Derivations

Question 1

derive var(x) = E(x^2) - [E(x)]^2

Answer 1

var(x) = E[(x-µ^2]

= E[x^2-2µx+µ^2]

= E(x^2) - 2µE(x) + µ^2

E(x^2) - 2µ^2 + µ^2

= E(x^2) - [E(x)]^2

Question 2

show that the inverse Fourier transform of a Dirac delta function aδ(0) is a constant.

Answer 2

definition of inverse Fourier transform

h(t) = aδ(0)

h(t) =(∞ ∫ -∞) aδ(0)exp[2πift]df

h(t) = a

Question 3

derive the natural logarithm for a normal distribution

Answer 3

L = (n Π i=1) p(x(i)|x)

insert p(x) for a normal distirbution

used the multiplicity of Π

L = [1/√2πσ^2]^n exp[-(x(1)-µ)^2 -(x(2)-µ)^2 -…-(x(n)-µ)^2/2σ^2]

becomes a sum index

then take the log l = ln(L)

gives -n/2 ln(2πσ^2) - (n Σ i = 1) (x(i)-µ)^2/2σ^2

Question 4

show that the maximum likelihood value is equal to the sample mean

Answer 4

l ∝ (x(1)^2-2x(1)µ +µ^2) - (x(2)^2-2x(2)µ +µ^2) -…-(x(n)^2-2x(n)µ +µ^2)

l ∝ -nµ^2 + 2µ(x(1)+x(2)+…+x(n)) - (x(1)^2+x(2)^2+x(3)^2)

take the derivative and set it equal to zero

x(hat) = 2(x(1)+x(2)+..+x(n)/2n

= sample mean

Question 5

derive cov(ax,by) = abcov(x,y)

Answer 5

E[(ax-aµ(x))(by-bµ(y))]

= E[ab(x-µ(x))(y-µ(y))]

ab cov(x,y)

Question 6

cov(X,Y) = E(XY) - µ(x) µ(y)

Answer 6

cov(X,Y) = E[(X-µ(x))(Y-µ(y))]

= E[XY]-Xµ(y) - Yµ(x) + µ(x) µ(y)]

= E[XY] - E[X] µ(y) - E[Y] µ(x) + E[µ(x) µ(y)]

= E[XY] - µ(x) µ(y) - µ(y) µ(x) + µ(x) µ(y)

hence E[XY] - µ(x) µ(y)

Question 7

cov(x,x+y) = var(x) + cov(x,y)

using cov(x,y) = E(xy) - µ(x)µ(y)

Answer 7

= E[x(x+y)] - E(x)E(x+y)

= E(x^2) + E(xy) - E(x)^2 - E(x)E(y)

= var(x) + cov(x,y)

Question 8

cov(x+y,x-y) = var(x) - var(y)

Answer 8

= E[(x+y)(x-y)] - E(x+y) E(x-y)

= E(x^2-y^2) - E(x)^2 + E(y)^2

= E(x^2) - E(x)^2 - [E(y^2) - E(y)^2]

= var(x) - var(y)

Question 9

p[ax,by] =

Answer 9

cov(ax, by)/[√var(ax)√var(bx)]

= ab cov(x,y)/[a√var(x) b√var(y)]

= p(x,y)

independent of scale

Question 10

verify the orthogonality properties of sine and cosine functions

Question 11

prove H(-f) = [H(f)]* if h(t) is real

Answer 11

H(-f) = (∞ ∫ -∞) h(t) exp[2πift] dt

= (∞ ∫ -∞) h(t) cos(2πft) dt + i (∞ ∫ -∞) h(t) sin(2πft) dt

= [H(f)]*

Question 12

prove H(-f) = -[H(f)]* if h(t) is imaginary

Answer 12

h(t) = i h(i)(t)

H(-f) = ∫ h(t) exp[2πift] dt

= ∫ h(t) cos(2πft) dt + i ∫ h(t) sin(2πft) dt

= i ∫ h(i)(t) cos(2πft) dt - ∫ h(i)(t) sin(2πft) dt

H(f)* = [i ∫ h(i)(t) cos(2πft) dt + ∫ h(i)(t) sin(2πft) dt]*

= - [i ∫ h(i)(t) cos(2πft) dt - ∫ h(i)(t) sin(2πft) dt]

H(-f) = -[H(f)]*

Question 13

if h(t) is even then H(f) is even

Answer 13

h(t) = h(-t)

H(-f) = (∞ ∫ -∞) h(t) exp[2πift] dt

= (∞ ∫ -∞) h(-y) exp[-2πify] dy

y = -t

= H(f)

Question 14

if h(t) is odd then H(f) is odd

Answer 14

H(-f) = (∞ ∫ -∞) h(t) exp[2πift] dt

= (∞ ∫ -∞) h(-y) exp[-2πify] dy

= -H(f)

Question 15

if h(t) is real and even then H(f) is real and even

Answer 15

H(-f) = [H(f)]*
H(-f) = [H(f)]

H(f)* = H(f)

Question 16

if h(t) is real and odd then H(f) is imaginary and odd

Answer 16

H(-f) = [H(f)]*

H(-f) = -H(f)

H(f) = -H(f)*

Question 17

if h(t) is imaginary and even, then H(f) is imaginary and even

Answer 17

H(-f) = H(f)

H(-f) = -[H(f)]*

H(f) = -H(f)*

Question 18

if h(t) is imaginary and odd then H(f) is real and odd

Answer 18

H(-f) = -H(f)

H(-f) = -H(f)*

H(f) = H(f)*

Question 19

fourier transform of h(at) = 1/|a| H(f/a)

Answer 19

Fourier transform

replace exp[-2πift] = exp[-2πi(f/a)at] and dt d(at)

for a > 0 gives as required for a < 0 gives as required

Question 20

fourier transform of h(t-t(0))

Answer 20

y = t - t(0)

(∞ ∫ -∞) h(y) exp[-2πify] exp[-2πift(0)] dy

= H(f) exp[-2πift(0)]

Question 21

fourier transform of h(t) exp[-2πif(0)t]

Answer 21

(∞ ∫ -∞) h(t) exp[-2πi(f+t(0)t] dt

= H(f+f(0))

Question 22

fourier transform of (g*h)(t)

Answer 22

(∞ ∫ -∞) (∞ ∫ -∞) g(s) h(y) ds exp[-2πif(y+s)] dy

y = t - s

(∞ ∫ -∞) g(s) exp[-2πifs] ds (∞ ∫ -∞) h(y) exp[-2πify] dy

= fourier g(t) fourier h(t)

Question 23

Fourier transform of corr(g,h)

Answer 23

(∞ ∫ -∞) (∞ ∫ -∞) g(s+t) h(s) ds exp[-2πift)] dt

(∞ ∫ -∞) g(y) h(s) ds exp[-2πify] exp[-2πifs] dy

(∞ ∫ -∞) g(y) exp[-2πify] dy (∞ ∫ -∞) h(s) exp[2πifs] ds

fourier g(t) [fourier h(t)]*

Question 24

if x and y are independent cov(x,y) = 0

Answer 24

cov(x,y) = ∫xyP(x,y)dxdy - ∫xP(x,y)dx ∫yP(x,y)dy

= ∫xP(x,y)dx ∫yP(x,y)dx - ∫xP(x,y)dx ∫yP(x,y)dy = 0

P(x,y) = P(x)P(y)

Question 25

show that the mean of the Poisson distribution is = λ

Answer 25

E(N) = (∞ Σ N=0) Np(N)

= (∞ Σ N=0) N [λ^N e^-λ]/N!

= 0 + (∞ Σ N=1) N [λ^N e^-λ]/N!

= λ (∞ Σ N=1) [λ^N-1 e^-λ]/(N-1)!

= λ (∞ Σ M=0) [λ^M e^-λ]/M!

= λ