By Aliakbar Montazer Haghighi

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Extra resources for Difference and Differential Equations with Applications in Queueing Theory
Example text
1), for a case for an arbitrary (continuous measurable) function of X, say g(X), where X is a bounded random variable with continuous pdf fX(x), will be: E ( g ( X )) = ∫ Ω g ( X ) dF = ∫ ∞ −∞ g ( x ) f ( x ) dx, provided that the integral converges absolutely. 2 The kth moments of the continuous random variable X with pdf fX(x), denoted by E[Xk], k = 1, 2, . . 3) x f ( x ) dx < +∞. 4), the kth of X exists if and only if the kth absolute moment of X, E(Xk), is finite. Notes: (1) It can be shown that if the kth, k = 1, 2, .
X =1 x =1 ∞ ∞ x−2 x −1 λ λ +λ = e − λ λ 2 ( x − 1)! x = 2 ( x − 2 )! x =1 ∑ ∑ x ∑ x ∑ = e − λ [λ 2 e λ + λ e λ ] = λ 2 + λ , Var ( X ) = λ 2 + λ − λ 2 = λ . 4. CONTINUOUS RANDOM VARIABLES So far we have been discussing discrete random variables, discrete distribution functions, and some of their properties. We now discuss continuous cases. , an interval or a union of intervals). The set consisting of all subsets of real numbers R is extremely large and it will be impossible to assign probabilities to all of them.
2. Markov’s Inequality Let X be a nonnegative discrete random variable with a finite mean, E(X). Let a be a fixed positive number. Hence, P {X ≥ a} ≤ E (X ) . 3) Proof: By definition, E (X ) = ∑ xf ( x) x = ∑ xf ( x ) + 0≤ x < a ∞ ∑ xf ( x ). 4) is positive. Hence, E (X ) ≥ ∞ ∑ xf ( x). 5) we have: E (X ) ≥ ∞ ∞ x=a x=a ∑ af ( x) = a∑ f ( x). 6) functionS of a random Variable 51 Hence, E ( X ) ≥ aP {X ≥ a} . 3) follows. 3. Chebyshev’s Inequality Let X be a nonnegative random variable with finite mean μ and variance σ2.