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If Y Exponential(8), a > 0 and b > 0, show that P(Y > a+b[Y > a) = e-b/B.

If Y Exponential(8), a > 0 and b > 0, show that P(Y > a+b[Y > a) = e-b/B.

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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If Y ~ Exponential(3), a > 0 and 6 > 0, show that P(Y > a+b|Y > a) = e-b/B.
Transcribed Image Text:If Y ~ Exponential(3), a > 0 and 6 > 0, show that P(Y > a+b|Y > a) = e-b/B.
Expert Solution
Step 1

P[A|B] = P[A∩B]/P[B]

So, P[Y>a+b | Y>a] = P[Y>a+b ∩ Y>a] / P[Y>a]

for a>0, b>0, a+b>a and this implies that :

{Y>a+b} ⊂{Y>a} which implies that {Y>a+b}∩{Y>a} = {Y>a+b}

Hence,    P[Y>a+b | Y>a] = P[Y>a+b] / P[Y>a]

Step 2

Now, for an exponential distribution, its probability density function is given by :

f(x)= 1βe-x/β0                             , otherwise , x>0, β>0

So, P[X>c] = ∫c∞f(x) dx = ∫c∞ 1βe-x/β dx 

                 = 1βe-x/β-1βc∞ = e(-c/β)

 

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