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A Continuous random variable X has probability density function $f(x)=ae^{-ax}$

where I found $a=0.5ln2$

I Found that the mean of this distribution occurs at X=2.

Now, I was then asked what is: P(X<3) given P(X>1)

Can someone explain why this is equal to P(X<2). Does it have to do with the symmetry of the graph? Can someone proof this relationship please, or at least say why.

Thanks a bunch!

user2250537
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  • How about integrating the pdf? – Misguided Apr 27 '15 at 13:53
  • Ye I got the answer, but my method was way longer (I did it by integrating), instead the answer said: "Award full marks for P(X<3 given X>1) = P(X<2) = 0.5 or quoting properties of exponential distribution." Which I do not understand, where does this come from? – user2250537 Apr 27 '15 at 14:13
  • "Does it have to do with the symmetry of the graph?" The graph of the PDF has no symmetry whatsoever. – Did Apr 28 '15 at 17:17

1 Answers1

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Notice that $X$ has distribution $Exp(0.5ln(2))$. Hence:

$P(X<3|X>1) = 1 - P(X>=3|x>1) = 1 - P(X>3|X>1) = 1 - P(X > 2+1|X>1) = 1 - P(X>2)$

As per the memorylessness of the exponential distribution.

Misguided
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  • how did u go from 1−P(X>2+1|X>1)=1−P(X>2) ? – user2250537 Apr 27 '15 at 16:26
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    The memorylessness property states: P(X > s + t | X > t) = P(X > s). Proving it for the exponential distribution shouldn't be much hard. Further information @ https://en.wikipedia.org/wiki/Memorylessness – Misguided Apr 27 '15 at 16:42