Continuous Random Variable and Probability Density Function MCQ Quiz - Objective Question with Answer for Continuous Random Variable and Probability Density Function - Download Free PDF

Concept: 

Autocorrelation function [R (τ)] of the power signal is given as:

$R\left(\tau \right)=\underset{t\to \mathrm{\infty }}{\mathrm{l}\mathrm{t}}\frac{\mathrm{\perp }}{T}{\int }_{-T/2}^{T/2}x\left(t\right)x\left(t-\tau \right)dt$

Calculation:

Given:

x(t) = A sin [π/2 – (2πft + θ)]

$ACF:R\left(\tau \right)=\underset{t\to \mathrm{\infty }}{\mathrm{l}\mathrm{t}}\frac{\mathrm{\perp }}{T}{\int }_{-T/2}^{T/2}x\left(t\right)\cdot x\left(t-\tau \right)dt$

$\underset{t\to \mathrm{\infty }}{\mathrm{L}\mathrm{t}}\frac{\mathrm{\perp }}{T}{\int }_{-T/2}^{T/2}\left[Asin[\pi /2–(2\pi ft+\theta )]\cdot Asin[\pi /2–(2\pi f(t-\tau )+\theta )])\right]\cdot dt$

Solving the above, we get:

$R\left(\tau \right)=\frac{A^2}{2}\cos 2\pi ft\times \tau$

∴ The maximum amplitude of the autocorrelation function of this signal is A²/2

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

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}_X\left(x\right)dx$

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{k}{x^2}\left(x-k\right)dx$

$=\underset{k}{\overset{\mathrm{\infty }}{\int }}\frac{k}{x^2}dx$

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

= 1

Y = X²

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

So, we have:

$E\left[X^n\right]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^n{f}_X\left(x\right)dx=0$      ---(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

$\overline{Y}={\overline{X}}^2+{\sigma }_X^2={\sigma }_X^2$

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)]

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

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

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

$=0-{\sigma }_X^2\times 0=0$

Thus, the correlation coefficient ρ = 0

Therefore, the random variables are uncorrelated.

Concept:

For a function f_x(x) to be a valid PDF, it must satisfy the following conditions:

1) f_x(x) ≥ 0

2) $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}_x\left(x\right)dx=1$

Calculation:

For the function given in option (1):

$f_x\left(x\right)=\frac{1}{\pi }\left(\frac{1}{1+x^2}\right)$

We observe that:

f_x(x) ≥ 0 for all x.

Also $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}_x\left(x\right)dx=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\pi }\left(\frac{1}{1+x^2}\right)dx=\frac{1}{\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{1+x^2}dx$

$={\frac{1}{\pi }{\tan }^{-1}\left(x\right)|}_{-\mathrm{\infty }}^{\mathrm{\infty }}=\frac{1}{\pi }[{\tan }^{-1}\left(\mathrm{\infty }\right)-{\tan }^{-1}\left(-\mathrm{\infty }\right)]$

$=\frac{1}{\pi }\left[{\tan }^{-1}\left(\mathrm{\infty }\right)+{\tan }^{-1}\left(\mathrm{\infty }\right)\right]=\frac{2}{\pi }\cdot \frac{\pi }{2}=1$

Since both the conditions are satisfied.

So, the given function is a valid PDF function.

For the function in Option (2):

$f_x\left(x\right)=\{\begin{array}{c}\left|x\right|,\left|x\right|<1\\ 0,otherwise\end{array}$

f_x(x) ≥ 0 for all x.

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}_x\left(x\right)dx=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\left|x\right|dx$

$=\underset{-1}{\overset{0}{\int }}\left(-x\right)dx+\underset{0}{\overset{1}{\int }}xdx$

$={\frac{-x^2}{2}|}_{-1}^0+{\frac{x^2}{2}|}_0^1$

$=\frac{-1}{2}\left(-1\right)+\frac{1}{2}\left(1-0\right)$

$=\frac{1}{2}+\frac{1}{2}=1$

Since this function also satisfies both the conditions.

So, f_x(x) in Option (2) is also a valid PDF.

Concept:

The probability that a random variable ‘x’ takes a value in the interval [a, b] is given by the integral of a function called the probability density function f_x(x):-

$\Rightarrow P(a\le x\le b)={\int }_a^b{f}_x\left(x\right)dx$

Calculation:

Given a random variable with the probability density function f(z) defined as f(z) = e^-z, o < z < ∞, the probability of 0 < z < 2 will be obtained as:

$\Rightarrow P(a\le z\le b)={\int }_0^2{f}_z\left(z\right)dz$

$={\int }_0^2{e}^{-z}dz$

$={-e^{-z}|}_0^2$

= - [e^-2 - 1]

= 1 – e^-2

= 1 – 0.134

Concept:

The probability density function of a zero-mean Gaussian variable is as shown:

Mathematically, the density function of a Gaussian Random Variable is defined as:

$f\left(x\right)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}$

Given distribution has zero mean i.e. μ = 0, so the above distribution can be written as:

$f\left(x\right)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{x^2}{2{\sigma }^2}}$

The Fourier Transform of a Gaussian distribution is as shown:

$F\left(\omega \right)=\sqrt{2{\sigma }^2\pi }{e}^{\frac{2{\sigma }^2{\omega }^2}{4}}$

Clearly, the phase spectrum is constant and with the same standard deviation for the given two random signals, the phase is uniform from -π to +π.

Concept:

The received signal is said to an error signal if we receive 1 but transmitted signal was 0, or if we receive 0 but transmitted signal was 1.

$\therefore P\left(e\right)=P\left(0\right)P\left(\frac{1}{0}\right)+P\left(1\right)P\left(\frac{0}{1}\right)$

where,

P(e) = Probability of error

P(0) = Probability of transmission of ‘0’

P(1) = Probability of transmission of ‘1’

P(0/1) = Probability of reception of ‘0’ on the transmission of ‘1’

P(1/0) = Probability of reception of ‘1’ on the transmission of ‘0’

The area under the curve of probability density function gives the probability.

Application:

There is no error in the reception of ‘0’.

But there is an error in reception of ‘1’

$\therefore P\left(e\right)=P\left(0\right)P\left(\frac{1}{0}\right)+P\left(1\right)P\left(\frac{0}{1}\right)$

$=\left(\frac{1}{2}\right)\left(0\right)+\frac{1}{2}\times \frac{1}{2}\times 1\times 0.25$

$P(e)=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{4}$

$P\left(e\right)=\frac{1}{16}$

Concept: 1. $PDF=\frac{d\left(CDF\right)}{dz}$ 

                2. $P\left(\frac{A}{B}\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$

                3. CDF gives the probability i.e. Fx(x) = P (X ≤ x)

Calculation:

Method – I:

mean = 1

$CDF=F_z\left(x\right)=\{\begin{array}{cc}1-e^{-x}& x\ge 0\\ 0& x<0\end{array}$

$PDF=\frac{d\left(CDF\right)}{dz}$

$f_z\left(x\right)=\frac{d}{dz}\left(1-e^{-x}\right)=e^{-x}$

$f_z\left(x\right)=\{\begin{array}{cc}{e}^{-x}& x\ge 0\\ 0& x<0\end{array}$

From concept of Probability:

$P\left(\frac{A}{B}\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$

$P\left(\frac{z>2}{z>1}\right)=\frac{P\left(z>2\right)\cap P\left(z>1\right)}{P\left(z>1\right)}$

$=\frac{P\left(z>2\right)}{P\left(z>1\right)}$

Area under Pdf gives probability

$P\left(z>2\right)=\underset{2}{\overset{\mathrm{\infty }}{\int }}{e}^{-x}dx$

$\Rightarrow {\left[-e^{-x}\right]}_2^{\mathrm{\infty }}$

$\Rightarrow -\left[e^{-\mathrm{\infty }}-e^{-2}\right]$

= e^-2

$P\left(z>1\right)=\underset{1}{\overset{\mathrm{\infty }}{\int }}{e}^{-x}dx$

= e^-1 (Simply replace ‘2’ by ‘1’ in above result)

$P\left(\frac{z>2}{z>1}\right)=\frac{e^{-2}}{e^{-1}}=\frac{1}{e}=0.367$

Method – 2:

$P\left(\frac{z>2}{z>1}\right)=\frac{P\left(z>2\right)}{P\left(z>1\right)}$

CDF gives the probability

i.e. F_x(x) = P (X ≤ x)

P(z > 2) = 1 – P(z ≤ 2)

= 1 – CDF(z = 2)

= 1 – (1 – e^-2)

= e^-2

P(z > 1) = 1 – P(z ≤ 1)

= 1 – CDF (z = 1)

= 1 – (1 – e^-1)

= e^-1

$P\left(\frac{z>2}{z>1}\right)=\frac{e^{-2}}{e^{-1}}=\frac{1}{e}$

= 0.367

Concept: 

Autocorrelation function [R (τ)] of the power signal is given as:

$R\left(\tau \right)=\underset{t\to \mathrm{\infty }}{\mathrm{l}\mathrm{t}}\frac{\mathrm{\perp }}{T}{\int }_{-T/2}^{T/2}x\left(t\right)x\left(t-\tau \right)dt$

Calculation:

Given:

x(t) = A sin [π/2 – (2πft + θ)]

$ACF:R\left(\tau \right)=\underset{t\to \mathrm{\infty }}{\mathrm{l}\mathrm{t}}\frac{\mathrm{\perp }}{T}{\int }_{-T/2}^{T/2}x\left(t\right)\cdot x\left(t-\tau \right)dt$

$\underset{t\to \mathrm{\infty }}{\mathrm{L}\mathrm{t}}\frac{\mathrm{\perp }}{T}{\int }_{-T/2}^{T/2}\left[Asin[\pi /2–(2\pi ft+\theta )]\cdot Asin[\pi /2–(2\pi f(t-\tau )+\theta )])\right]\cdot dt$

Solving the above, we get:

$R\left(\tau \right)=\frac{A^2}{2}\cos 2\pi ft\times \tau$

∴ The maximum amplitude of the autocorrelation function of this signal is A²/2

Concept:

The received signal is said to be an error signal if we receive 1 but the transmitted signal was 0, or if we receive 0 but the transmitted signal was 1.

$\therefore P\left(e\right)=P\left(0\right)P\left(\frac{1}{0}\right)+P\left(1\right)P\left(\frac{0}{1}\right)$

where,

P(e) = Probability of error

P(0) = Probability of transmission of ‘0’

P(1) = Probability of transmission of ‘1’

P(0/1) = Probability of reception of ‘0’ on the transmission of ‘1’

P(1/0) = Probability of reception of ‘1’ on the transmission of ‘0’

The area under the curve of the probability density function gives the probability.

Analysis:

Optimum threshold = point of intersection of both bits. We can find it to be $=\frac{4}{5}$ using eq. of line.

Alternatively,

Let the threshold be $\mathrm{x}$ and let it be between 0 and 1, then:

$P_e=\frac{1}{2}\left[\frac{1}{2}\times \left(1-x\right)\times \left(-x+1\right)\right]+\frac{1}{2}\left[\frac{1}{2}\times x\times \frac{x}{4}\right]$

$P_e=\frac{1}{2}\left[\frac{{\left(x-1\right)}^2}{2}+\frac{x^2}{8}\right]$

For the minimum probability of error:

$\frac{d{P}_e}{dx}=\frac{1}{2}\left[x-1+\frac{x}{4}\right]=0$

$5x=4$

$x=\frac{4}{5}$

Concept:

The probability that a random variable ‘x’ takes a value in the interval [a, b] is given by the integral of a function called the probability density function f_x(x):-

$\Rightarrow P(a\le x\le b)={\int }_a^b{f}_x\left(x\right)dx$

Calculation:

Given a random variable with the probability density function f(z) defined as f(z) = e^-z, o < z < ∞, the probability of 0 < z < 2 will be obtained as:

$\Rightarrow P(a\le z\le b)={\int }_0^2{f}_z\left(z\right)dz$

$={\int }_0^2{e}^{-z}dz$

$={-e^{-z}|}_0^2$

= - [e^-2 - 1]

= 1 – e^-2

= 1 – 0.134

Concept: 

1) The probability density function or PDF of a continuous random variable gives the relative likelihood of any outcome in a continuum occurring.

2) Unlike the case of discrete random variables, for a continuous random variable, any single outcome has a probability zero of occurring.

3) We always define the range of occurrence of any probability in the case of a continuous variable. 

​Observation:

For a continuous distributed random variable probability of at a single point cannot be defined

∴ P(x = 4) = 0

Concept:

The probability density function of a zero-mean Gaussian variable is as shown:

Mathematically, the density function of a Gaussian Random Variable is defined as:

$f\left(x\right)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}$

Given distribution has zero mean i.e. μ = 0, so the above distribution can be written as:

$f\left(x\right)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{x^2}{2{\sigma }^2}}$

The Fourier Transform of a Gaussian distribution is as shown:

$F\left(\omega \right)=\sqrt{2{\sigma }^2\pi }{e}^{\frac{2{\sigma }^2{\omega }^2}{4}}$

Clearly, the phase spectrum is constant and with the same standard deviation for the given two random signals, the phase is uniform from -π to +π.

Concept:

Properties of CDF

(a) CDF is non-decreasing and right continuous

(b) CDF goes to 1 as x tends to positive infinity

(c) F(x) goes to 0 as x tends to negative infinity

Calculations:

In option (1) and (2) value exceeds ‘1’, hence not valid CDF

In option (3) as x increase, the value F_x(x) decrease. Hence invalid CDF

Concept:

The received signal is said to an error signal if we receive 1 but transmitted signal was 0, or if we receive 0 but transmitted signal was 1.

$\therefore P\left(e\right)=P\left(0\right)P\left(\frac{1}{0}\right)+P\left(1\right)P\left(\frac{0}{1}\right)$

where,

P(e) = Probability of error

P(0) = Probability of transmission of ‘0’

P(1) = Probability of transmission of ‘1’

P(0/1) = Probability of reception of ‘0’ on the transmission of ‘1’

P(1/0) = Probability of reception of ‘1’ on the transmission of ‘0’

The area under the curve of probability density function gives the probability.

Application:

There is no error in the reception of ‘0’.

But there is an error in reception of ‘1’

$\therefore P\left(e\right)=P\left(0\right)P\left(\frac{1}{0}\right)+P\left(1\right)P\left(\frac{0}{1}\right)$

$=\left(\frac{1}{2}\right)\left(0\right)+\frac{1}{2}\times \frac{1}{2}\times 1\times 0.25$

$P(e)=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{4}$

$P\left(e\right)=\frac{1}{16}$