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Error Propagation: Fitting - Lecture Notes | AST 5765, Study notes of Astronomy

University of Central Florida (UCF)Astronomy

Material Type: Notes; Class: ADV ASTRONOMICAL DATA ANALYSIS; Subject: Astronomy; University: University of Central Florida; Term: Fall 2009;

Typology: Study notes

Pre 2010

Uploaded on 02/25/2010

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UCF Physics: AST 5765/4762: (Advanced) Astronomical Data Analysis
Fall 2009 Lecture Notes: 10. Error Propagation; Fitting
1 Check In: 10:30 — 10:35:, 5 min
•Questions before we start?
2 Go over HW3: 10:35 — 10:55, 20 min
3 Error in Addition: 10:55 — 11:00, 5 min
•This is important enough to memorize: spending extra class time on it!
•For addition, f=u+v, and all the partial derivatives are 1, so:
σ2
f≃σ2
u+σ2
v...(+2σ2
uv...)(1)
•“Add in quadrature”
•What is it for subtraction? Same, since the -1 in the partial derivative squares to 1.
•What if it’s just f=u+a, with constant a? Same as if σv= 0:
σ2
f≃σ2
u(2)
σf≃σu(3)
4 Error in Multiplication: 11:00 — 11:05, 5 min
•For multiplication, f=uv:
σ2
f≃v2σ2
u+u2σ2
v...(+2uvσ2
uv...)(4)
since f=uv,
σ2
f
f2≃σ2
u
u2+σ2
v
v2...(+2σ2
uv
uv ...)(5)
•How about multiplication by a constant, f=au? Then v=aand σv= 0
σ2
f≃a2σ2
u(6)
σf≃aσu(7)
•How about division, f=au/v? Then v′=v−1, and there’s a -1 in front of ∂ f
∂v . It gets
squared to 1 in the straight term but not in the cross term:
σ2
f
f2≃σ2
u
u2+σ2
v
v2...(−2σ2
uv
uv ...)(8)
1
pf3
pf4

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Download Error Propagation: Fitting - Lecture Notes | AST 5765 and more Study notes Astronomy in PDF only on Docsity!

UCF Physics: AST 5765/4762: (Advanced) Astronomical Data Analysis

Fall 2009 Lecture Notes: 10. Error Propagation; Fitting

1 Check In: 10:30 — 10:35:, 5 min

  • Questions before we start?

2 Go over HW3: 10:35 — 10:55, 20 min

3 Error in Addition: 10:55 — 11:00, 5 min

  • This is important enough to memorize: spending extra class time on it!
  • For addition, f = u + v, and all the partial derivatives are 1, so: σ f^2 ≃ σ u^2 + σ^2 v ...(+2σ^2 uv...) (1)
  • “Add in quadrature”
  • What is it for subtraction? Same, since the -1 in the partial derivative squares to 1.
  • What if it’s just f = u + a, with constant a? Same as if σv = 0: σ f^2 ≃ σ u^2 (2) σf ≃ σu (3)

4 Error in Multiplication: 11:00 — 11:05, 5 min

  • For multiplication, f = uv: σ^2 f ≃ v^2 σ u^2 + u^2 σ v^2 ...(+2uvσ uv^2 ...) (4)

since f = uv, σ^2 f f 2

σ^2 u u^2

σ^2 v v^2

σ uv^2 uv

  • How about multiplication by a constant, f = au? Then v = a and σv = 0 σ^2 f ≃ a^2 σ^2 u (6) σf ≃ aσu (7)
  • How about division, f = au/v? Then v′^ = v−^1 , and there’s a -1 in front of ∂f ∂v. It gets squared to 1 in the straight term but not in the cross term: σ f^2 f 2

σ^2 u u^2

σ^2 v v^2

σ uv^2 uv

5 Error in Mean and Wrapup: 11:05 — 11:10, 5 min

  • How about the error in the mean of many measurements?

¯x =

N

∑ xi (9)

That’s a sum and a multiplication. If σxi = σx, then

σ^2 ¯x ≃

N

) 2 N σ^2 x =

σ^2 x N

σ¯x ≃

σx √ N

  • The error in the mean improves with

N.

  • NOTE: This is not the scatter in the measurements, it is the error in the mean estimated from the measurements.
  • Remember: errors add in quadrature, sometimes with a weighting.
  • It’s easiest to work with variances rather than uncertainties.
  • Recall that in the Poisson distribution, the variance is the value!
  • Chapters 3 and 4 show how to average according to weights calculated from uncertainties.
  • They also discuss pragmatic reasons for doing this vs. discarding old data, and the limitations to taking more data to improve σ through large N.

6 Computational Error Analysis (Monte Carlo): 11:10 — 11:15,

5 min

• MODERN LIFE IS EASIER:

  • Program the formula into a computer
  • Generate samples of fake measurements distributed according their uncertainties.
  • Evaluate the function for each draw in the sample.
  • Calculate the uncertainty.
  • Note: This gives an answer, but no information on which parameters it’s sensitive to.
  • To do that, vary each σi and do the above many times.
  • Look at how σf changes with changes in each σi.

∂a

χ^2 =

∂a

∑

[ 1 σi

(yi − a − bxi)

] 2 (15)

∑

[ 1 σ i^2

(yi − a − bxi)

] = 0 (16)

∂ ∂b

χ^2 =

∂b

∑

[ 1 σi

(yi − a − bxi)

] 2 (17)

∑

[ xi σ i^2

(yi − a − bxi)

] = 0 (18)

(19)

  • Rearrange:

∑ (^) yi σ^2 i

= a

σ^2 i

  • b

∑ (^) xi σ^2 i

∑ (^) xiyi σ i^2

= a

∑ (^) xi σ^2 i

  • b

∑ (^) x^2 i σ^2 i

OR: C = aD + bE (22) F = aE + bG (23)

  • Solve to get:

a =

(GC − EF ) (24)

b =

(DF − EC) (25)

∆ = DG − E^2 (26)