A Poisson process is memoryless

Question

Suppose we have the Poisson process $\{N(t), t\geq 0\}$. I want to prove the following identity: $$ \mathbb{P}(N(s+t) = k | N(s) = j, \{N(u), 0 \leq u \leq s\}) = \mathbb{P}(N(s+t) = k | N(s) = j) $$ which seems intuitively clear. The events after time $s$ only depent on what happens between time $s$ and $t+s$, it does not depent on the events beforehand. But trying to prove this, I came along the following reasoning which seems to disprove the statement: \begin{align*} &\mathbb{P}(N(s+t) = k | N(s) = j, \{N(u), 0 \leq u \leq s\})\\[1mm] &= \frac{\mathbb{P}(N(s+t) = k, N(s) = j, \{N(u), 0\leq u\leq s\})}{\mathbb{P}(N(s) = j, \{N(u), 0\leq u\leq s\})}\\[1mm] &= \frac{\mathbb{P}(N(s+t) - N(s) = k-j, N(s) = j, \{N(u), 0\leq u\leq s\})}{\mathbb{P}(N(s) = j, \{N(u), 0\leq u\leq s\})}\\[1mm] &= \frac{\mathbb{P}(N(s+t) - N(s) = k-j)\cdot \mathbb{P}(N(s) = j, \{N(u), 0\leq u\leq s\})}{\mathbb{P}(N(s) = j, \{N(u), 0\leq u\leq s\})}\\[1mm] &= \mathbb{P}(N(s+t) - N(s) = k-j)\\[1mm] &> \mathbb{P}(N(s+t) = k, N(s) = j). \end{align*} The third equality holds because $N(s+t) - N(s)$ is independent of $\{N(u), 0 \leq u \leq s\}$, by definition of a Poisson process. The last inequality holds because if $N(s+t) = k$ and $N(s) = j$, then $N(s+t) - N(s) = k-j$, but the reverse is not necessarily true (so $ \{N(s+t) =k \cap N(s) = j\}\subset \{N(s+t) - N(s) = k-j\} $). Is there a mistake in this argument, or is a Poisson process not memoryless?

Answer 1

Everything is good up to your inequality. You have correctly shown that $$ P(N(s+t)=k | N(s)=j, \{N(u), 0<u<s\}) = P(N(s+t)-N(s)=k-j). $$

Now start with only $P(N(s+t)=k | N(s)=j)$, without the $\{N(u), 0<u<s\}$ and show that it is also equal to $P(N(s+t)-N(s)=k-j)$.

Then you are done!