†
l
†
†
Question: If we have a set of measurements of the same quantity: u x
1
± s
1 x
2
± s
2
... x n
± s n
What's the best way to combine these measurements?
u
How to calculate the variance once we combine the measurements?
u u u
Assuming Gaussian statistics, the Maximum Likelihood Methods combine the measurements as: n
 x i
/ s i
2 x = i = 1 n
 1/ weighted average s i
2 i = 1
If all the variances ( ) are the same: x =
The variance of the weighted average can be calculated using propagation of errors: s
2 x
1 n
= n
 i = 1 x i n
 i = 1
È
Í
Î
∂
∂ x i x
˘
˙
˚
2 s i
2 unweighted average
= n 1/
 i = 1
È
Î n
 i = 1
1/ s s i
4 i
2
˘
˚
2 s i
2
=
È
Í n
Â
Î i = 1
1/
1 s i
2
˘
˚
2 n
 i = 1
1/ s i
2 s
2 x
=
1 n
 1/ s i = 1 i
2 s x
is the error in the weighted mean
K.K. Gan L7: Some Advanced Topics 1
†

† u
If all the variances are the same:
+ s
2 x
= n
 s i
2
= 1/[ n / s i = 1
The error in the mean ( s x
2
] = s
2
Lecture 4 n
) gets smaller as the number of measurements ( n ) increases.
n
Don't confuse the error in the mean ( s x
) with the standard deviation of the distribution ( s )!
n
If we make more measurements
+
+ the standard deviation ( s ) of the distribution remains the same the error in the mean ( s x
) decreases
l
In Lec 5, we discussed how we can fit our data points to a linear function (straight line) and get the
"best" estimate of the slope and intercept. However, we did not discuss two important issues: u
How to estimate the uncertainties on our slope and intercept obtained from a LSQF?
l u
How to apply the LSQF when we have a non-linear function?
Estimation of Errors from a LSQF u
Assume we have data points that lie on a straight line: n y = a + b x
Assume we have n measurements of y 's.
n
For simplicity, assume that each y measurement has the same error s .
n
Assume that x
+ is known much more accurately than ignore any uncertainty associated with x .
y.
n
Previously we showed that the solution for a and b is: a = n
 i = 1 y i n
 n
 x i i = 1 n x n x i
2 i
2
( x y i n
 i = 1 n
 i
) i = 1
2 x i
and b = n
 i = 1 n i y i
n
 i = 1 n x i n x i
2
( x n
 i = 1
 i
)
2 y i
K.K. Gan i = 1 i = 1 i = 1 i = 1
L7: Some Advanced Topics 2
†

† n
Since a and b are functions of the measurements ( y i
's)
+
H use the Propagation of Errors technique to estimate s s
2
Q
= s
2 x
Ê
Á
Ë
∂ Q
∂ x
ˆ
˜
¯
2
+ s
2 y
Ê
Á
Ë
∂ Q
∂ y
ˆ
˜
¯
2
+ 2 s xy
Ê
Á
Ë
∂ Q
∂ x
ˆ
˜
¯
Ê
Á
Ë
∂ Q
∂ y
ˆ
˜
¯ a
and s b
.
Assumed that each measurement is independent of each other: s
2
Q
= s
2 x
Ê
Á
∂ Q
Ë ∂ x
ˆ
˜
¯
2
+ s
2 y
Ê ∂ Q
Á
Ë ∂ y
ˆ
˜
¯
2 s
2 a
= n
 s i = 1
2 y i
Ê
∂a
Á
Ë ∂ y i
ˆ
˜
¯
2
= s
2 n
 i = 1
Ê
∂a
Á
Ë ∂ y i
ˆ
˜
¯
2
∂a
∂ y i s
2 a
=
∂
∂ y i n
 i = 1 y i n
 j = 1 x
2 j
n
 i = 1 x i y i n
 i
2
( n
 x i
)
2 i = 1 i = 1 n
 j = 1 x j
= s
2 n
 i = 1
Ê
Á
Á
Á n
 j = 1 x
2 j
x i n
 j = 1 x
Á
Ë n
 i = 1 i
2
( n
 i = 1 x i
) j
2
˜
˜
˜
¯
ˆ
˜
2
= n
 x j
2
x i j = 1 n
 j = 1 x j n
 i
2
( n
 x i
)
2 i = 1 i = 1
= s
2 n
 i = 1
Á
Á
Ê
Á
( n
 j = 1 x j
2
)
2
Á
Ë
+ x i
2
( n
 x j
)
2 j = 1
2 x i
( n n x i = 1 i
2
( n
 ) i = 1 x i
2
)
2 n
 j = 1 x j n
 j = 1 x j
2
ˆ
˜
˜
˜
˜
¯
†
K.K. Gan L7: Some Advanced Topics 3

†
H s
2 a
= s
2
= s
2 n ( n
Â
2 j
)
2 j = 1 x + n
 i = 1 x i
2
( n
 j = 1 x j
)
2
2( n
 j = 1 x j
)
2
( n
 i
2
( n
 x i
) i = 1 i = 1
2
)
2 n
 j = 1 x j
2 n
 j = 1 x j
2 n n
 j = 1 x
2 j
( n
 j = 1 x j
)
2
( n
 i = 1 i
2
( n
 i = 1 x i
)
2
)
2
= s
2 n ( n
Â
2 j
)
2 j = 1 x n
 i = 1 x i
2
( n
 j
) j = 1 x
2
( n
 i
2
( n
 x i
)
2 i = 1 i = 1
)
2 n
 j = 1 x
2 j s
2 a
= s
2 n
 i
2
( n
 x i
)
2 variance in the intercept i = 1 i = 1
We can find the variance in the slope ( b ) using exactly the same procedure: s
2 b
= n
 i = 1 s
2 y i
Ê
∂b
Á
Ë ∂ y i
ˆ
˜
¯
2
= s
2 n
 i = 1
Ê
∂b
Á
Ë ∂ y i
ˆ
˜
¯
2
= s
2 n
 i = 1
Ê
Á
Á
Á
Ë
∂
∂ y i n n x i = 1 n i = 1 i n x y i
2 i
-
( n
 i = 1 n
 i = 1 x i x i n
 i = 1
)
2 y i
ˆ
˜
2
˜
˜
¯
= s
2 n
 i = 1
Á
Á
Ë
Ê
Á
Á n
 nx i = 1 i i
2
-
( n
 j = 1 x j n
 ) i = 1 x i
2
ˆ
˜
˜
2
˜
˜
¯
K.K. Gan
= s
2 n
2 n
 x j
2
+ n ( n
 x j
)
2 j = 1
2 n
 i n
 x j n
2 n
 x j
2
n ( n
 x
( n
 i = 1 j = 1 i
2
( n
 x )
2 i = 1
)
2 j = 1
= s
2 i i = 1
L7: Some Advanced Topics j = 1 j = 1
( n
 i = 1 i
2
( n
 i = 1 x i
)
2 j
)
)
2
2
4
†

s
2 b
= n s
2 n
 i
2
( n
 x i
)
2 i = 1 i = 1 variance in the slope n
† n
†
†
If we don't know the true value of s,
+ estimate variance using the spread between the measurements ( y i
H s
2
ª
1 n 2 n
 i = 1
( y i
y i fit
)
2
=
1 n 2 n
 i = 1
( y i n 2 = number of degree of freedom
a b x i
)
2
’s) and the fitted values of y :
= number of data points number of parameters ( a, b ) extracted from the data
If each y i s
2 a
measurement has a different error s i
=
1
D n x
 i = 1 s i
2 i
2
: s
2 b
=
1
D n
 i = 1 s
1 i
2 weighted slope and intercept
H
D = n
 i = 1 s
1 i
2 n x
 i = 1 s i
2 i
2
( n
 i = 1 s x i i
2
)
2
The above expressions simplify to the “equal variance” case.
r
Don't forget to keep track of the “ n ’s” when factoring out s . For example: n
 i = 1 s
1 i
2
= s n
2 not s
1
2
L7: Some Advanced Topics 5

l
LSQF with non-linear functions: u
For our purposes, a non-linear function is a function where one or more of the parameters that we are trying to determine (e.g. a , b from the straight line fit) is raised to a power other than 1.
n
Example: functions that are non-linear in the parameter t : y = A + x / t y = A + x t
2
† y = Ae
x / t
H
These functions are linear in the parameters A .
u
The problem with most non-linear functions is that we cannot write down a solution for
† the parameters in a closed form using, for example, the techniques of linear algebra (i.e. matrices).
n
Usually non-linear problems are solved numerically using a computer.
n
Sometimes by a change of variable(s) we can turn a non-linear problem into a linear one.
H
Example: take the natural log of both sides of the above exponential equation: n n r ln y = ln A x / t = C Dx
A linear problem in the parameters C and D !
r
In fact its just a straight line!
u
†
Example: Decay of a radioactive substance. Fit the following data to find N
0
and t : n
N
+
= N
To measure the lifetime t (Lab 6) we first fit for D and then transform D into t .
0 e
t / t
N represents the amount of the substance present at time t.
N
0
is the amount of the substance at the beginning of the experiment ( t is the lifetime of the substance.
t = 0).
K.K. Gan L7: Some Advanced Topics 6

† i 1 2 3 4 5 6 7 8 9 t i
N i
0
106
15
80
30
98
45
75
60
74
75
73
90
49
105
38
120
37 n y i
= ln N i
4.663
4.382
4.585
D = b = n n x i = 1 n i = 1 i y i i
2
n
 y i n
 i = 1 n
( x i i = 1 i = 1
 )
2 x i
4.317
4.304
4.290
3.892
= -
10 ¥ 2560.41
40.773
¥ 675
10 ¥ 64125 (675)
2
3.638
= 0.01033
3.611
t = 1/ D = 96.80 sec
The intercept is given by: C = 4.77 = ln A or A = 117.9
y = 4.7746 + -0.010331x R= 0.93518
5
10
135
22
3.091
4.5
4
3.5
K.K. Gan
3
-20 0 20 40 60 80 100 120 140 x t
L7: Some Advanced Topics 7

†
†
† u
Example : Find the values A and t taking into account the uncertainties in the data points.
n n n n
The uncertainty in the number of radioactive decays is governed by Poisson statistics.
The number of counts N i m = N i
= Variance
in a bin is assumed to be the average ( m ) of a Poisson distribution:
The variance of y s
2 y a =
= n s
 i = 1 s n
2
N y i i
2
 i = 1 s
(
1 i
2 n
 i = 1 i
∂ y / ∂ N x s
(= ln N i n
 i = 1 x s i
2 i
2
i
2 i
2
)
2
n
Â
) can be calculated using propagation of errors:
= ( N ) i = 1
( x s n i
 i = 1 y i
2 i x s
( i n
∂ ln x
 i = 1 s i
2
)
2 i i
2
N / ∂ N
and
)
2 b =
= ( N ( N )
2
= 1/ N
The slope and intercept from a straight line fit that includes uncertainties in the data points: n
 i = 1 s n
1 i
2
 i = 1 s
1 n
 i = 1 i
2 n x s
 i = 1 i y i
2 x s i i
2 i
2
-
n
 i = 1 s
( n
 x i = 1 i i
2 s x i
 i = 1 s i
2 n
)
2 y i i
2
Taylor P. 198 and Problem 8.9
n
H
If all the s 's are the same then the above expressions are identical to the unweighted case.
a = 4.725 and b = 0.00903
t = -1/ b = 1/0.00903 = 110.7 sec
To calculate the error on the lifetime, we first must calculate the error on n 1 b : s
2 b
= n
 i = 1 s
1 i
2 n
 i = 1 s
 i = 1 s x i
2 i
2
i
2
( n
 i = 1 s x i i
2
)
2
=
652
652 ¥ 2684700 (33240)
2
= 1.01
¥ 10
6 s
2 t
= s
2 b
( ∂t / ∂b )
2
= s b
(
1/ b
2 )
=
1.005
¥ 10
3 fi s t
(9.03
¥ 10
3
)
2
+
The experimentally determined lifetime is t = 110.7 ± 12.3 sec.
K.K. Gan L7: Some Advanced Topics
= 12.3
8
†