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I'm stuck on an exercise, which goes as follows: Let $ X $ be an exponentially distributed random variable. Show that \begin{equation} P(X\leq s+t\mid X>s)=P(X\leq t) \end{equation} for all $ s,t\geq0 $.

I know that $ P(X\leq s+t|X>s)=\frac{P((X\leq s+t)\cap(X>s))}{P(X>s)} $. But I don't know how to further rearrange the numerator.

I have seen proofs for a similar equation, but never for this one. Can someone help me or give a hint, how I could show this?

Bernard
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silenceofthreeparts
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  • What happens if you subtract both sides from $1$? – Sangchul Lee Nov 26 '20 at 17:42
  • Well, $ 1-P(X\leq t)=P(X>t) $. I'm not sure about the other side, though. Can I just do a proof by contradiction and suppose, that $ P(X\leq s+t|X>s)=P(X>t) $? – silenceofthreeparts Nov 26 '20 at 17:53
  • $\mathbb{P}(\cdot \mid X>s)$ is another probability law, so this enjoys all the property that $\mathbb{P}(\cdot)$ has. For instance, you still have $$1-\mathbb{P}(X\leq s+t\mid X>s)=\mathbb{P}(X>s+t\mid X>s).$$ Also, I am not sure about what you mean by 'proof by contradiction'. If you want to employ that proof technique, you will need to derive a contradiction from the assumption $$\mathbb{P}(X\leq s+t\mid X>s)\neq\mathbb{P}(X\leq t),$$ and I see no obvious (and non-redundant) way to do so. – Sangchul Lee Nov 26 '20 at 18:00

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