Question

1 pts Question 3 Var(x2) 5, and E(x4)= 14. What is var(X) (the answer is an integer)? You are told that E (X) = 1.

Answer 1

Solution

Back-up Theory

Variance of X = Var(X) = σ² = E(X²) – {E(X)}²……………………………………………..(1)

Now, to work out the solution,

Just for easy understanding, let y = X² .

Then, vide (1),

Var(y) = E(y²) – {E(y)}² .................................................................................................. (2)

Substituting the given values, Var(y) = Var(X²) = 5, E(y²) = E(x⁴) = 14, in (2), we have:

5 = 14 - {E(y)}²

Or, {E(y)}² = 9

=> E(y) = ± 3

y being a square, (i.e., non-negative), E(y) cannot be – 3.

So, E(y) = 3 i.e., E(X²) = 3 ............................................................................................. (3)

Now, vide (1),

Var(X) = E(X²) – {E(X)}²

= 3 – 1² [vide (3) and given E(X) = 1]

= 2 Answer

DONE