Question

For the random variable X with the given probability mass function as below, find the mean and variance.

`f(x) = {{:(1/2 "e"^(x/2),  "for"  x > 0),(0,  "otherwise"):}`

Answer 1

Mean: `mu = "E"("X")`

= `int_0^oo x f(x)  "d"x`

= `int_0^oo x 1/2 "e"^((-x)/2)  "d"x`

= `1/2 [- 2x"e"^((-x)/2) ]_0^oo + int_0^oo 2"e"^((-x)/2)  "d"x`

= `1/2[- 2x"e"^((-x)/2) + (2"e"^((-x)/2))/(- 1/2)]_0^oo`

= `1/2 [- 2x"e"^((-x)/2) - 4"e"^((-x)/2)]_0^oo`

= `1/2 [0 - 0 - (0 - 4)]`

= 2

Variance: `"E"("X"^2)`

= `int_0^oo x^2 f(x) "d"x`

`int u "dv" = "uv" - int "v" "du"`

u = `x int "dv" = int "e"^((-x)/2)  "d"x`

du =dx, v = `"e"^((-x)/2)/(- 1/2)`

v = `- 2"e"^((-x)/2)`

Bernoulli's formula

`int"u" "dv" = "uv" - "u'v"_1 + "u''v"_2 - ......`

u = x², `int"dv" = - int"e"^((-x)/2) "d"x`

u' = 2x, v = `- 2"e"^((-x)/2)`

u'' = 2, v₁ = `- 4"e"^((-x)/2)`

u'' = 0, v₂ = `- 8"e"^((-x)/2)`

`"E"("X"^2) = 1/2 int_0^oo x^2"e"^((-x)/2)  "d"x`

= `1/2[-2x^2 "e"^((-x)/2) - 8x"e"^((-x)/2) - 16"e"^((-x)/2)]_0^oo`

= `1/2 [0 - (- 16"e"^0)]`

= `1/2 xx 16`

= 8

Var(X) = E(X²) – [E(X)]²

= 8 – 4

= 4