Signal and systems miscellaneous Easy Questions and Answers | Page - 15

Signal and systems miscellaneous

The value of C for which joint probability density function of random variables x and y as
f_XY(x, y) = C (1 – x – y)

1. 2
1. 4
1. 6
1. 8

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F_XY(x, y) = ^∞ ^∞ C(1 – x – y)dx dy = 1
_–∞ _–∞

1 = ¹ ^1-x C(1 – x – y)dx dy
₀ ₀

∵ Variation of y = 0 to 1 – x, and x = 0 to 1 1

1 = ¹ C (1 – x).y – y ² ^1-x dx
₀ 2 ₀

1 = C ¹ (1 – x)(1 – x) – (1 - x)² dx
₀ 2

1 = C ¹ (1 - x)² dx
₀ 2

1 = C x + x³ - 2 x² 1
2 3 2 ₀

1 = C 1 + 1 - 1
2 3

or C = 6

Correct Option: C

F_XY(x, y) = ^∞ ^∞ C(1 – x – y)dx dy = 1
_–∞ _–∞

1 = ¹ ^1-x C(1 – x – y)dx dy
₀ ₀

∵ Variation of y = 0 to 1 – x, and x = 0 to 1 1

1 = ¹ C (1 – x).y – y ² ^1-x dx
₀ 2 ₀

1 = C ¹ (1 – x)(1 – x) – (1 - x)² dx
₀ 2

1 = C ¹ (1 - x)² dx
₀ 2

1 = C x + x³ - 2 x² 1
2 3 2 ₀

1 = C 1 + 1 - 1
2 3

or C = 6

E [2x + 3] in —
f_XY(x, y) = X + 2Y ; X = 0, 1
14
0; otherwise

1. 15
  14
1. 25
  14
1. 35
  14
1. 85
  14

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_∞ _∞
X(z) = (2X + 3Y) f_XY(x, y)
^{X = – ∞} ^{Y = – ∞}

₁ ₂
X(z) = (2X + 3Y).(X + 2Y)
^{X = 0} ^{Y = 1} 14

= 85
14

Correct Option: D

_∞ _∞
X(z) = (2X + 3Y) f_XY(x, y)
^{X = – ∞} ^{Y = – ∞}

₁ ₂
X(z) = (2X + 3Y).(X + 2Y)
^{X = 0} ^{Y = 1} 14

= 85
14

E[Y] in
f_XY(x, y) = X + 2Y ; X = 0, 1
14
0; otherwise

1. 11
  7
1. 9
  14
1. 13
  14
1. 23
  14

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_∞ _∞
E[y] = y.f _XY(x, y)
^{X = – ∞} ^{Y = – ∞}

₁ ₂
E[y] = (X + 2Y)
^{X = 0} ^{Y = 1} 14

₂
= [Y(0 + 2Y) + Y(1 + 2Y)]
^{Y = 1} 14

or E(y) = 1.2.1.(1 + 2.1) + 2.2.2 + 2(1 + 2.2.)
14

E(y) = 2 + 3 + 8 + 10
14

= 23
14

Correct Option: D

_∞ _∞
E[y] = y.f _XY(x, y)
^{X = – ∞} ^{Y = – ∞}

₁ ₂
E[y] = (X + 2Y)
^{X = 0} ^{Y = 1} 14

₂
= [Y(0 + 2Y) + Y(1 + 2Y)]
^{Y = 1} 14

or E(y) = 1.2.1.(1 + 2.1) + 2.2.2 + 2(1 + 2.2.)
14

E(y) = 2 + 3 + 8 + 10
14

= 23
14

The joint probability function of two discrete random variable x and y given as
f_XY(x, y) = X + 2Y ; X = 0, 1
14
0; otherwise

then E [x]

1. 1
  7
1. 2
  7
1. 3
  7
1. 4
  7

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_∞ _∞
E[X] = X.f _XY(x, y)
^{X = – ∞} ^{Y = – ∞}

₁ ₂
= x. (X + 2Y)
^{X = 0} ^{Y = 1} 14

= [1.(1 + 2) + 1.(1 + 4)]
14

= 8
14

= 4
7

Correct Option: D

_∞ _∞
E[X] = X.f _XY(x, y)
^{X = – ∞} ^{Y = – ∞}

₁ ₂
= x. (X + 2Y)
^{X = 0} ^{Y = 1} 14

= [1.(1 + 2) + 1.(1 + 4)]
14

= 8
14

= 4
7

The mean of a random variable, whose pdf is
f_x(x) = x ; 0 < x < 4
4
0; otherwise

1. 1
  3
1. 2
  3
1. 5
  3
1. 16
  3

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Given f_x(x) = x ; 0 < x < 4
4
0; otherwise

Mean, u_x = ^∞ xf_x(x)dx
_{– ∞}

= ⁴ x· x dx
₀ 0

= ⁴ x² dx
₀ 4

= x³ ⁴
12 ₀

= 64
12

= 16
3

Hence, alternative (D) is the correct choice.

Correct Option: D

Given f_x(x) = x ; 0 < x < 4
4
0; otherwise

Mean, u_x = ^∞ xf_x(x)dx
_{– ∞}

= ⁴ x· x dx
₀ 0

= ⁴ x² dx
₀ 4

= x³ ⁴
12 ₀

= 64
12

= 16
3

Hence, alternative (D) is the correct choice.