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Solutions to various probability problems, including exact and chebyshev bounds, linear transformations, and markov bounds. It covers topics such as pmf, cdf, and pdf of random variables, as well as geometric progressions and integrals.
Typology: Exercises
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X m P X m P P Z Q σ
X m X m P X m k σ P X m k σ P X m k σ P k P k σ σ
= P Z [ > k (^) ] + P Z [ < − k (^) ] = Q k ( (^) ) + 1 − Q (^) ( − k (^) ) = Q k ( (^) ) + 1 − (^) ( 1 − Q k ( (^) )) = 2 Q k ( )
2
3
5
7
k
k
k
k
k
−
−
−
−
1
3
5
7
k
X m k P X m k P k P Z k Q k k
k
σ σ
−
−
−
−
x f (^) X x e
The quantizer is like the one in figure 3.15 page 120.
Problem 2: [GAR]3.
Problem 1: [GAR]3.
d x d
d d x d
d d x d
d d x
Y d x d
d
d x d
d d x d
d x d
So Y is a discrete r.v. with
Y
d d d d d d d d S
Because of symmetry
d d^ d^ x d P Y P Y e dx e P X d
− (^) α α − α
−∞
∫
{ } [^ ]
(^2 2 )
3
d (^) x d d
d
d d P Y P Y e dx e e P d X d
− (^) α α − α − α
−
∫
{ } [^ ]
2 2
d (^) x d d
d
d d P Y P Y e dx e e P d X d
− α α − α − α
−
∫
{ } [^ ]
x d d
d d P Y P Y e dx e P d X
α (^) α − α
−
∫
Y = aX + b is a linear transformation of X.
If X is Gaussian then Y is Gaussian too with
E Y [ (^) ] = aE X [ (^) ]+ b = am + b = m ′
[ ] [ ]
2 2 2 2 Var Y = a Var X = a α = α ′ a
α
α
Problem 3: [GAR]3.
a)
PMF:
( )
(^1 ) 1 1
k (^) x k k k P Y k PY k P k X k (^) k e dx e e e e
λ λ λ λ λ λ
( )
1
0 0 0 geometric progression
k k k k i i Y Y i i i
k e F k P k e e e e e e
λ λ λ λ λ λ λ
− + − − − − − − = = =
( 1 ) 1 1
k k e
− λ + = + ^ −
0
k Y k
f x e e x k
λ λ δ
∞ − −
=
0
k Y k
F x e e u x k
λ λ
∞ − −
=
b) Clearly 0 ≤ ε < 1 ε = X − X
0 means & 1 0 1
k
X k k X k X k
P ε δ P X X δ P X X δ X k P X k
∞
=
= ≤ < + ⇒ ≤ − <
by the theorem on total probability
0 from the pmf of in part a
k
k Y
P X k X k e e
λ λ δ
∞ − −
=
( ) [ ] ( ) ( )
( 1 )
0 0 0
k k k^ x k k k k k k k
e P k X k e e e e dx e e e e
λ λ λ λ δ^ λ λ λ λ λ δ λ
∞ ∞ (^) + ∞ − − − − − − − − − +
= = =
∑ ∑ (^) ∫ ∑
( ) ( ) ( ) ( )( )
2 2 2 0 0
k k
k k
e e e e e e e e e
λδ λ λ λδ λ λ λδ λ λ
∞ ∞ − − − − − − − − − = =
∑ ∑
a)
P Y [ ≤ y ] (^) = FY (^) ( y )
SY = [ 0,2)
P Y [ ≤ (^0) ] = 0
y
y
y y P Y y P X y P y X y dx −
∫
y 3 3
y P Y y P X y P y X dx y −
≤ ≤ ⇒ ≤ = ≤ = − ≤ ≤ = (^) ∫ = +
( Note that X ∈ −( 2,1 (^) ])
Y Y
y (^) y
y y (^) y F y f y y y (^) y
y
y^2
b)
Problem 7:
3 3 3 2 1 2 1 2
z z e dz e e
∞ ∞ (^) − − − = (^) ∫ = − =
z e dz e
− (^) −
−∞
= (^) ∫ = (Note that here z < 0 ⇒ z = − z )
( ) ( ( ))
3 2 (^1) 3 2 error 1 2 2
e P e p p
− − ⇒ = + − =
Please see the attached file.
1 Z F X U
[ ] ( ) ( )
( )
1 0
F X x P Z x P FX U x P U FX x dt FX x
− ≤ = ^ ≤ ^ = ≤ = = ∫
So: P Z [ ≤ x ] (^) = FZ (^) ( x (^) ) = FX (^) ( x )
Which means Z has the same CDF as X
Problem 9:
Problem 10: