a) ∞ 𝐸(𝑋) = ∫ 𝑥𝑓(𝑥)𝑑𝑥 0 ∞ = ∫ 𝑥𝜆𝑒 −𝜆𝑥 𝑑𝑥 0 Let 𝜆𝑥 = 𝑧 then 𝜆𝑑𝑥 = 𝑑𝑧 so the above reduces to, 1 ∞ −𝑧 1 ∞ 2−1 −𝑧 𝐸(𝑋) = ∫ 𝑧𝑒 𝑑𝑧 = ∫ 𝑧 𝑒 𝑑𝑧 𝜆 0 𝜆 0 And this is integral Gamma(2) or 1!. So, 1 1 𝐸(𝑋) = 1! = 𝜆 𝜆 b) We know q(p) is the pth quantile function if, F(q(p)) = p Now for exponential distribution, F(x) = 1- 𝑒 −𝜆𝑥 So for pth quantile, 𝐹(𝑞(𝑝)) = 1 − 𝑒 −𝜆𝑞(𝑝) = 𝑝 1 => 𝑞(𝑝) = − ln(1 − 𝑝) 𝜆 1 1 1 1 ∫0 𝑞(𝑝)𝑑𝑝 = − 𝜆 ∫0 ln(1 − 𝑝) 𝑑𝑝 = − 𝜆 ∗ (−1) = 1 𝜆 = 𝐸(𝑋) Note: Online widget to calculate definite integral http://www.wolframalpha.com/widgets/view.jsp?id=8ab70731b1553f17c11a3bbc87e0b605 c) As for positive continuous function the inverse function 𝐹 −1 (𝑝) always exists. ∞ 1 𝐸(𝑋) = ∫0 𝑥𝑓(𝑥)𝑑𝑥 = ∫0 𝑞(𝑝)𝑓(𝑞(𝑝))𝑑𝑝 ; we can always write this by replacing x with q(p) and as q(p) is a function of p so dq(p) can be converted into dp which changes the limit to [0,1] Now for positive continuous function, 𝑓(𝑥) = 𝑑 𝐹(𝑥) 𝑑𝑥 So, 𝑑 𝑓(𝑞(𝑝)) = 𝑑𝑝 𝐹(𝑞(𝑝)) ; as q(p) is a function of p so the differentiate is taken over p. 𝑑 = 𝑑𝑝 𝐹(𝑞(𝑝)) = 𝑑 𝑑𝑝 𝑝 = 1 ; As 𝐹(𝑞(𝑝)) = 𝑝 So, ∞ 1 𝐸(𝑋) = ∫ 𝑥𝑓(𝑥)𝑑𝑥 = ∫ 𝑞(𝑝)𝑑𝑝 0 0