Question

The probability density function of a continuous random variable X is
f(x) = `{{:("a" + "b"x^2",",  0 ≤ x ≤ 1),(0",",  "otherwise"):}`
where a and b are some constants. Find a and b if E(X) = `3/5`

Answer 1

Let X be due continuous variable of density function

`int_(-oo)^oo "f"(x)  "d"x` = 1

Here `int_0^1 ("a" + "b" x^2) "d"x` = 1

`["a"x + ("b"x^3)/3]` = 1

`["a"(1) + ("b"(1)^3)/3] - ["a"(0)  + ("b"(0))/3]` = 1

`"a" + "b"/3` = 1

⇒ 3a + b = 3 → (1)

Given that E(x) = `3/5`

`int_0^1 x "f"(x)  "d"x = 1 = 3/5`

`int_0^1 x("a" + "b"x^2)  "d"x = 3/5`

`["a" x^2/2 + "b" x^4/4]_0^1 = 3/5`

`["a"(1/2) + "b"(1/4)] - [0]] = 3/5`

`"a"/2 + "b"/4 = 3/5`

`(2"a"+ "b")/4 = 3/5`

⇒ `2"a" + "b" = 12/5` → (2)

Equation (1) - Equation (2) 

⇒  3a + b =   3
     2a + b = 12/5
       a = 3 - 12/5  

a = `(15 - 12)/5`

∴ a = `3/5`

Substitute the value of a = `3/5` in equation

`3(3/5) + "b"` = 3

`9/5 + "b"` = 3 

⇒ b = `3 - 9/5`

b = `(15 - 9)/5`

∴ b = `6/5`