week 91 example a device containing two key components fails when and only when both components...
TRANSCRIPT
week 9 1
Example
• A device containing two key components fails when and only when both components fail. The lifetime, T1 and T2, of these components are independent with a common density function given by
• The cost, X, of operating the device until failure is 2T1 + T2. Find the density function of X.
otherwise
tetf
t
T0
0
week 9 2
Convolution
• Suppose X, Y jointly distributed random variables. We want to find the probability / density function of Z=X+Y.
• Discrete case
X, Y have joint probability function pX,Y(x,y). Z = z whenever X = x and Y = z – x. So the probability that Z = z is the sum over all x of these jointprobabilities. That is
• If X, Y independent then
This is known as the convolution of pX(x) and pY(y).
x
YXZ xzxpzp .,,
x
YXZ xzpxpzp .
week 9 3
Example
• Suppose X~ Poisson(λ1) independent of Y~ Poisson(λ2). Find the
distribution of X+Y.
4
Convolution - Continuous case• Suppose X, Y random variables with joint density function fX,Y(x,y). We want to
find the density function of Z=X+Y.Can find distribution function of Z and differentiate. How?The Cdf of Z can be found as follows:
If is continuous at z then the density function of Z is given by
• If X, Y independent then
This is known as the convolution of fX(x) and fY(y).
z
v x
YX
x
z
v
YX
x
xz
y
YXZ
dxdvxvxf
dvdxxvxf
dydxyxfzYXPzF
.,
,
,
,
,
,
x
XY dxxvxf ,
x
XYZ dxxzxfzf ,
x
YXZ dxxzfxfzf
week 9 5
Example
• X, Y independent each having Exponential distribution with mean 1/λ. Find the density for W=X+Y.
week 9 6
Some Recalls on Normal Distribution
• If Z ~ N(0,1) the density of Z is
• If X = σZ + μ then X ~ N(μ, σ2) and the density of X is
• If X ~ N(μ, σ2) then
zezz
Z ,2
1 2
2
xexf
x
X ,2
1 2
2
2
2
.1,0~ NX
Z
week 9 7
More on Normal Distribution
• If X, Y independent standard normal random variables, find the density of W=X+Y.
week 9 8
In general,
• If X1, X2,…, Xn i.i.d N(0,1) then X1+ X2+…+ Xn ~ N(0,n).
• If , ,…, then
• If X1, X2,…, Xn i.i.d N(μ, σ2) then
Sn = X1+ X2+…+ Xn ~ N(nμ, nσ2) and
2111 ,~ NX 2
222 ,~ NX 2,~ nnn NX
.,~ 221121 nnn NXXX
.,~2
nN
n
SX n
n
9
Sum of Independent χ2(1) random variables
• Recall: The Chi-Square density with 1 degree of freedom is the
Gamma(½ , ½) density.
• If X1, X2 i.i.d with distribution χ2(1). Find the density of Y = X1+ X2.
• In general, if X1, X2,…, Xn ~ χ2(1) independent then
X1+ X2+…+ Xn ~ χ2(n) = Gamma(n/2, ½).
• Recall: The Chi-Square density with parameter n is
otherwise
xxenxf
nx
n
X
0
0
2
2
11
22
1
2/
week 9 10
Cauchy Distribution
• The standard Cauchy distribution can be expressed as the ration of two Standard Normal random variables.
• Suppose X, Y are independent Standard Normal random variables.
Let . Want to find the density of Z. X
YZ
week 9 11
Change-of-Variables for Double Integrals• Consider the transformation , u = f(x,y), v = g(x,y) and suppose we are
interested in evaluating .
• Why change variables?In calculus: - to simplify the integrand.
- to simplify the region of integration.In probability, want the density of a new random variable which is a function of other random variables.
• Example: Suppose we are interested in finding .
Further, suppose T is a transformation with T(x,y) = (f(x,y),g(x,y)) = (u,v). Then,
• Question: how to get fU,V(u,v) from fX,Y(x,y) ?
• In order to derive the change-of-variable formula for double integral, we need the formula which describe how areas are related under the transformation T: R2 R2 defined by u = f(x,y), v = g(x,y).
Dxy
xydAyxF ,
.,,A
YX dxdyyxfAP
.,,AT
VU dudvvufAP
week 9 12
Jacobian
• Definition: The Jacobian Matrix of the transformation T is given by
• The Jacobian of a transformation T is the determinant of the Jacobian matrix.
• In words: the Jacobian of a transformation T describes the extent to which T increases or decreases area.
yx
vu
y
g
x
gy
f
x
f
yxJT ,
,,
week 9 13
Change-of-Variable Theorem in 2-dimentions
• Let x = f(u,v) and y = g(u,v) be a 1-1 mapping of the region Auv onto Axy with f, g having continuous partials derivatives and det(J(u,v)) ≠ 0 on Auv. If F(x,y) is continuous on Axy then
where
AuvAxy
dudvvuJvugvufFdxdyyxF ,,,,,
yxJv
y
u
yv
x
u
x
vuJ,
1,
week 9 14
Example
• Evaluate where Axy is bounded by y = x, y = ex, xy = 2 and xy = 3.Axy
xydxdy
week 9 15
Change-of-Variable for Joint Distributions• Theorem
Let X and Y be jointly continuous random variables with joint density
function fX,Y(x,y) and let DXY = {(x,y): fX,Y(x,y) >0}. If the mapping T given
by T(x,y) = (u(x,y),v(x,y)) maps DXY onto DUV. Then U, V are jointlycontinuous random variable with joint density function given by
where J(u,v) is the Jacobian of T-1 given by
assuming derivatives exists and are continuous at all points in DUV .
otherwise
DvuifvuJvuyvuxfvuf VUYX
VU0
,,,,,, ,,
,
v
y
u
yv
x
u
x
vuJ
,
week 9 16
Example• Let X, Y have joint density function given by
Find the density function of
otherwise
yxifeyxf
yx
YX0
0,,,
.YX
XU
week 9 17
Example
• Show that the integral over the Standard Normal distribution is 1.
week 9 18
Density of Quotient
• Suppose X, Y are independent continuous random variables and we are interested in the density of
• Can define the following transformation .
• The inverse transformation is x = w, y = wz. The Jacobian of the inverse transformation is given by
• Apply 2-D change-of-variable theorem for densities to get
• The density for Z is then given by
.X
YZ
xwx
yz ,
wwz
z
y
w
yz
x
w
x
zwJ
01
,
wwzfwfwwzwfzwf YXYXZW ,, ,,
dwwwzfwfzf YXZ
week 9 19
Example
• Suppose X, Y are independent N(0,1). The density of is X
YZ
week 9 20
Example – F distribution
• Suppose X ~ χ2(n) independent of Y ~ χ2
(m). Find the density of
• This is the Density for a random variable with an F-distribution with parameters n and m (often called degrees of freedom). Z ~ F(n,m).
./
/
mY
nXZ
week 9 21
Example – t distribution
• Suppose Z ~ N(0,1) independent of X ~ χ2(n). Find the density of
• This is the Density for a random variable with a t-distribution with parameter n (often called degrees of freedom). T ~ t(n)
.
nX
ZT
week 9 22
Some Recalls on Beta Distribution
• If X has Beta(α,β) distribution where α > 0 and β > 0 are positive parameters the density function of X is
• If α = β = 1, then X ~ Uniform(0,1).
• If α = β = ½ , then the density of X is
• Depending on the values of α and β, density can look like:
• If X ~ Beta(α,β) then and
otherwise
xxxxf X
0
101 11
otherwise
xforxxxf X
0
101
1
XE
.12
XV
week 9 23
Derivation of Beta Distribution
• Let X1, X2 be independent χ2(1) random variables. We want the density of
• Can define the following transformation
21
1
XX
X
21221
11 , XXY
XX
XY