Conditional probability Question: $G\sim \operatorname{Geom}(p)$ Show that for any $k, l\ge 0, P(G > k + l|G > k) = P(G > l).$

Question

$G\sim \operatorname{Geom}(p)$. Show that for any $k, ℓ ≥ 0$, $$ P(G > k + ℓ|G > k) = P(G > ℓ). $$ So, for this I equated both formulas but replaced the powers but I don't know what happens to the $p$ in the formula. $$ ((1-p)^{k+l}/(1-p)^k) = (1-p)^l $$ that eventually leads me to $(1-p)^l = p(1-p)^l$

Can anyone help?

Answer 1

Note that $G>n$ if and only if you start with $n$ failures, so $P(G>n)= (1-p)^n.$ So your current calculation is fine, there is no missing $p.$