Difference between revisions of "5.2b Divyanshu Kamboj ECE302Fall2008sanghavi" - Rhea

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) }

@@ Line 1: / Line 1: @@
-<nowiki>Insert non-formatted text here</nowiki>
+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.
-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 }
+'''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;
-E1, 1+ 2+ 3+……. n = min { E1,1, E1,2, E1,3, ……….., E1,n }
+'''E(1, 1+ 2+ 3+……. n) = min { E(1,1), E(1,2), E(1,3), ……….., E(1,n) }'''