If x is a discrete random variable and let a, b be two constants, show

1.	If x is a discrete random variable and let a, b be two constants, show that 1. E(a) = a 2. E(ax) = aE(x) 3. E(ax + b) = aE(x) + b 4. Var (a) = 0 5. Var (ax) = a2 Var (x)
Answer» Proof: From the definitions of probability mass function and mathematical expectation 1. E(a) = Σa .P(a) = a. ΣP(x) = a.1 = a. ∵ ΣP(x) = 1 2. E(ax) = Σax P(x) = aΣx P(x) = a. E(x) ∵Σx P(x) = E(x) 3. E(ax + b) = E(ax + b). P(x) Removing bracket = Σax. P(x) + Σb.P(x) = aΣx . P(x) + bΣP(x) = a. E(x) + b. 1 = aE(x) + b 4. Var (a) = E[a – E(a)]² ; ∴ V(x) = E[x – E(x)]² = E[a – a]² = E(0) = 0 5. Var (an) = E[ax – E(ax)]² = E[ax – aE(x)]² ; ∴E(ax) = aE(x) = a²E(x – E(x)]² = a²Var (x).

If x is a discrete random variable and let a, b be two constants, show that 1. E(a) = a 2. E(ax) = aE(x) 3. E(ax + b) = aE(x) + b 4. Var (a) = 0 5. Var (ax) = a2 Var (x)

Answer»

Proof: From the definitions of probability mass function and mathematical expectation

1. E(a) = Σa .P(a) = a. ΣP(x) = a.1 = a.

∵ ΣP(x) = 1

2. E(ax) = Σax P(x) = aΣx P(x) = a. E(x)

∵Σx P(x) = E(x)

3. E(ax + b) = E(ax + b). P(x) Removing bracket

= Σax. P(x) + Σb.P(x) = aΣx . P(x) + bΣP(x)

= a. E(x) + b. 1 = aE(x) + b

4. Var (a) = E[a – E(a)]² ;

∴ V(x) = E[x – E(x)]²

= E[a – a]² = E(0) = 0

5. Var (an) = E[ax – E(ax)]² = E[ax – aE(x)]² ;

∴E(ax) = aE(x)

= a²E(x – E(x)]² = a²Var (x).