Question

If random variable X has probability distribution function.
f(x) = `c/x`, 1 < x < 3, c > 0, find c, E(x) and Var(X)

Answer 1

∴ X has probability distribution function
∴  `int_1^3 c/x dx = 1`

`[ c log x]_1^3` = 1

c [ log3 - log 1 ] = 1
c log 3 = 1

c = `1/log 3`  

E(X) = `int_1^3 xf(x)`

        = `int_1^3 x . c/x dx`

        = `c int_1^3 1 dx`

        = `c | x |_1^3`

        = c | 3 - 1|
        = 2c
E(X) = `2/log 3`

V(X) = `E( "X"^2 ) - E( "X"^2 )`

V(X) = `int_1^3 x^2f(x) dx - [ int_1^3 x x f(x) dx ]^2`

= `int_1^3 x^2 c/x dx - [ 2/log 3 ]^2`

= `int_1^3  cx dx - [ 2/log 3 ]^2`

= `c[ x^2/2 ]_1^3 - [ 2/log 3 ]^2`

= `1/log 3 xx [ (9 - 1)/2] - 4/[(log 3)^2]`

V(X) = = `4/log 3- 4/[(log 3)^2]`

Answer 2

∴ X has probability distribution function
∴  `int_1^3 c/x dx = 1`

`[ c log x]_1^3` = 1

c [ log3 - log 1 ] = 1
c log 3 = 1

c = `1/log 3`  

E(X) = `int_1^3 xf(x)`

        = `int_1^3 x . c/x dx`

        = `c int_1^3 1 dx`

        = `c | x |_1^3`

        = c | 3 - 1|
        = 2c
E(X) = `2/log 3`

V(X) = `E( "X"^2 ) - E( "X"^2 )`

V(X) = `int_1^3 x^2f(x) dx - [ int_1^3 x x f(x) dx ]^2`

= `int_1^3 x^2 c/x dx - [ 2/log 3 ]^2`

= `int_1^3  cx dx - [ 2/log 3 ]^2`

= `c[ x^2/2 ]_1^3 - [ 2/log 3 ]^2`

= `1/log 3 xx [ (9 - 1)/2] - 4/[(log 3)^2]`

V(X) = = `4/log 3- 4/[(log 3)^2]`

Answer 3

∴ X has probability distribution function
∴  `int_1^3 c/x dx = 1`

`[ c log x]_1^3` = 1

c [ log3 - log 1 ] = 1
c log 3 = 1

c = `1/log 3`  

E(X) = `int_1^3 xf(x)`

        = `int_1^3 x . c/x dx`

        = `c int_1^3 1 dx`

        = `c | x |_1^3`

        = c | 3 - 1|
        = 2c
E(X) = `2/log 3`

V(X) = `E( "X"^2 ) - E( "X"^2 )`

V(X) = `int_1^3 x^2f(x) dx - [ int_1^3 x x f(x) dx ]^2`

= `int_1^3 x^2 c/x dx - [ 2/log 3 ]^2`

= `int_1^3  cx dx - [ 2/log 3 ]^2`

= `c[ x^2/2 ]_1^3 - [ 2/log 3 ]^2`

= `1/log 3 xx [ (9 - 1)/2] - 4/[(log 3)^2]`

V(X) = = `4/log 3- 4/[(log 3)^2]`

Answer 4

∴ X has probability distribution function
∴  `int_1^3 c/x dx = 1`

`[ c log x]_1^3` = 1

c [ log3 - log 1 ] = 1
c log 3 = 1

c = `1/log 3`  

E(X) = `int_1^3 xf(x)`

        = `int_1^3 x . c/x dx`

        = `c int_1^3 1 dx`

        = `c | x |_1^3`

        = c | 3 - 1|
        = 2c
E(X) = `2/log 3`

V(X) = `E( "X"^2 ) - E( "X"^2 )`

V(X) = `int_1^3 x^2f(x) dx - [ int_1^3 x x f(x) dx ]^2`

= `int_1^3 x^2 c/x dx - [ 2/log 3 ]^2`

= `int_1^3  cx dx - [ 2/log 3 ]^2`

= `c[ x^2/2 ]_1^3 - [ 2/log 3 ]^2`

= `1/log 3 xx [ (9 - 1)/2] - 4/[(log 3)^2]`

V(X) = = `4/log 3- 4/[(log 3)^2]`