I think there is a little mistake in this exercise about the memoryless of Geometric Distribution An exercise of Jacod and Protter: Let $X$ be Geometric. Show that for $i, j > 0$,
$$P(X > i + j | X > i) = P(X > j)$$ I did it and I got a different asnwer:
$$P(X > i + j | X > i) = P(X \geq j)$$ 
(The exercise is quite simple using Bayes Theorem) 
¿Am I doing it wrong or what?
Idea: 
\begin{align*}
P(X > i + j | X > i) &= P(X \geq i + j + 1 | X \geq i + 1) \\ \\
&=  \frac{P(X \geq i + 1 | X \geq i + j + 1) \: P(X \geq i + j + 1) }{P(X \geq i + 1) } \\ \\
&= \frac{1 \cdot (1-p)^{i+j+1}}{(1-p)^{i+1}} = (1-p)^{j} = P(X \geq j)
\end{align*}
 A: For a geometric random variable with parameter $p$ that counts the number of trials until the first occurrence of a success (and hence takes on values $1,2,3, \ldots$),
$$P(X>n) = P(\text{first $n$ trials ended in failure}) = (1-p)^n.$$
Hence,
\begin{align}P(X>i+j  \mid X > i) &= \frac{P\big((X>i+j)\cap(X>i)\big)}{P(X>i)}\\&= \frac{P(X>i+j)}{P(X>i)} \\&= (1-p)^j \\&= P(X>j).\end{align}
On the other hand, some people prefer to count the number of failures that have occurred before the first success and so they define a geometric random variable (call it $Y$ to distinguish it from the $X$ above) as one that takes on values $0, 1,2, \ldots$. For these folks, $$P(Y > n)= P(\text{first $n+1$ trials ended in failure}) = (1-p)^{n+1}$$ as the OP says in a comment.  Hence, (and without invoking Bayes' theorem at all), we have that
\begin{align}
P(Y>i+j  \mid Y > i) &= \frac{P\big((Y>i+j)\cap(Y>i)\big)}{P(Y>i)}
\\&= \frac{P(Y>i+j)}{P(Y>i)}
\\&= \frac{(1-p)^{i+j+1}}{(1-p)^{i+1}}\\
&= (1-p)^j\\
&= P(Y > j-1) = P(Y \geq j)
\end{align}
as the OP correctly computed in his work in the question above.  So it certainly seems as if Jacod and Protter were using one definition in the text and the other in the exercise (or that everywhere in the exercise, they mistyped $>$ when they meant to write $\geq$)
