Let X be a random variable which is de
fined over the interval [0, 2]. Find E[X], E[X^2], and E[ 1/
X ] for the following scenarios:
(a) The density function for X is given by: f(x) = ax for x is the set [0, 2] and f(x) = 0 otherwise. Here A
is a constant to be determined by you.
(b) The density function for X is given by: f(x) = bx^2 for x is the set [0, 2] and f(x) = 0 otherwise. Here B is a constant to be determined by you.
(c) Find V ar(X) for both part a) & b).
I dont know where to start. Help please
X ] for the following scenarios:
(a) The density function for X is given by: f(x) = ax for x is the set [0, 2] and f(x) = 0 otherwise. Here A
is a constant to be determined by you.
(b) The density function for X is given by: f(x) = bx^2 for x is the set [0, 2] and f(x) = 0 otherwise. Here B is a constant to be determined by you.
(c) Find V ar(X) for both part a) & b).
I dont know where to start. Help please
-
a) Integral(0,2) ax dx = 1 gives
a[x^2/2](0,2) = 1
2a = 1
a = 1/2
E(X) = Integral(0,2) (1/2)x^2 dx = (1/2)[x^3/3](0,2) = 8/6
E(X^2) = Integral(0,2) (1/2)x^3 dx = (1/2)[x^4/4](0,2) = 2
V(X) = E(X^2) - [E(X)]^2 = 2 - (8/6)^2 = 0.2222
a[x^2/2](0,2) = 1
2a = 1
a = 1/2
E(X) = Integral(0,2) (1/2)x^2 dx = (1/2)[x^3/3](0,2) = 8/6
E(X^2) = Integral(0,2) (1/2)x^3 dx = (1/2)[x^4/4](0,2) = 2
V(X) = E(X^2) - [E(X)]^2 = 2 - (8/6)^2 = 0.2222