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So i know that the exponential Distribution is memoryless so $P(X>x+t\mid X>x)=P(X>t)$. However, when we have $P(X<x+t\mid X>x)$ we cannot apply this right? So we have to use that definition of the conditional to get $\frac{P(x<X<x+t)}{P(X>x)}$. For example we want the first goal in a game to happen in time $x+t$ and we know that it did not happen in the first $x$ time. We would apply what i wrote out to calculate the probability of that, right?

Thanks

Michael Hardy
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Sorfosh
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    Observe that $P(X<x+t|X>x)=1-P(X>x+t|X>x),$ since $X$ is a continuous random variable, and using properties of conditional probability. – RideTheWavelet Oct 27 '17 at 16:38
  • @RideTheWavelet That makes a lot of sense, thanks, this makes calculations a lot easier! However, my method would yield the same result, right? – Sorfosh Oct 27 '17 at 16:39

2 Answers2

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$$ \Pr(X<x+t\mid X>x) = 1 - \Pr(X>x+t\mid X>x). $$

Michael Hardy
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The Exponential cumulative distribution function is so easy-going, that applying Bayes rule to find the conditional statement is trivial - and one will again get lack of memory,

$$P(X<x+t \mid X>x) = P(X<t)$$

Alecos Papadopoulos
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