Probability question about geometric distributionLet X have a geometric distribution. Show that P ( X &gt; k + j | X &gt; k ) = P ( X &gt; j ), where k and j are nonnegative integers.

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Probability question about geometric distributionLet X have a geometric distribution. Show that   P  (  X  &amp;gt;  k  +  j      |    X  &amp;gt;  k  )  =  P  (  X  &amp;gt;  j  ), where k and j are nonnegative integers.

Kyle Delacruz · Accepted Answer

Step 1I will use the formula of conditional probability:   P  (  A      |    B  )  =            P      (      A      ∩      B      )              P      (      B      )      .Thus   P  (  X  &amp;gt;  k  +  j      |    X  &amp;gt;  k  )  =  P  (  X  &amp;gt;  (  k  +  j  )     and     X  &amp;gt;  k  )      /    P  (  X  &amp;gt;  k  )  =  P  (  X  &amp;gt;  (  k  +  j  )  )      /    P  (  X  &amp;gt;  k  )Step 2As we are working with a geometric distribution,   P  (  X  &amp;gt;  (  k  +  j  )  )  =      q          k      +      j       and   P  (  X  &amp;gt;  k  )  =      q    k    .Then   P  (  X  &amp;gt;  (  k  +  j  )  )      /    P  (  X  &amp;gt;  k  )  =      q          k      +      j            /        q    k    =      q    j    =  P  (  X  &amp;gt;  j  )  =  P  (  X  &amp;gt;  j  ) when q is the probability of failure.

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