This Graduate project will discuss the topic of order statistics. Order statistics can provide efficient linear unbiased estimates of parameters, such as mean and standard deviation. Thus it can be used various aspects of life such as health care, finance and sports. This project will include a historical review of order statistics, some definitions as well as some theoretical properties to help further elaborate on this topic. This project will also included an application with respect to basketball data, and a simulation.
Rotations, play calling
Leader boards in points, rebounds and assists
Allocation of funds on a roster
Consider a set \(X_1,X_2,...,X_n\) of independent and identically distributed (continuous) random variables
Let \(X_{(r)}\) denote the \(r^{th}\) smallest of \(X_1,X_2,...,X_n\) . The random variables \(X_{(1)},X_{(2)},...,X_{(n)}\) are called the order statistics and \(X_{(r)}\) the \(r^{th}\) order statistic, \(r=1,2,...,n\)
For the order statistics we thus have that \(\\\)
\(\\\)
Let us consider the set of numbers
\(\\\)
In general, they are represented by \(x_i\):
\(\\\)
Then, the order statistics of this data would be:
It is of natural interest to find the joint probability distribution of these ordered random variables, and we will begin by finding the marginal probability distributions of the extremes, that is
\(\\\)
From the common distribution function \(F(x)\) it follows that
\(F_{(1)}(x)\)\(=Pr\{X_{(1)} \leq x\}\)
\(=Pr[min(X_1,X_2,...,X_n)\leq x]\)
\(=1-Pr[min(X_1,X_2,...,X_n)> x]]\)
\(=1-(Pr(X_1> x,X_2> x,...,X_n> x))\)
\(=1-(Pr(X_1> x)Pr(X_2> x)\cdots Pr(X_n> x))\)
\(=1-(1-F(x))^n\)
From the common distribution function \(F(x)\) it follows that
\(F_{(n)}(x)\)\(=Pr\{X_{(n)} \leq x\}\)
\(=Pr(max(X_1,X_2,...,X_n)\leq x)\)
\(=Pr(X_1\leq x,X_2\leq x,...,X_n\leq x)\)
\(=Pr(X_1\leq x)Pr(X_2\leq x)\cdots Pr(X_n\leq x)\)
\(=[Pr(X_1\leq x)]^n\)
\(=F^n(x)\)
We now generalize and look at the marginal probability distribution of \(X_{(r)}\) ;i.e the \(r^{th}\) order statistic, \(r=1,2,..n\)
Consider a set \(X_{(1)},...,X_{(n)}\) of independent and identically distributed (continuous) random variables with density function \(f(x)\) and distribution function \(F(x)\). For \(r=1,2,..,n\) the density of \(X_{(r)}\) is given by \(\\\)
Consider a set \(X_1,X_2,...,X_n\) of independent and identically distributed (continuous) random variables with density function \(f(x)\) and distribution function \(F(x)\). The joint density of the extremes is given by
\(\\\)
The density of \(R_n\) is then given by
for \(r\geq 0\)
Consider a set \(X_1,X_3,...,X_n\) of independent and identically distributed (continuous) random variables with density function f(x) and distribution function \(F(x)\). The joint density of the order statistic is given by
for \(-\infty<y_1\leq y_2<\cdots\leq y_n<\infty\)
Based on the previous assumptions, the joint distribution of \(X_{(r)}\) and \(X_{(s)}\) \((1\leq r < s \leq n)\) is given by
\(f_{rs}(x,y)=C_{rs}F^{r-1}(x)f(x)[F(y)-F(x)]^{s-r-1}f(y)[1-F(y)]^{n-s}\)
\(\\\) where\(\\\)
In a basketball league they have tracked free throw attempts of each player over their career. A player can shoot thousands free throws over this time. Each player is independent of one another and are modeled by a continuous uniform distribution on (0,10)
5 random players are selected randomly.
What is the probability that the minimum number of free throw attempts is between 2000-6000?
What is the expected value for the maximum amount of free throw attempts?
Distribution of \(X_1\)
The cdf is \(F(x)=\frac{x}{10}\) where \(0<x<10\)
\(F_{X_{(1)}}(x)\)\(=Pr(X_{(1)} \leq x)\)
\(=1-(1-F(x))^5\)
\(=1-(1-\frac{x}{10})^5\), \(0<x<10\)
\(P(X_{(1)}\) is between 2 and 6\()=P(2<X_{(1)}<6)\)
\(=P(X_{(1)} \leq 6)-P(X_{(1)} \leq 2)\)
\(=[1-(1-\frac{6}{10})^5]-[1-(1-\frac{2}{10})^5]\)
\(=(1-\frac{2}{10})^5-(1-\frac{6}{10})^5\)
\(=0.8^5-0.4^5\)
\(=0.32\)
A 32% probability the minimum free attempts will be between 2000 and 6000
\(F_{X_{(5)}}(x)\)=\(P(X_{(5)} \leq x )\)
\(=F^5(x)\)
\(=(\frac{x}{10})^5\)
\(f_{X_{(5)}}(x)\)\(=5(\frac{x}{10})^4\frac{1}{10}\)
\(=\frac{x^4}{2000^4}\)
\(E(X_5)=\)\(\int_0^{10}\frac{1}{2*10^4}x^4dx\)
\(=8.33\)
The expected max number of free throws attempts is 8330.