Q.6.34 - A First Course in Probability

Question

Let X1, ... , Xn be independent uniform (0, 1) random variables. Let R = X(n) − X(1) denote the range and M = [X(n) + X(1)]/2 the midrange of X1, ..., Xn. Compute the joint density function of R and M. 

Step-by-Step Solution

Answer

$f(R,M)=\left\{\begin{array}{l}1 ;0\le (r,m)\le 1\\ 0 ; \mathrm{otherwise}\end{array}\right.$

Step 1: Introduction

Let $X_1,\dots ,X_n$  be independent uniform  random variables.
The range of the uniform distribution is defined as $R=X_{\left(n\right)}-X_{\left(1\right)}$

The Midrange of the uniform distribution is defined as 

$M=\frac{\left[X_{\left(n\right)}+X_{\left(1\right)}\right]}{2}$

Step 2: Explanation

Compute the joint density function of R and M
The probability density function of the Uniform distribution is given by, 

$$\begin{gathered} f_i\left(x_i\right)=\left\{\begin{array}{l}1 ;0\le x_i\le 1\\ 0 ;Otherwise\end{array}\right. \\ Since all the variables are independent to each other, \\ the joint distribution is given by, \\ f\left(x_1{x}_n\right)=f\left(x_1\right)f\left(x_n\right)=1 \\ Consider \\ R=X_{\left(n\right)}-X_{\left(1\right)} M=\frac{X_{\left(n\right)}+X_{\left(1\right)}}{2} \end{gathered}$$

Step 3: Explanation

Let use say x, y are the minimum and maximum values in a sample which is taken from the given distribution. That is  $X_{\left(n\right)}=y$ and $X_{\left(1\right)}=x$
By using the transformation, 

$$\begin{gathered} r=y-x\Rightarrow x=y-r \\ m=\frac{y+x}{2} \\ \Rightarrow y=2m-x \\ \Rightarrow y=2m-(y-r) \\ \Rightarrow 2y=2m+r \\ \Rightarrow y=m+\frac{r}{2} \\ And, \\ x \&=y-r \\ =m+\frac{r}{2}-r \\ =m-\frac{r}{2} \\ Thus, \\ x=m-\frac{r}{2} y=m+\frac{r}{2} \end{gathered}$$

Step 4: Explanation

Use Jacobian transformation to find the Jacobian coefficient. That is,

$$\begin{gathered} J =\left|\begin{array}{cc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial m}\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial m}\end{array}\right| \\ = \left|\begin{array}{cc}\frac{\partial }{\partial r}\left(m-\frac{r}{2}\right)& \frac{\partial }{\partial m}\left(m-\frac{r}{2}\right)\\ \frac{\partial }{\partial r}\left(m+\frac{r}{2}\right)& \frac{\partial }{\partial m}\left(m+\frac{r}{2}\right)\end{array}\right| \\ =\left|\begin{array}{cc}-\frac{1}{2}& 1\\ \frac{1}{2}& 1\end{array}\right| \\ =-\frac{1}{2}\left(1\right)-\frac{1}{2}\left(1\right) \\ =-1 \end{gathered}$$

Step 5: Explanation

$$\begin{gathered} Now, the joint distribution of R and M is given by, \\ f(R, M) =f (x_1,x_n)\times \frac{ 1 }{\left|J\right|} \\ =1\times \frac{1}{|-1|} \\ =1 \\ Therefore, the joint density function of R and M is, \\ f(R,M)=\left\{\begin{array}{l}1 ;0\le (r,m)\le 1\\ 0 ; \mathrm{otherwise}\end{array}\right. \end{gathered}$$