Probability Distribution Mean (Expectation), Variance :: Problems

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Prove that: (i) var(2) = 0, (ii) var(3x) = 9 (iii) var (x + 4x) = 16 var (x)

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Solution  
 

We know, var (x) = E(x2) − (E (x))2
= Σ px2 − (Σ px)2

Therefore,
(i)
var (2) = Σ p × (2)2 − (Σ p × (2))2
= Σ p × 4 − (2 Σ p)2
= 4 Σ p − (2 × 1)2
= 4 × 1 − (2)2
= 4 − 4
= 0
(ii)
var (3x) = Σ p × (3x)2 − (Σ p × (3x))2
= Σ p × 9x2 − (3 Σ px)2
= 9 Σ p x2 − [ (3)2 × (Σ px)2]
= 9 Σ p x2 − [9 (Σ px)2]
= 9 {Σ p x2 − (Σ px)2}
= 9 {E (x2) − (E (x))2}
= 9 var (x)
(iii)
var (3 + 4x) = Σ p (3 + 4x)2 − ( Σ p (3 + 4x))2
= Σ p {(3)2 + 2(3)(4)x + ((4)x)2)} − (Σ p × 3 + p × 4x)2
= Σ p (9 + 24x + 16x2) − {Σ (3p + 4px)2}
= Σ (9p + 24px + 16px2) − (Σ 3p + Σ 4px)2
= {Σ 9p + Σ 24px + Σ 16px2} − {3 Σ p + 4 Σ px)2}
= {9 Σ p + 24 Σ px + 16 Σ px2} − {3 × 1 + 4 Σ px)2}
= {9 × 1 + 24 Σ px + 16 Σ px2} − {3 + 4 Σ px)2}
= {9 + 24 Σ px + 16 Σ px2} − { (3)2 + 2 (3)(4 Σ px) + (4 Σ px)2}
= {9 + 24 Σ px + 16 Σ px2} − { 9 + 24 Σ px + [(4)2 × (Σ px)2]}
= {9 + 24 Σ px + 16 Σ px2} − { 9 + 24 Σ px + 16 (Σ px)2}
= 9 + 24 Σ px + 16 Σ px2 − 9 − 24 Σ px − 16 (Σ px)2
= 16 Σ px2 − 16 (Σ px)2
= 16 [Σ px2 − (Σ px)2]
= 16 [E(x2) − (E(x))2]
= 16 var (x)

Credit : Vijayalakshmi Desu

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