Question

A PAM source generates four symbols 3 V, 1 V, -1 V and -3 V with probability of p(3) = p(-3) = 0.2 and p(1) = p(-1) = 0.3 respectively. The variance for the source will be
1. 4.2 V
2. 3.2 V
3. 3.6 V
4. 4.6 V

Answer 1

Correct Answer - Option 1 : 4.2 V

Concept:

For discrete random variable x:

Mean = E[x] = ∑_I x_i p (x_i)

Mean square value (MSQ) = E[x²] = ∑_I x_i² p (x_i)

Variance (σ²) = E[x²] – {E(x)}²

Standard deviation (σ) \( = \sqrt {variance} \)

Calculation:

Given data: for the given PAM source

xi	-3	-1	1	3
P(xi)	0.2	0.3	0.3	0.2

 

Mean = E[x] = ∑_i x_i p (x_i)

= (-3)(0.2) + (-1)(0.3) + (1)(0.3) + (3)(0.2)

= 0

MSQ = E[x²] = ∑_I x_i² p (x_i)

= (-3)²(0.2) + (-1)²(0.3) + (1)²(0.3) + (3)²(0.2) V

= 4.2 V

Variance = E[x²] – {E[x]}²

= 4.2 – 0

= 4.2 V

Important point:

For continuous random variable x:

Mean = E[x] \( = \mathop \smallint \nolimits_{ - \infty }^\infty f\left( x \right)dx\)

∵ f(x) = probability density function of C.R.V (X)

Mean square value = E[x²] \( = \mathop \smallint \nolimits_{ - \infty }^\infty {x^2}f\left( x \right)dx\)

Variance (σ²) = E[x²] – {E[x]}²