13 A review of RVs thus far

We’ve been exploring and accumulating univariate probability models:


Model Type \(E(X)\) \(Var(X)\)
Bin(\(n,p\)) discrete \(np\) \(np(1-p)\)
Pois(\(\lambda\)) discrete \(\lambda\) \(\lambda\)
Hypergeo(\(w, b, n\)) discrete \(n \frac{w}{w+b}\) \(n \frac{w}{w+b}\frac{b}{w+b}\frac{w+b-n}{w+b-1}\)
Geo(\(p\)) discrete \(\frac{1}{p}\) \(\frac{1-p}{p^2}\)
Unif(\(a,b\)) continuous \(\frac{a+b}{2}\) \(\frac{(b-a)^2}{12}\)
Exp(\(\lambda\)) continuous \(\frac{1}{\lambda}\) \(\frac{1}{\lambda^2}\)
N(\(\mu,\sigma^2\)) continuous \(\mu\) \(\sigma^2\)

For each scenario below:

  • specify the model of \(X\) and any parameters upon which it depends
  • provide a formula for the PDF/PMF of \(X\) (don’t forget the support!)
  • calculate \(E(X)\) and \(Var(X)\) using the table above



  1. \(X\) = the number of times we roll a fair die until our first “6”


    Solution

    \[\begin{split} X & \sim Geo(1/6) \\ p_X(x) & = (1/6)(5/6)^{x-1} \;\; x \in \{1,2,3,...\} \\ E(X) & = \frac{1}{1/6} = 6 \\ Var(X) & = \frac{1-1/6}{(1/6)^2} = 30 \\ \end{split}\]



  1. \(X\) = the number of “6”s we get in 1 roll of a fair die


    Solution

    \[\begin{split} X & \sim Bin(1, 1/6) \\ p_X(x) & = (1/6)^x(5/6)^{1-x} \;\; x \in \{0,1\} \\ E(X) & = 1/6 \\ Var(X) & = 1/6(1-1/6) = 5/36 \\ \end{split}\]



  1. \(X\) = value produced by a random number generator that randomly selects numbers between 0 and 0.25


    Solution

    \[\begin{split} X & \sim Unif(0, 0.25) \\ f_X(x) & = 4 \;\; x \in [0,0.25] \\ E(X) & = \frac{0+0.25}{2} = \frac{1}{8} \\ Var(X) & = \frac{(0.25-0)^2}{12} = \frac{1}{192} \\ \end{split}\]



  1. \(X\) = the number of “6”s we get in 50 rolls of a fair die


    Solution

    \[\begin{split} X & \sim Bin(50, 1/6) \\ p_X(x) & = \left(\begin{array}{c} 50 \\ x \end{array} \right)(1/6)^x(5/6)^{50-x} \;\; x \in \{0,1,...,50\} \\ E(X) & = 50/6 \approx 8.333\\ Var(X) & = 50*1/6(1-1/6) \approx 6.944 \\ \end{split}\]



  1. \(X\) = the number of red dice that we get when pulling 10 dice out of a bag that contains 12 red dice and 8 yellow dice


    Solution

    \[\begin{split} X & \sim Hypergeo(12, 8, 10) \\ p_X(x) & = \frac{\left(\begin{array}{c} 12 \\ x \end{array} \right)\left(\begin{array}{c} 8 \\ 10-x \end{array} \right)}{\left(\begin{array}{c} 20 \\ 10 \end{array} \right)} \;\; \text{ for } x \in \{2,3,...,10\} \\ E(X) & = 10*\frac{12}{12+8} = 6\\ Var(X) & = 10*\frac{12}{12+8}*\frac{8}{12+8}*\frac{12+8-10}{12+8-1} \approx 1.263 \\ \end{split}\]