paper scheme

1. JNTUK R20 II-I Sem RVSP R2021044
1(A) Define conditional distribution function and list its properties
Let A,B are two events, if A= {𝑋 ≤ 𝑥} for the random variable X then the conditional distribution
function of X when the event B is known is denoted as 𝐹
𝑋
(𝑥 𝐵
⁄ )
Defination of Conditional distribution function 𝐹
𝑋
(𝑥 𝐵
⁄ )
= 𝑃{𝑋 ≤ 𝑥|𝐵} [3 Marks]
List of the Properties:
(1) 𝐹
𝑋
(∞ 𝐵
⁄ )
= 1
(2) 𝐹
𝑋
(−∞ 𝐵
⁄ )
= 0
(3) 0 ≤ 𝐹
𝑋
(𝑥 𝐵
⁄ )
≤ 1
(4) 𝐹
𝑋
(𝑥1 𝐵
⁄ )
≤ 𝐹
𝑋
(𝑥2 𝐵
⁄ )
𝑖𝑓 𝑥1 < 𝑥2
(5) 𝑃{𝑥1 < 𝑋 < 𝑥2} = 𝐹𝑋
(𝑥2 𝐵
⁄ )
− 𝐹
𝑋
(𝑥1 𝐵
⁄ )
(6) 𝐹
𝑋
(𝑥∗ 𝐵
⁄ )
= 𝐹
𝑋
(𝑥 𝐵
⁄ )
[4 Marks]
1(B) Define Gaussian Random variable X, Draw the PDF of X for two different values of
𝝈, 𝒊𝒆. , 𝝈𝟏𝒂𝒏𝒅 𝝈𝟐 𝒂𝒔𝒔𝒊𝒈𝒏𝒊𝒏𝒈 𝝈𝟏 < 𝝈𝟐 what is your observation
Gaussian density function is a bell shaped curve with mean and vaiance as the parameters
𝑓𝑥
(𝑥)
=
1
√2𝜋𝜎𝑥
2
𝑒
−
(𝑥−𝜇𝑥)
2
2𝜎𝑥
2
for all x values [2 Marks]
Gaussian Graph: [2 Marks]
𝜎 𝑖𝑠 𝑛𝑎𝑚𝑒𝑑 𝑎𝑠 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑖𝑡 𝑑𝑒𝑠𝑐𝑟𝑖𝑏𝑠 𝑎𝑏𝑜𝑢𝑡 𝑠𝑝𝑟𝑒𝑎𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑢𝑠𝑠𝑖𝑎𝑛
𝑮𝒂𝒖𝒔𝒔𝒊𝒂𝒏 𝒇𝒐𝒓 𝒕𝒘𝒐 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕 𝒗𝒂𝒍𝒖𝒆𝒔 𝝈𝟏𝒂𝒏𝒅 𝝈𝟐 [2 Marks]
As the standard deviation increases the spread of the Gaussian increases and covers large number of
samples in to consideration [1 Mark]

2. 2(a) Explain the Following (i) Point Conditioning (ii) Interval conditioning
(i) Point Conditioning [3 Marks]
The distribution of one random variable x conditioning by the fact that a second rancom
variable Y has some specific value y called point conditioning and it can be handeled
such problem as defining an event 𝐵 = {𝑦 − ∆𝑦 < 𝑦 ≤ 𝑦 +
∆𝑦}, 𝑤ℎ𝑒𝑟𝑒 ∆𝑦 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑚𝑎𝑙𝑙 𝑞𝑢𝑎𝑙𝑡𝑖𝑡𝑦 𝑎𝑝𝑟𝑜𝑎𝑐ℎ 𝑡𝑜 0.
𝐹
𝑋
(𝑥 𝐵
⁄ )
= 𝑃{𝑋 ≤ 𝑥|𝐵} = 𝐹
𝑋
(𝑥 𝐵
⁄ )
= 𝑃{𝑋 ≤ 𝑥|𝑦 − ∆𝑦 < 𝑦 ≤ 𝑦 + ∆𝑦}
𝑃{𝑋 ≤ 𝑥|𝑦 − ∆𝑦 < 𝑦 ≤ 𝑦 + ∆𝑦}=
∫ ∫ 𝑓𝑥𝑦
(𝑥,𝑦)
𝑑𝑥 𝑑𝑦
𝑥
−∞
𝑦+∆𝑦
𝑦−∆𝑦
∫ 𝑓𝑦
(𝑦)
𝑦+∆𝑦
𝑦−∆𝑦 𝑑𝑦
(ii) Interval conditioning
B is defined in the intervals as such problem as defining an event 𝐵 = {𝑦1 < 𝑦 ≤ 𝑦2},
Then the conditional distribution of the 𝑃{𝑋 ≤ 𝑥|𝑦1 < 𝑦 ≤ 𝑦2} is named as Interval
conditioning
𝑃{𝑋 ≤ 𝑥|𝑦1 < 𝑦 ≤ 𝑦2} =
∫ ∫ 𝑓𝑥𝑦
(𝑥,𝑦)
𝑑𝑥 𝑑𝑦
𝑥
−∞
𝑦2
𝑦1
∫ 𝑓𝑦
(𝑦)
𝑦2
𝑦1
𝑑𝑦
[4 Marks]
2(b) Suppose there is an error probability of 0.05 per word in typing using an electronic type writer
machine. What is the probability that there will be more than one error in a page of 120 words?
Use poisons density function
𝑃(𝑋 = 𝑘) =
𝑒−𝜆𝜆𝑘
𝑘!
[3 marks]
k=6
more than one error probability 𝑃(𝑋 > 1) = 1 − 𝑃(𝑋 ≤ 1)
𝑃(𝑋 ≤ 1) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) [2 Marks]
𝑒−660
0!
+
𝑒−661
1!
= (7𝑒−6
)= 0.01735
𝑃(𝑋 > 1) = 1 − 0.01735 = 0.9826 [2 Marks]
3(a) if Y= ax+b where a and b are real and constant find variance of Y
𝑣𝑎𝑟(𝑌) = 𝐸[𝑌2] − [𝐸(𝑌)]2
[2Marks]
𝐸[𝑌2] = 𝐸[(𝑎𝑋 + 𝑏)2] = 𝐸[𝑎2
𝑥2
+ 2𝑎𝑏𝑥 + 𝑏2]
= [𝑎2
𝐸[𝑥2
] + 2𝑎𝑏𝐸[𝑥] + 𝑏2] [2 Marks]
[𝐸(𝑌)]2
= [𝐸(𝑎𝑥 + 𝑏)]2
[𝐸(𝑎𝑥 + 𝑏)]2
= (𝑎𝐸(𝑥) + 𝑏)2
[𝐸(𝑌)]2
= (𝑎𝐸(𝑥) + 𝑏)2
= 𝑎2
[𝐸(𝑥)]2
+ 2𝑎𝑏𝐸(𝑥) + 𝑏2
[2 Marks]

3. 𝑣𝑎𝑟(𝑌) = 𝐸[𝑌2] − [𝐸(𝑌)]2
= [𝑎2
𝐸[𝑥2
] + 2𝑎𝑏𝐸[𝑥] + 𝑏2] − {𝑎2[𝐸(𝑥)]2
+ 2𝑎𝑏𝐸(𝑥) + 𝑏2}
𝑣𝑎𝑟(𝑌) = [𝑎2
𝑣𝑎𝑥(𝑥)] [ 1 Mark]
3(b) Characteristic function of a random variable is given as 𝝓𝒙
(𝒘)
=
𝟏
𝟏−𝒋𝒘
find (i) Mean value of X,
(ii) variance of X.
(i) Mean calculation using Characteristic function 𝑚1 = (−𝑗)
𝑑𝜙𝑋
(𝑤)
𝑑𝑤
/ 𝑤 = 0 [3 Marks]
𝑚1 = (−𝑗)
𝑑
1
1 − 𝑗𝑤
𝑑𝑤
= (−𝑗)
−1(𝑗)
(1 − 𝑗𝑤)2
=
1
(1 − 𝑗𝑤)2
Substitute w=0 we get m1=1
(ii) Variance of the variable X [4 Marks]
𝑣𝑎𝑟(𝑋) = 𝐸[𝑋2] − [𝐸(𝑋)]2
= 𝑚2 − (𝑚1)2
𝐸[𝑋2] = 𝑚2 = (−𝑗)2
𝑑2
𝜙𝑋
(𝑤)
𝑑𝑤2
/ 𝑤 = 0
𝑚2 =
2
(1 − 𝑗𝑤)2
Substitute w=0 m2=2
𝑣𝑎𝑟(𝑋) = 𝐸[𝑋2] − [𝐸(𝑋)]2
= 𝑚2 − (𝑚1)2
= 2 − 1 = 1
4(a) Define the following (i) Skew (ii) Coefficient of Skew (iii) nth central moment
(i) Skew: 3rd
central moment of the random variable X it is given by 𝜇3 = 𝐸[(𝑥 − 𝑥̅)3
]. It is
used to measure asymmetry of density function 𝑓𝑥
(𝑥)
[ 2 Marks]
(ii) Coefficient of Skew: Normalized 3rd
enteral moment or ration of 3rd
central moment to
the cube of standard deviation named as coefficient to skew
𝝁𝒙
𝝈𝒙
𝟑 =
𝐸[(𝑥−𝑥̅)3]
𝐸[(𝑥−𝑥̅)3]3/2 [3 Marks]
(iii) nth central moment: 𝜇𝑛 = 𝐸[(𝑥 − 𝑥̅)𝑛
] [2 Marks].
4(b) A random variable X is uniformly distributed in the interval of (0,1) find the PDF of new
random variable 𝒀 = 𝑿𝟐
X is uniform density function with density function 𝑓𝑥
(𝑥)
=
1
𝑏−𝑎
= 1 [2 Marks]
As per random variable transformation approach 𝑓𝑌
(𝑦)
= 𝑓𝑥
(𝑥) 𝑑𝑥
𝑑𝑦
[2 Marks]
𝑤𝑒 𝑛𝑜𝑤 𝑌 = 𝑋2
dy=2xdx

4. 𝑑𝑥
𝑑𝑦
=
1
2𝑥
𝑓𝑌
(𝑦)
= 𝑓𝑥
(𝑥)
(
1
2𝑥
) = (
1
2√𝑦
) [3 Marks]
5(b) Define the following (i) Joint moment about origin (ii) Joint central moment (iii) Correlation
coefficient
(i) Joint moment about Origin: [2 Marks]
(n+k)th moment about origin is defined as 𝑚𝑛𝑘 = 𝐸[𝑥𝑛
𝑦𝑘] =
∫ ∫ 𝑥𝑛
𝑦𝑘
𝑓𝑥𝑦
(𝑥,𝑦)
𝑑𝑦 𝑑𝑥
∞
𝑦=−∞
∞
𝑥=−∞
0th
moment, 1st
moment, 2nd
moment about origin by changing n,k values as sum of n+k
(ii) Joint central moment: [2Marks]
(n+k)th moment about origin is defined as 𝜇𝑛𝑘 = 𝐸[(𝑥 − 𝑥̅)𝑛
(𝑦 − 𝑦
̅)𝑘] =
∫ ∫ (𝑥 − 𝑥̅)𝑛
(𝑦 − 𝑦
̅)𝑘
𝑓
𝑥𝑦
(𝑥,𝑦)
𝑑𝑦 𝑑𝑥
∞
𝑦=−∞
∞
𝑥=−∞
0th
moment, 1st
moment, 2nd
moment about mean by changing n,k values as sum of n+k
(iii) Correlation coefficient: Normalized second order moment is named Correlation
coefficient. [3 Marks]
𝝆 =
𝜇11
√𝜇𝟐𝟎𝜇𝟎𝟐
= 𝑬[
(𝑥 − 𝑥̅)1
(𝑦 − 𝑦
̅)1
𝜎𝑥𝜎𝑦
𝒄𝒐𝒓𝒓𝒆𝒍𝒂𝒕𝒊𝒐𝒏 𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒊𝒔 𝒊𝒏 𝒕𝒉𝒆 𝒓𝒂𝒏𝒈𝒆 𝒐𝒇 − 𝟏 𝒕𝒐 + 𝟏
6(a) Define joint characteristic function, how the joint moments are calculated from the joint
characteristic function [7 Marks]

5. 6(b) Joint PDF of X,Y is given as 𝒇𝒙𝒚(𝒙, 𝒚) = {
𝟒𝒙𝟐
, 𝟎 < 𝒙, 𝒚 < 𝟏
𝟎, 𝑬𝒍𝒔𝒆𝒘𝒉𝒆𝒓𝒆
find (i) E[xy] and (ii) 𝒇𝒚(𝒚)
(i) E[xy]
(ii) 𝒇𝒚(𝒚)

6. 7(a) A random process 𝑿(𝒕) = 𝑨 𝒄𝒐𝒔(𝒘𝒐𝒕 + 𝜽)𝑨𝒂𝒏𝒅 𝑾 𝒂𝒓𝒆 𝒓𝒆𝒂𝒍 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒔,
𝒂𝒏𝒅 𝜽 𝒊𝒔 𝒖𝒏𝒊𝒇𝒐𝒓𝒎𝒍𝒚 𝒅𝒊𝒔𝒕𝒓𝒊𝒃𝒖𝒕𝒆𝒅 𝒐𝒗𝒆𝒓 (−𝝅, 𝝅)𝒗𝒆𝒓𝒊𝒇𝒚 𝒕𝒉𝒂𝒕 𝑿(𝒕)𝒊𝒔 𝑾𝑺𝑺?

7. 7(b) if X(t) is WSS verify |𝑹𝒙𝒙(𝝉)| ≤ 𝑬[𝒙𝟐(𝒕)]|
|𝑹𝒙𝒙(𝝉)| ≤ 𝑬[𝒙𝟐(𝒕)]|
8(a) Prove that |𝑹𝒙𝒚(𝝉)| ≤
𝟏
𝟐
[𝑹𝒙𝒙(𝟎) + 𝑹𝒚𝒚(𝟎)]

8. 8(b) Derive the relation between Auto correlation function and auto covariance function of a
random processes.

9. 9(a) A WSS white noise process of PSD
𝑵𝒐
𝟐
is input of a first order RC Low pass filter find the
output variance
9(b) List the properties
(i) Power spectrum density of a random process X(t)
(ii) Cross power spectrum density of random process X(t) and Y(t)

10. (i) Properties of Power spectrum of a random process X(t)
(iii) Cross power spectrum density of random process X(t) and Y(t)
10 Derive the relation between PSD of input and PSD of Output of an LTI system