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Sir Tanika Mukopadhyay taught us Probability at Homi Bhabha National Institute. He gave us assignments so that we can practice what we learned in form of problems. Here is solution to those problems. Its main emphasis is on following points: Analysis, Stochastic, Systems, Guassian, Power, Easily, Shown, leads, Consequently, Gaussian, Variables
Typology: Exercises
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Problem 1
3.51 P Y 3.5d P Y 3.5d e^ dx
3d x = = = - =
a a 2
= 1 -3^ α^ d 2 e
P Y 2.5d P Y 2.5d e 2
dx 3d
2d x = = = - =
a a
Similarly,
P Y 1.5d P Y -1.5d 1 e e 2
P Y 0.5d P Y 0.5d 1 2
P Y 4 d 2 e 2
dx e 4 d
x (^4) d > = =
a a^ a
2
Problem 2
3.53 When X is Gaussian
fi Y = a X + b is also Gaussian where:
m' = E Y = a E X + b = a m +b
b g^ s '^2 =^ a^2 s^2 fi^ a^ =^ s s
b = m' - ms s
Problem 3
3.56 Power = P =R X^2
F y P R X y
F y / R F y / R y 0
p^2
X X
b g
e (^) j e (^) j
fi = +
f y
f y / R 2R y / R
f y / R 2R y / R
f y / R f y / R 2 R y
p
x x
x x
b g
e j e j
e j e j
fi f y 1 2 R y
p 2 e
y 2^2 R b g =^
/ s
4
Problem 5
3.59 (a) For y £ 0 P Y £ y = 0
y > 0
( )
( ) ( )
X
X
Y X
P Y y P e y P X ln (y)
F ln (y)
0 y 0 F y F ln (y) y 0
fi " y > 0
f y d d y
F y
d d y
F ln y d d y
ln y
f ln y 1 y
f ln y y
Y Y
X
X
x
b g b g
d c b ghi c b gh
c b gh
c b gh
(b) For a Gaussian RV X,
f y
0 y 1 2 y
Y (^) e ln y^ m y 0 b g = c b g h^2 2
R S
|
T|^
p s
/ s
Problem 6
E X kP x k
k n
n n 1 2n
n 1 2
k 1
n
k 1
n
=
=
b g
5
sx^2 = E X^2 - (^) cE X h^2
E X k n
n 1 2n 1 6
2 k 1
n 2 = = +^ + =
b gb g
fi sx^2 = - n^2 12
Problem 7
3.67 (a) E X k k!
e k 1!
e k 0
k k 1
k 1 = = = -
l (^) l (^) l l l b g
= l el e-l = l
E X k k!
e k k 1!
e
j 1 j!
e where j k 1
1
2 2 k k 1
k 1
j 0
j
2
=
l (^) l l
l l
l l l l
l l
l
b g
b g
l q
fi s (^) x^2 = l^2 + l - l^2 = l
(b) E X = (^) barrival rate g b∑ time (^) g = lt
Problem 8
3.71 For non-negative random variables one can use the following formula for E X :
[ ] ( (^) X(^ )) 0
E X 1 F x dx
∞ = (^) ∫ −
7
c^ E Y h^2 =^ cE A h^2 Cos^2 b gwt^ +^ 2C Cos^ b gwt E A^ +C^2
fi s (^) Y^2 = Cos^2 b gwt E A^2 - cE A h^2
s (^) Y^2 = s (^) A^2 Cos^2 b gwt
Problem 11
3.76 g X b g =ba X
E g X ba k!
e b a k!
e
be e be
k 0
k k k 0
k
a a 1
b g
b g
b g
=
l (^) l l l
l l l
Problem 12
3.78 E Y[ ] g x f( ) (^) X (^) ( x (^) )dx
∞
− ∞
= (^) ∫
( ) (^ )^ (^ )^ ( ) (^ )
a (^) a X X X a a
x a f x dx 0 f x dx x a f x dx
− (^) ∞
− ∞ −
= (^) ∫ + + (^) ∫ + (^) ∫ −
[ ] ( ) ( ) (^) ( ) ( ( ))
a X X X X a
E Y x f a dx x f x dx aF a a 1 F a
− (^) ∞
− ∞
= (^) ∫ + (^) ∫ + − − −
For Laplacian RV, f x 2 X^ b g^ =^ a^ e-ax
Since fX (^) b gx is symmetric around zero,
a X X a
x f x dx x f x dx
− (^) ∞
− ∞
⇒ (^) ∫ = −∫
8
fi = - - -
= -^ - - =
E Y aF a a 1 F a a 2
e a 2
e 0
X X a a
a a
a 2 2 2 2 X X a
E Y x a f x dx x a f x dx
− (^) ∞
− ∞
a a 2 2 X X X X a a
x f x dx x f x dx 2 a x f x dx x f x dx
− (^) ∞ − ∞
− ∞ − ∞
For Laplacian, this leads to:
a 2 2 2 x 2 x a
E Y x e dx x e dx 2 2
− (^) ∞
σ = = α^ + α
using x 2 e x^ dx e^ x 2 x 2
x 3
a a 2 2 a
− ∞ ∞
x -a x 2 2 2 2 2 (^3) - 3 a
σ e^ x 2 x 2 e x 2 x 2 2
2 e^ a^2 a^2 2a^2 e^ a^2 a^2 2a^2 a
s a^ a a
2 2 2 a 2 = a^ +^ 2a^ +^2 e-a
