expression:
    • $P(X=x)=\left\{\begin{array}{cc}\frac{1}{4}& x=0\\ \frac{1}{2}& x=1\\ \frac{1}{4}& x=2\end{array}\right\}$
    • The output is the Support of a Random Variable.
  • Graph:
    • These are point masses

• Example: Store

  • Consider a store which sells electronics $0,1,2,3$ or $4$ TVs daily with probabilities $k,3k,3k,2.5k,0.5k$ respectively
  • Solve for $k$, write and graph the mass function
  • Table:
    • $\begin{array}{cccccc}X=x& 0& 1& 2& 3& 4\\ P(X=x)& k& 3k& 3k& 2.5k& 0.5k\end{array}$
    • Sum up all the $P(X=x)$ to be $1$
    • $k+3k+3k+2.5k+0.5k=1$
    • $10k=1$
    • $k=\frac{1}{10}=0.1$
    • $\begin{array}{cccccc}X=x& 0& 1& 2& 3& 4\\ P(X=x)& 0.1& 0.3& 0.3& 0.25& 0.05\end{array}$
  • Expression:
  • Graph:
  • What's the probability of selling at least $1$ TV?
    • $P(X\ge 1)=P(X=1)+P(X=2)+P(X=3)+P(X=4)$
      • $P(X{\ge }_1)=0.3+0.3+0.25+0.05$
      • $P(X\ge 1)=0.9$
      • $90\mathrm{\%}$
    • Or use the complement of selling $0$ TVs
      • $P(X\ge 1)=1-P(x<1)$
      • $P(X\ge 1)=1-0.1$
      • $P(X\ge 1)=0.9$
  • What is the probability of selling $2.37$ TVs
    • Doesn't support this random variable.
    • This is a Discrete Random Variable. So it accepts specific values.
  • What is the probability of selling between $1.53$ and $3.62$ TVs
    • Grab the numbers in the support.
    • $P(1.53\le X\le 3.62)=P(X=2)+P(X=3)$
    • $P(1.53\le X\le 3.62)=0.3+0.25$
    • $P(1.53\le X\le 3.62)=0.55$
  • What is the probability of selling at least $3$ TVs given the store sold at least $2$ TVs
    • $P(X\ge 3|X\ge 2)=\frac{P(X\ge 3\cap X\ge 2)}{P(X\ge 2)}$
    • $P(X\ge 3|X\ge 2)=\frac{P(X\ge 3\cap X\ge 2)}{0.3+0.25+0.05}$
      • $P(X\ge 3\cap X\ge 2)=P(X\ge 3)$
    • $P(X\ge 3|X\ge 2)=\frac{0.25+0.05}{0.3+0.25+0.05}$
    • $P(X\ge 3|X\ge 2)=\frac{0.3}{0.6}$
    • $P(X\ge 3|X\ge 2)=\frac{0.3}{0.6}=0.5=\frac{1}{2}$