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| + | From the memoryless property of Exponential Distribution function: | ||
| + | Suppose E1,λ and E1,μ are independent, then; | ||
| + | P[min{ E1,λ , E1,μ } > t] = P[E1,λ > t] . P[E1,μ } > t] | ||
| + | = eˉλt . eˉμt | ||
| + | = eˉ(λ + μ)t | ||
| + | which shows that minimum of E1,λ and E1,μ is exponentially distributed. | ||
So, | So, | ||
| − | + | E1, λ1+ λ2+ λ3+……. λn = min { E1,λ1, E1,λ2, E1,λ3, ……….., E1,λn } | |
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Here, if we put λ = 1, then; | Here, if we put λ = 1, then; | ||
| − | + | E1, 1+ 2+ 3+……. n = min { E1,1, E1,2, E1,3, ……….., E1,n } | |
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Revision as of 17:48, 6 October 2008
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From the memoryless property of Exponential Distribution function: Suppose E1,λ and E1,μ are independent, then; P[min{ E1,λ , E1,μ } > t] = P[E1,λ > t] . P[E1,μ } > t] = eˉλt . eˉμt = eˉ(λ + μ)t which shows that minimum of E1,λ and E1,μ is exponentially distributed. So, E1, λ1+ λ2+ λ3+……. λn = min { E1,λ1, E1,λ2, E1,λ3, ……….., E1,λn } Here, if we put λ = 1, then; E1, 1+ 2+ 3+……. n = min { E1,1, E1,2, E1,3, ……….., E1,n }
