Transcript Error Propagation

Error Propagation
Uncertainty
• Uncertainty reflects the
knowledge that a measured
value is related to the mean.
• Probable error is the range from
the mean with a 50% certainty.
– For a normal distribution:
 0.675
Coefficient of Variation
• The coefficient of variation (CV)
is the ratio of the standard
deviation to the mean.
– Fractional standard deviation
cv 


• The CV for the Poisson
distribution is fixed by the mean.


1
cv  




• For Poisson or normal
distributions the measured count
n has a fixed estimate for .
– Significance is 0.683


1
  n n
n 1 





Multiple Uncertainties
• Measurements often involve
multiple variables.
• First order expansion:
Q
Q
( xi  x ) 
( yi  y )
x
y
1
1
Q   Qi  NQ( x , y )  Q( x , y )
N
N
Qi  Q( x , y ) 
• Assume two independent
variables x, y.
– N measurements of xi, yi
– Assume small differences
from means.
• The cross term (covariance)
vanishes for large N.
• The variance in Q is:
1
 Q2 
N
 Q

Q
(
x

x
)

(
y

y
)
  x i
i

y


2
 Q 
 Q 
 Q2     x2     x2
 x 
 y 
2
2
Count Rate Error
• Counting experiments are
expected to have Poisson
statistics.
– Assume precise time
measurement
• Gross count rate includes all
counts in a time interval.
– Number of counts ng
– Time tg
n
Pn 
n!
e
 g   g  ng
rg 
ng
tg
 gr 
g
tg

ng
tg

rg
tg
Significant Difference
Typical Problem
• Two technicians measure a
sample with a 35% efficiency
counter, Julie gets 19 counts in
1 min and Phil gets 1148 counts
in 60 min. What is the
confidence that the activity is
the stated value of 42.0 min-1?
Answer
• From the given value, the
expected count rate rg = 14.7
min-1.
• The difference in Julie’s rate from
expected is 4.3 cpm.
– By chance 27% of time
rJ  
J

nJ  t
4.3

 1.12
t
14.7
• The difference in Phil’s rate from
expected is 4.4 cpm.
– By chance 10-19 of time
rP  
P

nP  t 1148  882

 8.96
t
882
Net Count Rates
• Experiments typically have a
background counting rate.
– Net count rate subtracts
background
• This involves two variables
with errors.
– Use standard deviation of
gross and background rates
ng
nb
rn  rg  rb 

t g tb
 nr 
 nr 
 g2
t g2
ng

 b2
tb2
  gr2   br2
rg rb
nb
 2 

2
t g tb
t g tb
Background Subtraction
Typical Problem
• A counter with 28% efficiency
measures a background of 2561
counts in 90 min. A 10 min
measure of a sample is gets
1426 counts. How long must a
sample be measured to get the
activity within 5% with 95%
confidence?
Answer
• The activity will have the same
significance as the count rate.
– Scales with efficiency
• Find the count rates.
– rg = 142.6 min-1
– rb = 28.5 min-1
– rn = 114 min-1
– 0.05 rn = 5.7 min-1
• 95% is 1.96, nr = 2.91 min-1.
 nr 
tg 
rg
tg

rb
tb
rg
 nr2 
rb
tb
 17.5 min
Optimum Counting
• The time to measure can be
optimized.
– Based on background and
gross rates.
• A fixed total time can be
partitioned into optimum
segments.
– Variance can be used
instead of standard
deviation
T  t g  tb
r
d 2
d  rg
 nr 
 b
dt g
dt g  t g T  t g
rg

0


rg rb
rb
 2
 2  2 0
2
t g T  t g 
t g tb
tg
tb

rg
rb
Short Lives
• A rapidly decaying does not
obey Poisson statistics.
– Still Bernoulli process
– Use lt >> 1
– For large t,   0
• The equations can be adapted
for a fixed efficiency.
– At long times  limited by
efficiency
  Np  N 1  e lt 
  Npq  N 1  e lt e lt
  e lt
 c  N 1  e  lt 
 c  N 1  e lt 1    e lt 
 ct   N 1     c 1   
False Measurements
• Many measurements only seek to determine if any activity
is present above background.
• The minimum significant measurement determines the
conditions to assert that there is activity above background.
– Failure is false positive; type I error.
• The minimum detectable measurement determines the
conditions to assert that there is no activity.
– Failure is false negative; type II error.
Tails of the Curve
• The type I and type II errors are
due to statistical fluctuations.
• A type I error is set by a rate r1
– Measurements in curve area
a give a false positive.
• A type II error is set by a rate r2
– Measurements in curve area
b give a false negative.
Pn(rn)
a
b
r1
0
r2
rn
Type I Error
• The minimum significant
measurement, r1, depends on
the tolerance for error.
• Set the probability a.
– Area in tail of normal
distribution equal to a
– Equate to number of
standard deviations ka
r1  ka   
2
gr
ka2 ka
r1 

2t g 2
2
br
 ka
r1  rb rb

tg
tb
 t g  tb 
ka2

 4rb 
2
 t t 
tg
 gb 
for tg = tb,
r1 

ka
ka  ka2  8nb
2t

Background Error
• A type I error is a false positive.
– Can’t be made if a real
signal is present.
• With no signal, the only
measurement is background.
– For nb >> ka2,
k
r1  a 2nb
t
– For well measured
background, nb2 = B,
r1 
ka
t
B
Typical Problem
• A background reading at 29%
efficiency is 410 counts in 10
minutes. The maximum risk of
false positive is 5%. How
accurate is the approximation?
Answer
• From 5%, ka = 1.65.
• To compare, nb = 410.
• Times are equal, r1 = 4.9 with
the full formula or 4.7 with the
approximation.
Type II Error
• The minimum detectable
measurement also depends on
risk.
– Willingness to risk a miss
• The result is related to the
minimum significant
measurement.
rg  rb  r2  k b
• Set the probability b.
– Area in tail of normal
distribution equal to b
– Equate to number of
standard deviations kb
r2  rg  rb  k b
r2  r1  k b
rg
tg

rg  rb
tg
rb
tb
 rb
t g  tb
r1
 rb
tg
t g tb
t g  tb
t g tb
Background Counts
• As with type I, there are
approximations based on the
knowledge of the background.
– With a large background
count, nb >> ka2.
– The best case fixes the
background, nb2 = B.
r2 
r2 
ka  k b
t
ka
t
B
2nb
ka
t
B  r2t
k b2
k b2 
ka
B 
r2 
ka 
 kb 1 


t
2 B
B 4 B 

Noise
• Counts are often made at a
variety values for some
variable.
– Time, position, angle, mass,
frequency, energy, etc.
• Noise refers to the fluctuations
in the counts from point to point
in the variable.
– Statistical tests apply to one
or more bins.
Autocorrelation
• An autocorrelation function
tests for a periodic signal in the
presence of noise.
y (t )  s (t )  n(t )
• The autocorrelation of the
combination is based on mean
values compared at two points.
– Cross terms are zero for
long times.
 yy (t )  s(t )  n(t )s(t   )  n(t   )
 yy (t )  s (t ) s (t   )  s (t )n(t   )
 n(t ) s (t   )  n(t )n(t   )
 yy (t )  ss (t )  sn (t )  ns (t )  nn (t )
 yy (t )  ss (t )  nn (t )
Correlation Measure
• Autocorrelated noise should
peak strongly at 0.
• Sinusoidal signals would show
periodic peaks of auto
correlation.
• “This sample autocorrelation
plot shows that the time series
is not random, but rather has a
high degree of autocorrelation
between adjacent and nearadjacent observations.”
nist.gov