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Suppose '''E(1,λ) and E(1,μ)''' are independent, then; | Suppose '''E(1,λ) and E(1,μ)''' are independent, then; | ||
| − | P[min{ E(1,λ) , E(1,μ) } > t] = P[E(1,λ) > t] . P[E(1,μ) } > t] | + | P [min{ E(1,λ) , E(1,μ) } > t] = P [E(1,λ) > t] . P [E(1,μ) } > t] |
= exp (-λt) . exp (-μt) | = exp (-λt) . exp (-μt) | ||
Revision as of 17:46, 6 October 2008
From the memoryless property of Exponential Distribution function:
Suppose E(1,λ) and E(1,μ) are independent, then;
P [min{ E(1,λ) , E(1,μ) } > t] = P [E(1,λ) > t] . P [E(1,μ) } > t]
= exp (-λt) . exp (-μt)
= exp {-(λ + μ)t}
which shows that minimum of E1,λ and E1,μ is exponentially distributed.
So,
E(1, λ1+ λ2+ λ3+……. λn) = min { E(1,λ1), E(1,λ2), E(1,λ3), ……….., E(1,λn) }
Here, if we put λ = 1, then;
E(1, 1+ 2+ 3+……. n) = min { E(1,1), E(1,2), E(1,3), ……….., E(1,n) }'
