Derivations Flashcards
derive var(x) = E(x^2) - [E(x)]^2
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
show that the inverse Fourier transform of a Dirac delta function aδ(0) is a constant.
definition of inverse Fourier transform
h(t) = aδ(0)
h(t) =(∞ ∫ -∞) aδ(0)exp[2πift]df
h(t) = a
derive the natural logarithm for a normal distribution
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
show that the maximum likelihood value is equal to the sample mean
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
derive cov(ax,by) = abcov(x,y)
E[(ax-aµ(x))(by-bµ(y))]
= E[ab(x-µ(x))(y-µ(y))]
ab cov(x,y)
cov(X,Y) = E(XY) - µ(x) µ(y)
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)
cov(x,x+y) = var(x) + cov(x,y)
using cov(x,y) = E(xy) - µ(x)µ(y)
= 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)
cov(x+y,x-y) = var(x) - var(y)
= 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)
p[ax,by] =
cov(ax, by)/[√var(ax)√var(bx)]
= ab cov(x,y)/[a√var(x) b√var(y)]
= p(x,y)
independent of scale
verify the orthogonality properties of sine and cosine functions
prove H(-f) = [H(f)]* if h(t) is real
H(-f) = (∞ ∫ -∞) h(t) exp[2πift] dt
= (∞ ∫ -∞) h(t) cos(2πft) dt + i (∞ ∫ -∞) h(t) sin(2πft) dt
= [H(f)]*
prove H(-f) = -[H(f)]* if h(t) is imaginary
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)]*
if h(t) is even then H(f) is even
h(t) = h(-t)
H(-f) = (∞ ∫ -∞) h(t) exp[2πift] dt
= (∞ ∫ -∞) h(-y) exp[-2πify] dy
y = -t
= H(f)
if h(t) is odd then H(f) is odd
H(-f) = (∞ ∫ -∞) h(t) exp[2πift] dt
= (∞ ∫ -∞) h(-y) exp[-2πify] dy
= -H(f)
if h(t) is real and even then H(f) is real and even
H(-f) = [H(f)]*
H(-f) = [H(f)]
H(f)* = H(f)
if h(t) is real and odd then H(f) is imaginary and odd
H(-f) = [H(f)]*
H(-f) = -H(f)
H(f) = -H(f)*
if h(t) is imaginary and even, then H(f) is imaginary and even
H(-f) = H(f)
H(-f) = -[H(f)]*
H(f) = -H(f)*
if h(t) is imaginary and odd then H(f) is real and odd
H(-f) = -H(f)
H(-f) = -H(f)*
H(f) = H(f)*
fourier transform of h(at) = 1/|a| H(f/a)
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
fourier transform of h(t-t(0))
y = t - t(0)
(∞ ∫ -∞) h(y) exp[-2πify] exp[-2πift(0)] dy
= H(f) exp[-2πift(0)]
fourier transform of h(t) exp[-2πif(0)t]
(∞ ∫ -∞) h(t) exp[-2πi(f+t(0)t] dt
= H(f+f(0))
fourier transform of (g*h)(t)
(∞ ∫ -∞) (∞ ∫ -∞) 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)
Fourier transform of corr(g,h)
(∞ ∫ -∞) (∞ ∫ -∞) 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)]*
if x and y are independent cov(x,y) = 0
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)
show that the mean of the Poisson distribution is = λ
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
