Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
    
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. If you don't do this, people are less likely to answer your later questions. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, How does accept rate work?. –  Tom Oldfield Dec 21 '12 at 5:50
add comment

1 Answer 1

up vote 3 down vote accepted

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 $1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.