auto-correlation variance

Ask

If the variance $\sigma_x^2$ of  $d(n)= x (n) - x(n-1)$ is one-tenth the variance $\sigma_x^2 $ of a stationary zero mean discrete time signal x(n), then the normalized auto-correlation function

$\frac{R_{xx}(k) }{\sigma_x^2}$ at k = 1 is

(a) 0.95

(b) 0.90

(c) 0.10

(d) 0.05

Answer is ( a )

Concept

Let's understand the definition of variance $$\sigma^2 = Var [A] $$, where A is any random process

Then , $$\sigma^2 = E [ (A-E(A))^2] $$
                            $ = E[(A^2 - 2AE(A) + (E(A))^2)]$
                                               Taking Expectation E within the brackets:  
                                                          $ =E[A^2] -2E[A]E[A] + (E(A))^2 $
                                                          $ = E[A^2] -(E[A])^2$
A mnemonic for the above expression is "mean of square minus square of mean"

The process is defined by :
$d(n) = x(n) - x(n-1)$

And also given that process is zero-mean.E[x(n)]  =0 $\implies (E[x(n)-x(n-1)])^ 2 = 0 $
Hence,Variance of d(n) = $\sigma_d^2 = E[(x(n)-x(n-1))^2] - (E[x(n)-x(n-1)])^2$
$\sigma_d^2 = E[(x(n)-x(n-1))^2] - 0 $
$\sigma_d^2 = E[(x(n))^2 +(x(n-1))^2 - 2x(n)x(n-1)] $
$\sigma_d^2 = E[(x(n))^2] +E[(x(n-1))^2] - 2 E[x(n)x(n-1)] $
$$\sigma_d^2 = \sigma_x^2 \times \frac{1}{10}$$

The expression $ E[x(n)x(n-1)] $ is auto-correlation by lag 1.

In general , the autocorrelation function with lag k is defined by the following equation
$$R_{xx}(k) = E[x(n)x(n-k)]$$

$\sigma_x^2 \times \frac{1}{10}= \sigma_x^2 + \sigma_x^2 - 2R_{xx}(1)$
$\implies \frac{ R_{xx}(1)}{\sigma_x^2 } = 19/20 = 0.95$