Let f(x)=1/6, x=1,2,3,4,5,6, zero elsewhere, be the pmf of a distribution of the discrete type. Show that the pmf of the smallest observation of a random sample of size 5 from this distribution is g1(y1)=((7-y1)/6)^5 - ((6-y1)/6)^5, y1=1,2,...,6, zero elsewhere. Note that in this exercise the random sample is from a distribution of the discrete type. All formulas in the text were derived under the assumption that the random sample is from a distribution of the continuous type and are not applicable. Why?

QUESTION:

Let $f(x)=\frac{1}{6}, x=1,2,3,4,5,6,$ zero elsewhere, be the pmf of a distribution of the discrete type. Show that the pmf of the smallest observation of a random sample of size 5 from this distribution is 

$g_{1}(y_{1})=(\frac{7-y_{1}}{6})^{5}-(\frac{6-y_{1}}{6})^{5},y_{1}=1,2,...,6,$

zero elsewhere. Note that in this exercise the random sample is from a distribution of the discrete type. All formulas in the text were derived under the assumption that the random sample is from a distribution of the continuous type and are not applicable. Why?

SOLUTION:

Given

$\begin{aligned}f(x)&=\frac{1}{6}\\x&=1,2,3,4,5,6\\n&=Sample\;size=5\end{aligned}$

We need to determine the probability mass function of the smallest observation $Y_{1}=6$. When $Y_{1}=6$, then we require that all five data values are equal to 6.

$\begin{aligned}P(Y_{1}=6)&=P(Y_{1}=6,Y_{2}=6,Y_{3}=6,Y_{4}=6,Y_{5}=6)\\&=P(Y=6).P(Y=6).P(Y=6).P(Y=6).P(Y=6)\\&=f(6).f(6).f(6).f(6).f(6)\\&=(f(6))^{5}\\&=(\frac{1}{6})^{5}\\&=(\frac{7-6}{6})^{5}-(\frac{6-6}{6})^{5}\end{aligned} $

When $Y_{1}=5$, then we require that all five data values are at least 5 but not all five data values are equal to 6 $(Y_{1}=6)$.

$\begin{aligned}P(Y_{1}=5)&=P(Y_{i}\geq 5,i=1,2,3,4,5)-P(Y_{1}=6)\\&=(P(Y_{i}\geq 4))^{5}-P(Y_{1}=6)\\&=(\frac{2}{6})^{5}-(\frac{1}{6})^{5}\\&=(\frac{7-5}{6})^{5}-(\frac{6-5}{6})^{5}\end{aligned}$

When $Y_{1}=4$, then we require that all five data values are at least 4 but not $Y_{1}=5 \; nor\;Y_{1}=6$.

$\begin{aligned}P(Y_{1}=4)&=P(Y_{i}\geq 4,i=1,2,3,4,5)-P(Y_{1}=5)-P(Y_{1}=6)\\&=(P(Y_{i}\geq 4))^{5}-P(Y_{1}=5)-P(Y_{1}=6)\\&=(\frac{3}{6})^{5}-(\frac{2}{6})^{5}+(\frac{1}{6})^{5}-(\frac{1}{6})^{5}\\&=(\frac{3}{6})^{5}-(\frac{2}{6})^{5}\\&=(\frac{7-4}{6})^{5}-(\frac{6-4}{6})^{5}\end{aligned}$

When $Y_{1}=3$, then we require that all five data values are at least 3 but not $Y_{1}=4 \; nor\;Y_{1}=5 \; nor\;Y_{1}=6$.

$\begin{aligned}P(Y_{1}=3)&=P(Y_{i}\geq 3,i=1,2,3,4,5)-P(Y_{1}=4)-P(Y_{1}=5)-P(Y_{1}=6)\\&=(P(Y_{i}\geq 3))^{5}-P(Y_{1}=4)-P(Y_{1}=5)-P(Y_{1}=6)\\&=(\frac{4}{6})^{5}-(\frac{3}{6})^{5}\\&=(\frac{7-3}{6})^{5}-(\frac{6-3}{6})^{5}\end{aligned}$

When $Y_{1}=2$, then we require that all five data values are at least 2 but not $Y_{1}=3 \; nor\;Y_{1}=4 \; nor\;Y_{1}=5 \; nor\;Y_{1}=6$.

$\begin{aligned}P(Y_{1}=2)&=P(Y_{i}\geq 2,i=1,2,3,4,5)-P(Y_{1}=3)-P(Y_{1}=4)-P(Y_{1}=5)-P(Y_{1}=6)\\&=(P(Y_{i}\geq 2))^{5}-P(Y_{1}=3)-P(Y_{1}=4)-P(Y_{1}=5)-P(Y_{1}=6)\\&=(\frac{5}{6})^{5}-(\frac{4}{6})^{5}\\&=(\frac{7-2}{6})^{5}-(\frac{6-2}{6})^{5}\end{aligned}$

When $Y_{1}=1$, then we require that all five data values are at least 1 but not $Y_{1}=2 \; nor\;Y_{1}=3 \; nor\;Y_{1}=4 \; nor\;Y_{1}=5 \; nor\;Y_{1}=6$.

$\begin{aligned}P(Y_{1}=1)&=P(Y_{i}\geq 1,i=1,2,3,4,5)-P(Y_{1}=2)-P(Y_{1}=3)-P(Y_{1}=4)-P(Y_{1}=5)-P(Y_{1}=6)\\&=(P(Y_{i}\geq 1))^{5}-P(Y_{1}=2)-P(Y_{1}=3)-P(Y_{1}=4)-P(Y_{1}=5)-P(Y_{1}=6)\\&=(\frac{6}{6})^{5}-(\frac{5}{6})^{5}\\&=(\frac{7-1}{6})^{5}-(\frac{6-1}{6})^{5}\end{aligned}$

Thus we then note that $g_{1}(y_{1})=P(Y_{1}=y_{1})=(\frac{7-y_{1}}{6})^{5}-(\frac{6-y_{1}}{6})^{5}, \;for\;y_{1}=1,2,3,4,5,6$ (while zero otherwise as $y_{1}$ cannot take on any other values).