In this paper, we estimate probability P{X | Yugoslav Journal of Operations Research 24 (2014) Number 2, 283 - 291 DOI: ESTIMATION OF P{X 0 and β>0. We denote it with X: G( α , β). Its probability density function is given by: x f ( x; α , β ) = − 1 xα −1e β , x ≥ 0, α β Г(α ) where α is a shape parameter and β is a scale parameter. Let us suppose that random variable Y has exponential distribution with parameter λ, where λ>0. We denote it with Y: E(λ). Its probability density function is given by: g ( y; λ ) = 1 λ − e y λ , y ≥ 0, where λ is a scale parameter. It is known that E(λ) distribution is indeed a G(1, λ) distribution. . Maximum likelihood estimator of R Let X: G( α , β) and Y:E(λ), where X and Y are independent random variables. Therefore ∞ ∞ ∫ ∫ R = P{ X < Y } = dx f ( x; α , β ) g ( y; λ )dy 0 ∞ = 1 ∫ β α Г(α ) x 0 α −1 e − x ∞ x β dx 1 − y ∫ λ e λ dy x (1) 285 М. Jovanović, V. Rajić/ Estimation of P{X < Y} = ∞ 1 ∫ β α Г(α ) 0 xα −1e − x( 1 1 + ) β λ dx α ⎛ λ ⎞ =⎜ ⎟ . ⎝λ+β ⎠ Joint distribution for (X,Y) is given by h( x, y; α , β , λ ) = 1 λβ α Г(α ) xα −1e −( x y + ) β λ , ( X1 , Y1 ), ( X 2 , Y2 ),., ( X n , Yn ) be a random sample from that x ≥ 0 , y ≥ 0 . Let distribution. Therefore, the likelihood function and its ln are given by L(α , β , λ ) = n 1 (λβ α Г(α )) n (∏ xi )α −1 e −( 1 n 1 n x+ y) β ∑ i λ∑ i i =1 i =1 (2) i =1 n 1 i =1 β ln L(α , β , λ ) = −n(ln λ + α ln β + ln Г(α )) + (α − 1)∑ ln xi − n 1 n ∑ xi − ∑ yi , i =1 λ i =1 Taking partial derivatives of lnL with respect to α , β , and λ , we get n ∂ ln L Г′(α ) = − n(ln β + ) + ∑ ln xi , ∂α Г(α ) i =1 ∂ ln L −nα 1 = + 2 β ∂β β ∂ ln L −n 1 = + ∂λ λ λ2 n ∑ xi , i =1 n ∑ yi . i =1 Given the above identities to be equal to 0 and solving those equations, we obtain Г′(α ) 1 n = ∑ ln xi − ln β , Г(α ) n i =1 β= λ= 1 nα n ∑ xi , (3) (4) i =1 1 n ∑ yi n i =1 (5) Therefore, λˆ = Y . n From (3) and (4), we .
