Topics Part 1 – Single measurement

1. Basic stuff (Chapter 1 and 2)2. Propagation of uncertainties (Chapter 3)

Part 2 – Multiple measurements as independent results

1. Mean and standard deviation (Chapter 4)2. Basic on probability distribution function (not in text explicitly)3. The Binomial distribution (Chapter 10)4. The Poisson distribution (Chapter 11)5. Normal distribution (first half of Chapter 5)6. χ2 test – how well does the data fit the distribution model? (Chapter 12)

Part 3 – Multiple measurements as one sample

1. Central limit theorem (not in text explicitly)2. Normal distribution (second half of Chapter 5)3. Propagation of error (Chapter 3)4. Rejection of data (Chapter 6)5. Merging two sets of data together (Chapter 7)

Part 4 - Dependent variablesPart 4 Dependent variables

1. Curve fitting (Chapter 8)2. Covariance and correlation (Chapter 9)

The ProblemSuppose two persons A and B measure the same physical quantity.  A measure the quantity m times and obtain the results xA1, xA2, …., xAm.  B measure the quantity n times and obtain the results xB1, xB2, …., xBm.

After calculating the mean and the SDOM, A reports the value as x A ±σAand B reports the value as x B ±σB.p B  B

In combining the results of A and B, there is no problem if we know all the data points xA1, xA2, …., xAm, xB1, xB2, …., xBm.  We just need to calculate the mean and SDOM of these (n+m) numbersmean and SDOM of these (n+m) numbers.

The problem is, what should we do if we do not have the details of the original data sets, but only the reported values of A and B (i.e. x A ±σA and 

)? H h ld h l i ?x B ±σB)?  How should we merge these two values into one?  

Following the notation of the textbook, we will denote the merged result as x wav ±σwav .  “wav” stands for “weighed average”. 

A particular case: σA = σBA particular case: σA = σB

Since x and x have the same accuracy it is natural to expect the resultant valueSince x A and x B have the same accuracy, it is natural to expect the resultant value to be at the mid point between x A and x B:

xxx BAwav

+=

Since we have two numbers (x A and x B), so the SDOM should be improved by a factor of 1/√2:

2wav

2or

2 BA

wavσσσ =

Example for the particular case of σA = σB

Suppose two students measure the value of g and reported the result asSuppose two students measure the value of g and reported the result as gA= (9.7 ± 0.2) m/s2 and  gB= (10.0 ± 0.2) m/s2.

01079 2BAwav m/s 9.85

20.107.9

2gg g =

+=

+=∴

2A m/s100.2σσ

2

Awav

/)1099(

m/s1.0~22

==σ

2wav m/s)1.09.9(g ±=∴

General case: σA ≠ σBGeneral case: σA ≠ σB

The situation is more complicated if σ ≠ σ but we can still know some simpleThe situation is more complicated if σA ≠ σB, but we can still know some simple facts.  Let us assume σA < σB:

1 Since σA < σB xA is more accurate than xB The resultant value x should1. Since σA < σB, xA is more accurate than xB.  The resultant value xwav should be closer to xA than xB.

2. The SDOM of the resultant value should be smaller than both σA and σB :A Bσwav < σA and σwav < σB.                                  

General case: σA ≠ σBGeneral case: σA ≠ σBUsing Central Limit Theorem, the probability for A to obtain the particular value of x is: 2X)(value of xA is:

and the probability for B to obtain the particular value of xB is:

2A

2A

2X)(x-

AA e1 )P(x σ

σ

−

∝

and the probability for B to obtain the particular value of xB is:

2B

2B

2X)(x-

BB e1 )P(x σ

σ

−

∝

The probability for A to obtain xA and B to obtain xB at the same time:

)P()P()P(

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −+

−

∝

=

2B

2B

2A

2A

2X)(x

2X)(x-

BABA

e1

)P(x)P(x)x,P(x

σσ

σσ BAσσ

Principle of Maximum LikelihoodMaximizing P(xA, xB) with respect to X:

2B

2A

ABA 0X)(xX)(x 0 ) x,P(x

x σσ=

−+

−⇒=

∂∂

2B

B2

A

A2

B2

A

BA

xx11 X

x

σσσσ

σσ

+=⎟⎟⎠

⎞⎜⎜⎝

⎛+⇒

∂

2B

B2

A

A

11

xx

X σσ+

=⇒

wav

2B

2A

: xas X of estimatebest thiscall willWe

11σσ

+

2B

B2

A

A

wav 11

xx

x σσ

+

+=∴

2B

2A σσ

+

Uncertainty in Merged DataC id i f ti f d i ti fConsidering xwav as a function of xA and xB, using propagation of error::

2

Bwav

2

Awav2

2B

B2

A

A

xx11

xx

x ⎟⎟⎞

⎜⎜⎛ ∂

+⎟⎟⎞

⎜⎜⎛ ∂

=+

= σσσσσ

2

2B

2

2A

BB

AA

wav

2B

2A

wav

11

xx 11x

⎟⎟⎟⎞

⎜⎜⎜⎛

+⎟⎟⎟⎞

⎜⎜⎜⎛

=

⎟⎟⎠

⎜⎜⎝ ∂

+⎟⎟⎠

⎜⎜⎝ ∂+

σσσσ

σσσ

σσ

2

B

2

A

B

2B

2A

A

2B

2A

11

1111

⎟⎟⎞

⎜⎜⎛

+⎟⎟⎞

⎜⎜⎛

⎟⎟

⎠⎜⎜

⎝+

+⎟⎟

⎠⎜⎜

⎝+

=

σσ

σ

σσ

σ

σσ

22

2B

2A

B

2B

2A

A

11

1111

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎟⎟⎟

⎠⎜⎜⎜

⎝+

+⎟⎟⎟

⎠⎜⎜⎜

⎝+

=

σσ

σσσσ

2B

B2

A

A2

B

B2

A

A

2B

2A

2wav

1

xx

x11

xx

But x

11 1

σσσσ

σσσ

+=⇒

+=

+=∴

2

2

2B

2A

BA

1

11

⎞⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎠⎝⎠⎝=

σσ

σσ

2B

B2

A

A2wav

wav

2wav

wav

2B

2A

wav

xx x

1 x 11But x

σσσ

σσσ

+=∴

⇒+

2

2B

2A

11⎟⎟⎠

⎞⎜⎜⎝

⎛+σσ

Merging data together ‐ summaryMerging data together  summaryIf you have two results, x A ±σA and x B ±σB, you can combine these two results into one x ±σ :one x wav ±σwav :

111+= G t fi t2

B2

A2wav σσσ

+= Get σwav first

2B

B2

A

A2wav

wav xx xσσσ

+= Then use σwav to calculate x wav

I find these equations quite easy to remember, but the author has another way to do this.

Textbook’s Notation (Weighted Average)If you have two results, x A ±σA and x B ±σB, you can combine these two results into one x wav ±σwav :

1 wDefine 2=

1

w w w

wav

BAwav

2

=⇒

+=∴

σ

σ

xxx

ww

2B

2Awav

BA

+=

+

xwxwx

xwx w x w

BBAA

BBAAwavwav

2B

2A

2wav

+=⇒

+=⇒

+σσσ

)average" weighted(" ww

xwxw x

w x

BA

BBAAwav

wavwav

++

=⇒

=⇒

BA

The author said this form is easier to remember.

Generalization of ResultsGeneralization of ResultsNow assume you have many results x 1 ±σ1, x 2 ±σ2, …, x i ±σi, …. And you want to  combine these results into one x ±σ :combine these results into one x wav ±σwav :

22222

1 111 1 ∑=+++= LL 2ii

2i

22

21

2wav

xxxxx

σσσσσ ∑

2i

i

i2

i

i2

2

22

1

12wav

wav x x xx xσσσσσ ∑=+++= LL

Generalization of Results in Textbook’s Notation

Now assume you have many results x 1 ±σ1, x 2 ±σ2, …, x i ±σi, …. And you want to  combine these results into one x wav ±σwav :

w ww w w

1 wDefine

ii21wav

2

∑=+++=∴

=

LL

σ

w1 or

ii

wav

i

∑

∑

=σ

www

xwxwx w x ii2211wav

i

++++++

=LL

)average" weighted(" w

xw

www

i

iii

i21

∑∑

=

+++ LL

ii∑

ExampleThe particle Data Tables list the following four experimental measurements of the mean lifetime of the Ks meson with their uncertainties, in units of 10‐10s.  Find the weighted mean of the data and the uncertainty in the mean.

( )004408920.0

001608929.0

001408941.0

002108971.0/x

2222

2ii

i+++∑ σ

0.8971±0.0021  0.8941±0.0014   0.8929±0.0016   0.8920±0.0044

( )

0.89419 11792391054461

0044.01

0016.01

0014.01

0021.01

0044.00016.00014.00021.0/1

2222

2i

i

i

==

+++==

∑ σμ

1179239

( )00440

100160

100140

100210

11

/11

2222

2i

i

2

+++==

∑μ σσ

0.00092 108.48

108.48 1179239

1

0044.00016.00014.00021.0

7-

7-

i

=×=∴

×==

μσ

∴ The lifetime of Ks meson is (0.8942 ± 0.0009) × 10‐10 s