(New page: 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] = e...) |
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| + | From the memoryless property of''' Exponential Distribution''' function: | ||
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| + | Suppose '''E(1,λ) and E(1,μ)''' are independent, then; | ||
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| + | 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} | ||
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which shows that minimum of E1,λ and E1,μ is exponentially distributed. | which shows that minimum of E1,λ and E1,μ is exponentially distributed. | ||
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So, | So, | ||
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| + | '''E(1, λ1+ λ2+ λ3+……. λn) = min { E(1,λ1), E(1,λ2), E(1,λ3), ……….., E(1,λn) }''' | ||
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Here, if we put λ = 1, then; | Here, if we put λ = 1, then; | ||
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| + | '''E(1, 1+ 2+ 3+……. n) = min { E(1,1), E(1,2), E(1,3), ……….., E(1,n) }'''''' | ||
Revision as of 17:45, 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) }'
