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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
α (^) − α
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:
S X = [ 0, ∞ ⇒) SY ={0,1,2,3,... }
[ ] ( ) [ ]
( )
(^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
CDF:
( )
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 + = + ^ −
pdf of Y : ( ) ( ) ( )
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 )
− 2 < X ≤ 1 ⇒ 0 ≤ X < 2 ⇒
SY = [ 0,2)
P Y [ ≥ 2 ] = P X ≥ 2 = P X [ > 2 ] + P X [ < − 2 ] = 0 ⇒ P Y [ < 2 ] = 1
P Y [ ≤ (^0) ] = 0
If [ ] [ ]
y
y
y y P Y y P X y P y X y dx −
∫
If [ ] [ ] ( )
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)
Here for X < 0 ⇒ f X ( X )= 0 So Y = X = X
Problem 7:
P ( error ) = P ( error X = 0 ) P ( X = 0 ) + P ( error X = 1 ) P ( X = 1 )
= P ( error X = 0 ) p + P ( error X = 1 ) ( 1 − p )
P ( error X = 0 ) = P ( Y>1 2 X = 0 ) = P ( 0 + Z > 1 2) = P Z ( >1 2)
3 3 3 2 1 2 1 2
z z e dz e e
∞ ∞ (^) − − − = (^) ∫ = − =
P ( error X = 1 ) = P ( Y < 1 2 X = 1 ) = P ( 1 + Z < 1 2 ) = P Z ( < −1 2)
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:
