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