# How to prove the memoryless property $\Pr(X>s+t|X>s)=\Pr(X>t)$

How to prove the memoryless property?

Memoryless Property: Suppose $X \sim~ \mathrm{Exp}(\lambda)$. Then $\Pr(X>s+t|X>s)=\Pr(X>t)$ for $s,t\in\mathbb{R}$?

Attempts: $\Pr(X>s+t|X>s)=\frac{\Pr(X>s+t,X>s)}{P(X>s)}$ and I claim that $\Pr(X>s+t,X>s)=\Pr(X>s+t)$ but don't know whether that claim is correct or not.

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Write formally what events $\{X > s+t, X >s\}$ and $\{X>s+t\}$ are. Is it possible that one of them happens and the other doesn't? –  Yury Dec 21 '12 at 5:46
@Yury Ya, intuitively i have the picture but i am not sure whether i can immediately write that. –  Mathematics Dec 21 '12 at 5:48
The equality $\Pr(X>s+t,X>s)=\Pr(X>s+t)$ holds since conditions $\{X>s+t,X>s\}$ and $\{X>s+t\}$ define the same event when $t \geq 0$.
• Indeed, if $X> s+t$ and $X> s$ then $X > s+t$ thus $\{\omega:X(\omega)>s+t,X(\omega)>s\}\subseteq\{\omega:X(\omega)>s+t\}$.
• If $X > s+t$ then $X>s+t \geq s$. Thus $\{\omega:X(\omega)>s+t,X(\omega)>s\}\supseteq\{\omega:X(\omega)>s+t\}$.
The statement trivially holds when $t < 0$ since the probabilities on the left and on the right are equal to $1$.