Lecture 7
Some Advanced Topics using Propagation of Errors and Least Squares Fitting
Error on the mean (review)
Question: If we have a set of measurements of the same quantity:
x1  1 x2   2 ... xn   n
u
What's the best way to combine these measurements?
u How to calculate the variance once we combine the measurements?
u Assuming Gaussian statistics, the Maximum Likelihood Method says combine the measurements as:
n

2
 xi /  i
x  i1n
weighted average
2
1/  i
l
i1
If all the variances are the same:
1 n
x   xi
unweighted average
n i1

u The variance of the weighted average can be calculated using propagation of errors:
u

n
2 2 n 1/  i4
1
2
2
   x  i  


1/


i
i
2
2
n
i1xi 
i1n
2 
2  i1
1/  i 
1/  i 
i1

i1

1
 2x  n
x=error in the weighted mean
2
1/  i
 2x
n  
i1
1
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u
If all the variances are the same:
 x2

1
n
2
1 /  i
i 1

2
n
The error in the mean (x) gets smaller as the number of measurements (n) increases.
n Don't confuse the error in the mean (x)with the standard deviation of the distribution ( )!
n If we make more measurements:
the standard deviation () of the (gaussian) distribution remains the same,
BUT the error in the mean (x) decreases
More on Least Squares Fit (LSQF)
l
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?
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:
y = a + bx
n Assume we have n measurements of the y's.
n For simplicity, assume that each y measurement has the same error, .
n Assume that x is known much more accurately than y.
ignore any uncertainty associated with x.
n Previously we showed that the solution for a and b is:
n
n
n
n
i1
n
i1
n
i1
2
 yi  xi   xi yi  xi
a  i1
n  xi2  (  xi )2
i1
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i1
and b 
n
n
n
i1
n
i1
n
i1
n  xi yi   xi  yi
n  xi2  (  xi )2
i1
i1
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2
n
Since a and b are functions of the measurements (yi's)
We can use the Propagation of Errors technique to estimate a and b.
 Q2
2
2
Q Q 
2 Q 
2 Q 
  x     y    2 xy   
See lecture 4
x 
x y 
y 
Assume that: a) Each measurement is independent of each other (xy=0).
b) We can neglect any error () associated with x.
c) Each y measurement has the same , i.e. i=.

2
2
2
2  Q 
2  Q 
Q   x    y 
assumption a)
 x 
 y 
 Q2
 Q 
  y2  
a 
2
assumption b)
 y 
2  a
  yi 
i 1
 yi
n
n
a

yi
2
2
n  a 

   2   
i 1 yi 

2
n
n
2
y x  x y x
 i 1 i j 1 j i 1 i i j 1 j
n
n
yi
n  xi2  (  xi ) 2
i 1
i 1
assumption c)
n
n

n
2
 x j  xi  x j
j 1
j 1
n
n
n  xi2  (  xi ) 2
i 1
i 1
2
 n x2  x n x 
 ( n x 2 )2  x 2 ( n x )2  2 x n x n x 2 



  j
j
i
j 
i  j
i  j  j 
n
n
j 1
j 1
j 1 j 1
   2   j 1

 a2   2   j n1
n
n
n
2
2

i 1
i 1
(n  xi2  (  xi ) 2 ) 2
 n  xi  (  xi ) 


i 1
i 1
i 1
 i 1



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 a2   2
n
n
n
n
n
j1
i1
j1
j1
j1
n(  x 2j )2   xi2 (  x j )2  2(  x j )2  x 2j
n
n
2
(n  xi  (  xi )2 )2
i1
i1
n
n
2
n  x j  (  x j )2
n
j1
2
   x 2j j1
n
n
j1 (n x 2  ( x )2 )2
 i
 i
i1
i1
n
 x 2j
 a2   2 n j1 n
variance in
n  xi2  (  xi )2
i1
i1
2
n
n
n
j1
i1
n
j1
n(  x 2j )2   xi2 (  x j )2
n
(n  xi2  (  xi )2 )2
i1
i1
the intercept
We can find the variance in the slope (b) using exactly the same procedure:

n

2
2
2
 nxi   x j 
n
n
n




b
b
j1

 b2    2y i     2      2   n
n
yi 
i1
i1yi 
i1n x 2  ( x )2 
 i 
  i
i1
 i1

n
2
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n
n
n
n
2
2
 x j  n(  x j )  2n  xi  x j
j1
j1
i1 j1
n
n
(n  xi2  (  xi )2 )2
i1
i1
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P416/Lecture 7
n 2
2
b 
n
n
n
n  xi2
i1
n
 (  xi )
variance in the slope
2
i1
If we don't know the true value of , we can estimate variance using the spread between the
measurements (yi’s) and the fitted values of y:

1 n
1 n
2
fit 2
2
 
 (yi  yi ) 
 (yi  a  bxi )
n  2 i1
n  2 i1
n  2 = number of degree of freedom
= number of data points  number of parameters (a,b) extracted from the data
If each yi measurement has a different error i:

1 n xi2
2
a   2
D i1 i
 b2 
weighted slope
and intercept
1 n 1

D i1 i2
n
D 
1
n

xi2
2
2
i1 i i1 i
n
 (
xi
2
)
2
i1 i
The above expressions simplify to the “equal variance” case.
Don't forget to keep track of the “n’s” when factoring out equal ’s. For example:

n

1
2
i1 i

n
2
not
1
2
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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 :
y  A  x /
y  A  x 2
y  Ae  x / 
However, 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, on a computer.
n Sometimes by a change of variables we can turn a non-linear problem into a linear one.
Example: take the natural log of both sides of the above exponential equation:
lny=lnA-x/=C+Dx
u
= -1/D
A linear problem in the parameters C and D!
In fact its just a straight line!
To measure the lifetime  (Lab 6) we first fit for D and then transform D into .
Example: Decay of a radioactive substance. Fit the following data to find N0 and :
N  N0et / 
n
n
n

N represents the amount of the substance present at time t.
N0 is the amount of the substance at the beginning of the experiment (t = 0).
 is the lifetime of the substance.
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Lifetime Data
i
1
2
3
4
5
6
7
8
9
10
ti
0
15
30
45
60
75
90
105
120
135
Ni
106
80
98
75
74
73
49
38
37
22
yi = ln Ni
4.663
4.382
4.585
4.317
4.304
4.290
3.892
3.638
3.611
3.091
D
n
n
n
i 1
n
i 1
n
i 1
n  xi yi   yi  xi
n  xi2  (  xi ) 2
i 1

10  2560.41  40.773  675
 0.01033
2
10  64125  (675)
i 1
  1 / D  96.80 seconds
n
The intercept is given by: C = 4.77 = ln A or A = 117.9
5
y = 4 .77 46 + -0 .01 03 31 x R= 0.93 51 8
Y(x)
N(t)
4 .5
4
3 .5
3
-20
0
20
40
60
time, t
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8 0 1 00 1 20 1 40
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Example: Find the values A and  taking into account the uncertainties in the data points.
n The uncertainty in the number of radioactive decays is governed by Poisson statistics.
n The number of counts Ni in a bin is assumed to be the average (m) of a Poisson distribution:
m = Ni = Variance
n The variance of yi(= ln Ni) can be calculated using propagation of errors:
2
2
2
 2y   N2 y /N   (N) ln N /N   (N)1/ N  1/ N
n The slope and intercept from a straight line fit that includes uncertainties in the data points:
n y n x2
n x y n x
n 1 n x y
n x n y
i
i 
i i
i
i i 
 2 2  2  2
 2  2  i2  i2








Taylor P198 and
a  i1 i i1 i 2 i1 i i1 i and b  i1 i i1 i 2 i1 i i1 i

n 1 n x
n x
n 1 n x
n x
problem 8.9
2
i (
i )2
i (
 2 2  2
 2  2  i2 )
u
i1 i i1 i
i1 i
i1 i i1 i
i1 i
If all the 's are the same then the above expressions are identical to the unweighted case.
a=4.725 and b = -0.00903
 = -1/b = -1/0.00903 = 110.7 sec

n
To calculate the error on the lifetime (), we first must calculate the error on b:
n 1
 2
652
i1 i
6
 b2 


1.0110
n 1 n x2
n x
652 2684700 (33240)2
i (
i )2
 2 2  2
i1 i i1 i
i1 i
2
2
1.005103

 12.3
3 2
(9.0310 )
 
    b  / b       b 1/ b
2
2
The experimentally determined lifetime is:  = 110.7 ± 12.3 sec.
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
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