Memoryless property of the exponential distribution with a sum of inputs

Question

I have the following line in a proof;

Let T ~ Expon($\lambda$). Thus $\mathbb{P}(T>s+t | T > s) = \frac{\mathbb{P}(\{T>s+t\}\cap \{ T > s \})}{\mathbb{P}(T>s)} = \frac{\mathbb{P}(T>s+t)}{\mathbb{P}(T>s)} = ...$

I can't explain why $\mathbb{P}(\{T>s+t\}\cap \{ T > s \}) = \mathbb{P}(\{T>s+t\}$ for $t\leq 0$.

Can someone please explain?

---

My attempts at answering;

If $t \geq 0$, then $s + t \geq s$, and all the relevant information is contatined in $\{T>s+t\}$. To formalise this statement, I think $\{\omega \in \Omega: T(\omega)>s\} \cap \{\omega \in \Omega: T(\omega)>s + t\} = \{\omega \in \Omega: T(\omega)>s+t\}$.

However I can't explain why this is also true for $t<0$ as $s+t < s$, and by the argument above we get $\{T>s\}$ as the resulting set.

Answer 1

It does not hold when $t$ is negative.

If a number is greater than $s+\lvert t\rvert$ then it is greater than $s$.

Thus $\{T: T>s+\lvert t\rvert\}\subseteq \{T:T>s\}$.

Now, for any sets $A,B$ when $A\subseteq B$, then $A\cap B= A$.

So $\{T: T>s+\lvert t\rvert\}\cap \{T:T>s\} = \{T:T>s+\lvert t\rvert\}$, and because the probabilities for identical events are equal, thus: $$\mathsf P(T>s+\lvert t\rvert~\cap T>s)=\mathsf P(T>s+\lvert t\rvert)$$

And similarly:

$$\mathsf P(T>s-\lvert t\rvert~\cap T>s)=\mathsf P(T>s)$$

Answer 2

The memorylessness of the exponential distribution holds only for $s,t \ge 0$.

The proof must be assuming this, even if only implicitly.

Note that for $s > 0$ but $t \le 0$,

$$\Pr (T > s+ t, T > s) = \Pr (T > s)$$

Intuitively, the event that $\{ T> s \}$ is included in the event that $\{T > s+t\}$ when $t \le 0$.