| (5 intermediate revisions by the same user not shown) | |||
| Line 3: | Line 3: | ||
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) | ||
| Line 9: | Line 9: | ||
= exp {-(λ + μ)t} | = exp {-(λ + μ)t} | ||
| − | which shows that minimum of | + | which shows that minimum of E(1,λ) and E(1,μ) is exponentially distributed. |
So, | So, | ||
| Line 17: | Line 17: | ||
Here, if we put λ = 1, then; | 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) } | + | '''E(1, 1+ 2+ 3+……. n) = min { E(1,1), E(1,2), E(1,3), ……….., E(1,n) }''' |
Latest revision as of 17:50, 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 E(1,λ) and E(1,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) }
