Examples

Example 1 Distribution of the Sample Median

The sample median $M$ is defined as

$$M = \left\{ \begin{array}{l} Y_{\frac{n+1}{2}} \quad \mb... ...ac{n}{2}+1}]/2 \ \ \mbox{for $n$\ even}. \end{array}\right.$$ (5.14)

For the case of $n$ odd, replace $r$ by $(n+1)/2$ in (5.5) on page  .

For the case of $n$ even, let $n=2m$, and $U=[Y_m+Y_{m+1}]/2$. Then

$$f_{Y_m,Y_{m+1}}(y_m,y_{m+1}) = \frac{(2m)!}{[(m-1)!]^2} [F(y_m)]^{m- 1}[1-F(y_{m+1})]^{m-1}f(y_m)f(y_{m+1}).$$

Define $u$ and $v$ as follows

$$\begin{eqnarray*} u & = & (y_m + y_{m+1})/2\\ v & = & y_{m+1} \ . \end{eqnarray*}$$

Then

$$\begin{eqnarray*} y_m & = & 2u-v\\ y_{m+1} & = & v\\ \mbox{and} \ \ \ \ \vert J\vert & = & 2. \end{eqnarray*}$$

Thus

$$\begin{eqnarray*} f_{U,V}(u,v) & = & \frac{(2m)!}{[(m-1)!]^2} [F(2u-v)]^{m-1}[1... ...~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -\infty < u < v < \infty \end{eqnarray*}$$

and integrating with respect to $v$ we obtain the pdf of $U$ (the sample median for a sample size $n=2m$),

$$f_U(u) = \frac{2(2m)!}{[(m-1)!]^2} \int^\infty_u [F(2u-v)]^{m-1}[1- F(v)]^{m-1}f(2u-v).f(v)dv.$$ (5.15)

Example 2 Distribution of the Sample Midrange

For an ordered sample $Y_1<Y_2<\ldots<Y_n$, this is defined as $\frac{1}{2}(Y_1+Y_n)$, and its pdf can be found for a particular distribution, beginning with (5.11) and using the technique of bivariate transformation.

Example 3 Distribution of the Sample Range

For an ordered sample $Y_1<Y_2<\ldots<Y_n$, the sample range is $R=Y_n-Y_1$. Assuming that the sample is from a continuous distribution with pdf $f(x), a<x<b$ and cdf $F(x)$, the joint pdf of $Y_1$ and $Y_n$ is given by (5.11) on page . Finding the distribution of $R$ becomes a problem in bivariate transformations. Define

$$r=y_n-y_1 \ \mbox{ and } \ v=y_1.$$

The inverse relationship, which is one-to-one, is

$$y_1=v \ \mbox{ and } \ y_n=r+v,$$

with $\vert J\vert=1$. So we have

$$f_{R,V}(r,v)=f_{Y_1,Y_n}(y_1,y_n)\vert J\vert=n(n-1)[F(r+v)-F(v)]^{n-2} \, f(r+v) \, f(v).$$

To find the range space, we deduce from the fact that $a < y_1 < y_n < b$, $v>a$, $r>0$ and $r+v<b$ or $v<b-r$. So the range space is $a<v<b-r; 0<r<b-a$, and

$$f_R(r)=\int_a^{b-r}n(n-1)\left[F(r+v)-F(v)\right]^{n-2}f(v)f(r+v)dv, \, 0<r<b-a.$$

As a special case for $f(x)=\frac{1}{b-a}, a<x<b$, find $f_R(r)$.

We have

$$f(v)=\frac{1}{b-a}, a<v<b \mbox{ and } f(r+v)=\frac{1}{b-a}, a<r+v<b, \mbox{ i.e. } a<v<b-r.$$

Now

$$F(r+v)-F(v)=\int_v^{r+v}\frac{1}{b-a}dx=\frac{r}{b-a}$$

So

$$\begin{eqnarray*} f_R(r)&=&n(n-1)\int_a^{b-r}\left(\frac{r}{b-a}\right)^{n-2}\cd... ...(b- a)^2}\\ &=&\frac{n(n-1)r^{n-2}(b-r-a)}{(b-a)^n}, \, 0<r<b-a. \end{eqnarray*}$$

Example 4 Estimating Coverage

Figure 5.5: Coverage

Define the $i$th coverage as $U_i=F(Y_i)-F(Y_{i-1})$. We have data, $X_1 \ldots X_n$ but do not assume a parametric form to the distribution and rely on order statistics to estimate coverage.

From theorem 5.5, $Z=F(X) \sim Unif(0,1)$ and $Z_{(1)} \ldots Z_{(n)}$ is an ordered sample from $Unif(0,1)$. By definition, $U_i=Z_{(i)}-Z_{(i-1)}$. From theorem 5.5, $Z_{(i)} \sim \mbox{Beta}(i,n-i+1)$ and $Z_{(i-1)} \sim \mbox{Beta}(i-1,n-i+2)$. Theorem 5.5 gives the joint distribution of $Z_{(i-1)}, Z_{(i)}$ leading to the joint distribution of $Z_{(i-1)},U_{(i)}$. Integrating wrt $Z_{(i-1)}$ gives the distribution of $U_i$, $U_i \sim \mbox{Beta}(1,n)$. Therefore $E(U_i) = 1/(n+1)$ and $\mbox{var}(U_i)=n/\{(n+1)^2(n+1)\}$.