91Ó°ÊÓ

Let X have the geometric distribution with parameter p and let\(k \ge 0\).then for every integer\(t \ge 0\)

\(\begin{array}{c}{\rm P}\left( {X = k + t|X \ge k} \right) = \frac{{{\rm P}\left( {X = k + t|X \ge k} \right)}}{{{\rm P}\left( {X \ge k} \right)}}\\ = \frac{{{\rm P}\left( {X = k + t} \right)}}{{{\rm P}\left( {X \ge k} \right)}}\end{array}\)

For geometric distribution with parameter\(p\left( {0 < p < 1} \right)\)

\(f\left( {x|1,p} \right) = \left\{ \begin{array}{l}p{\left( {1 - p} \right)^x}\,\,\,for\,x = 0,1,2..\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,otherwise\end{array} \right.\)

\({\rm P}\left( {X = k + t} \right) = {\rm P}{\left( {1 - p} \right)^{k + t}}\)

\({\rm P}\left( {X \ge k} \right) = {\left( {1 - p} \right)^k}\)

Therefore,

\(\begin{array}{c}{\rm P}\left( {X = k + t|X \ge k} \right) = {\rm P}{\left( {1 - p} \right)^t}\\ = {\rm P}\left( {X = t} \right)\end{array}\)

\({\rm P}\left( {X = k + t|X \ge k} \right) = {\rm P}\left( {X = t} \right)\)

Hence proved.