Communications Test 7

Given the function,

\({f_X}\left( x \right) = \frac{k}{{{x^2}}}u\left( {x - k} \right)\)

Now, we check the validity of this function

\(\mathop \smallint \limits_{ - \infty }^\infty {f_X}\left( x \right)dx\)

\( = \mathop \smallint \limits_{ - \infty }^\infty \frac{k}{{{x^2}}}\left( {x - k} \right)dx\)

\( = \mathop \smallint \limits_k^\infty \frac{k}{{{x^2}}}dx\)

\(= k\left[ {\frac{{{x^{ - 1}}}}{{ - 1}}} \right]_k^\infty = \frac{k}{k}\)

= 1

Concept:

If ‘X’ is said to be normal Random Variable defined in - ∞ to ∞ with mean ‘μ’ and variance ‘ σ² ‘ then the Random variable is known as “Normal Random Variable”.

Its probability density function is defined as:

\(N\left( {X;\mu ,{\sigma ^2}} \right) = f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{1}{{\sigma \sqrt {2\pi } }}{e^{ - \frac{1}{2}(\frac{{{{\left( {x - \mu } \right)}^2}}}{{{\sigma ^2}}}}}}\\{ - \infty < x < \infty }\\{ - \infty < \mu < \infty }\\{0 < \sigma < \infty }\\{0;else}\end{array}} \right\}\)

Variance = σ2

Calculation:

From the above equation comparing with the question we get

2σ² = 8

σ² = 4

Y = X²

Since the probability density function f_X(x) of random variable X has even symmetry.

So, we have:

\(E\left[ {{X^n}} \right] = \mathop \smallint \limits_{ - \infty }^\infty {x^n}{f_X}\left( x \right)dx = 0\;\)      ---(1)

Where n is an odd integer.

Therefore, we have the mean of random variable X as

X̅ = E[X] = 0

Also, the mean value of random variable Y is given by

Y̅ = E[Y] = E[X²] = X̅²

\(\bar Y = {\bar X^2} + \sigma _X^2 = \sigma _X^2\) 

Now, we check the orthogonality and correlation for the given random variables.

Orthogonal:

Two random variables X and Y are orthogonal if E[XY] = 0

For the given problem, we have:

E[XY] = E[X X²] = E[X³]       ---(2)

Substituting n = 3 in equation (1), we get

E[X³] = 0

So, substituting this value in equation (2), we obtain

E[XY] = 0

Therefore, the variables X and Y are orthogonal

Correlation:

Two random variables X and Y are uncorrelated only if their correlation coefficient is zero; i.e.

\(\rho = \frac{{cov\left[ {X,\;\;Y} \right]}}{{{\sigma _X}{\sigma _Y}}} = 0\) 

Now, for the given variables, we obtain the covariance as

cov[X, Y] = E[(X – X̅)(Y – Y̅)]

\( = E\left[ {\left( {X - 0} \right)\left( {{X^2} - \sigma _X^2} \right)} \right]\) 

\(= E\left[ {X\left( {{X^2} - \sigma _X^2} \right)} \right]\) 

\(= e\left[ {{X^3}} \right] - \sigma _X^2E\left[ X \right]\) 

\(= 0 - \sigma _X^2 \times 0 = 0\) 

Thus, the correlation coefficient ρ = 0

Therefore, the random variables are uncorrelated.

Concept:

Variance of random variable X is given by

\(\sigma _x^2 = E\left[ {{X^2}} \right] - {\left[ {E\left[ X \right]} \right]^2}\)

here, X = cosθ

where, -π < θ < π

Calculation:

\(E\left[ {{{\cos }^2}\theta } \right] = \mathop \smallint \limits_{ - \pi }^{ + \pi } \frac{1}{{2\pi }}Co{s^2}\theta \;d\theta \;\)

\(= \frac{1}{{2\pi }}\mathop \smallint \limits_{ - \pi }^{ + \pi } \left( {\frac{{1 + \cos 2\theta }}{2}} \right)d\theta \)

\(\Rightarrow \frac{1}{{4\pi }}\left[ {\mathop \smallint \limits_{ - \pi }^{ + \pi } d\theta + \mathop \smallint \limits_{ - \pi }^{ + \pi } cos2\theta d\theta } \right]\)

\(\Rightarrow \frac{1}{{4\pi }}\;\left( {2\pi } \right) = \frac{1}{2}\)

\(E\left[ {{{\left( {{{\cos }^2}\theta } \right)}^2}} \right] = \frac{1}{{2\pi }}\mathop \smallint \limits_{ - \pi }^{ + \pi } {\left( {Co{s^2}\theta } \right)^2}d\theta\)

\(= \frac{1}{{2\pi }}\mathop \smallint \limits_{ - \pi }^{ + \pi } {\left( {\frac{{1 + \cos 2\theta }}{2}} \right)^2}d\theta \)

\(= \frac{1}{{8\pi }}\mathop \smallint \limits_{ - \pi }^{ + \pi } {\left( {Co{s^2}\theta } \right)^2}d\theta \)

= 3/8

Variance = \(\left( {\frac{3}{8}} \right) - {\left( {\frac{1}{2}} \right)^2}\)

\(= \frac{3}{8} - \frac{1}{4}\)

\(= \frac{1}{8}\)