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Uncertainties and error analysis

For a given rate of emitted photons, there's a probability function which gives the number of photons we detect, even assuming 100% detection efficiency, because of statistical uncertainties. In addition, there may also be instrumental uncertainties. Consequently, we now turn to the concepts of probability distributions, with particular interest in the distribution which applies to the detection of photons.

Distributions and characteristics thereof

Some definitions relating to values which characterize a distribution:

mean≡μ = $\displaystyle \int$xp(x)dx

variance≡σ2 = $\displaystyle \int$(x - μ)2p(x)dx

standarddeviation≡σ = $\displaystyle \sqrt{{variance}}$

median : mid-point value.

$\displaystyle {\int_{-\infty}^{x_{median}} p(x) dx \over \int_{-\infty}^{\infty} p(x) dx}$ = $\displaystyle {1\over 2}$

mode : most probable value

Note that the geometric interpretation of above quantities depends on the nature of the distribution; although we all carry around the picture of the mean and the variance for a Gaussian distribution, these pictures are not applicable to other distributions, but the quantities are still well-defined.

Also, note that there is a difference between the sample mean, variance, etc. and the population quantities. The latter apply to the true distribution, while the former are estimates of the latter from some finite sample (N measurements) of the population. The sample quantities are derived from:

sample mean : $\displaystyle \bar{x}$$\displaystyle {\sum x_i \over N}$

sample variance≡$\displaystyle {\sum(x_i - \bar x)^2 \over N-1}$ = $\displaystyle {\sum x_i^2 - (\sum x_i)^2/N \over N-1}$

The sample mean and variance approach the true mean and variance as N approaches infinity. But note, especially for small samples, your estimate of the mean and variance may differ from their true (population) values (more below)!

\textit{Understand the concept of probability distribution functi...
...he difference between population
quantities and sample quantities.}

The binomial distribution

Now we consider what distribution is appropriate for the detection of photons. The photon distribution can be derived from the binomial distribution, which gives the probability of observing the number, x, of some possible event, given a total number of events n, and the probability of observing the particular event (among all other possibilities) in any single event, p, under the assumption that all events are independent of each other:

P(x, n, p) = $\displaystyle {n! p^x (1-p)^{n-x} \over x! (n-x)!}$

For the binomial distribution, one can derive:

mean≡$\displaystyle \int$xp(x)dx = np

variance≡σ2$\displaystyle \int$(x - μ)2p(x)dx = np(1 - p)

The Poisson distribution

In the case of detecting photons, n is the total number of photons emitted, and p is the probability of detecting a photon during our observation out of the total emitted. We don't know either of these numbers! However, we do know that p < < 1 and we know, or at least we can estimate, the mean number detected:

μ = np


In this limit, the binomial distribution asymtotically approaches the Poisson distribution:

P(x, μ) = $\displaystyle {\mu^x \exp^{-\mu} \over x!}$

From the expressions for the binomial distribution in this limit, the mean of the distribution is μ, and the variance is

variance = $\displaystyle \sum_{x}^{}$[(x - μ)2p(x, μ)]

variance = np = μ

σ = $\displaystyle \sqrt{{\mu}}$

This is an important result.

Note that the Poisson distribution is generally the appropriate distribution not only for counting photons, but for any sort of counting experiment where a series of events occurs with a known average rate, and are independent of time since the last event.

What does the Poisson distribution look like? Plots for μ = 2, μ = 10, μ = 10000.

Understand the Poisson distribution and when it applies....
...e Poisson distribution
is related to the mean of the distribution.}

The normal, or Gaussian, distribution

Note, for large μ, the Poisson distribution is well-approximated around the peak by a Gaussian, or normal distribution:

P(x, μ, σ) = $\displaystyle {1 \over \sqrt{2\pi} \sigma}$e$\scriptstyle {-{(x-\mu)^2\over 2\sigma^2}}$

This is important because it allows us to use ``simple'' least squares techniques to fit observational data, because these generally assume normally distributed data. However, beware that in the tails of the distribution, and at low mean rates, the Poisson distribution can differ significantly from a Gaussian distribution. In these cases, least-squares may not be appropriate to model observational data; instead, one might need to consider maximum likelihood techniques instead.

The normal distribution is also important because many physical variables seem to be distributed accordingly. This may not be an accident because of the central limit theorem: if a quantity depends on a number of independent random variables with ANY distribution, the quantity itself will be distributed normally (see statistics texts). In observational techniques, we encounter the normal distribution because one important source of instrumental noise, readout noise, is distributed normally.

Know what a Gaussian (normal) distribution is, including...
...nces the
Poisson distribution is similar to a normal distribution.}

Importance of error distribution analysis

You need to understand the expected uncertainties in your observations in order to:

Confidence levels

For example, say we want to know whether some single point is consistent with expectations, e.g., we see a bright point in multiple measurements of a star, and want to know whether the star flared. Say we have a time sequence with known mean and variance, and we obtain a new point, and want to know whether it is consistent with known distribution?

If the form of the probability distribution is known, then you can calculate the probability of getting a measurement more than some observed distance from the mean. In the case where the observed distribution is Gaussian (or approximately so), this is done using the error function (sometimes called erf(x)), which is the integral of a gaussian from some starting value.

Some simple guidelines to keep in mind follow (the actual situation often requires more sophisticated statistical tests). First, for Gaussian distributions, you can calculate that 68% of the points should fall within plus or minus one sigma from the mean, and 95.3% between plus or minus two sigma from the mean. Thus, if you have a time line of photon fluxes for a star, with N observed points, and a photon noise σ on each measurement, you can test whether the number of points deviating more than 2σ from the mean is much larger than expected. To decide whether any single point is really significantly different, you might want to use more stringent criterion, e.g., 5σ rather than a 2σ criterion; a 5σ has much higher level of significance. On the other hand, if you have far more points in the range 2 - -4σ brighter or fainter than you would expect, you may also have a significant detection of intensity variations (provided you really understand your uncertainties on the measurements!).

Also, note that your observed distribution should be consistent with your uncertainty estimates given the above guidelines. If you have a whole set of points, that all fall within 1σ of each other, something is wrong with your uncertainty estimates (or perhaps your measurements are correlated with each other)!

For a series of measurements, one can calculate the χ2 statistic, and determine how probable this value is, given the number of points.

χ2 = $\displaystyle \sum$$\displaystyle {(observed(i) - model(i) )^2 \over \sigma_i^2}$

For a given value of χ2 and number of measurements, one can calculate the probability that the measurements are consistent with the model (and the uncertainties are correctly predicted). A quick estimate of the consistency of the model with the observed data points can be made using reduced χ2, defined as χ2 divided by the degrees of freedom (number of data points minus number of parameters), which should be near unity if the measurements are consistent wth the model.

Noise equation: how do we predict expected uncertainties?


Astronomers often describe uncertainties in terms of the fractional error, e.g. the amplitude of the uncertainty divided by the amplitude of the quantity being measured; often, the inverse of this, referred to as the signal-to-noise ratio is used. Given an estimate the number of photons expected from an object in an observation, we can calulate the signal-to-noise ratio:

$\displaystyle {S\over N}$ = $\displaystyle {S \over \sqrt{\sigma^2}}$

which is the inverse of the predicted fractional error (N/S).

Consider an object with observed photon flux (per unit area and time, e.g. from the count equation above), S, leading to a signal, S = STt where T is the telescope area and t is the exposure time. In the simplest case, the only noise source is Poisson statistics from the source, in which case:

σ2 = S = STt

$\displaystyle {S\over N}$ = $\displaystyle \sqrt{{S}}$ = $\displaystyle \sqrt{{S^\prime T t}}$

In other words, the S/N increases as the square root of the object brightness, telescope area, efficiency, or exposure time. Note that S is directly observable, so one can calculate the S/N for an observation without knowing the telescope area or exposure time! We've just broken S down so that you can specifically see the dependence on telescope area and/or exposure time.

Understand the concept of S/N and fractional error. Know how
S/N depends on the signal for the Poisson-limited case.}

Background noise

A more realistic case includes the noise contributed from Poisson statistics of ``background'' light (more on the physical nature of this later), B, which has units of flux per area on the sky (i.e. a surface brightness); note that this is also usually given in magnitudes.

B = $\displaystyle \int$$\displaystyle {B_\lambda\over{hc\over\lambda}}$qλ

The amount of background in our measurement depends on how we choose to make the measurement (how much sky area we observe). Say we just use an aperture with area, A, so the total observed background counts is


Again, BTt is the directly observable quantity, but we split it into the quantities on which it depends to understand what factors are important in determining S/N.

The total number of photons observed, O, is

O = S + AB = (S + AB)Tt

The variance of the total observed counts, from Poisson statistics, is:

σO2 = O = S + AB = (S + AB)Tt

To get the desired signal from the object only , we will need to measure separately the total signal and the background signal to estimate:

SSTt = O - A < B >

where < B > is some estimate we have obtained of the background surface brightness. The noise in the measurement is

σS2 $\displaystyle \approx$ σO2 = S + AB = (S + AB)Tt

where the approximation is accurate if the background is determined to high accuracy, which one can do if one measures the background over a large area, thus getting a large number of background counts (with correspondingly small fractional error in the measurement).

This leads to a common form of the noise equation:

$\displaystyle {S\over N}$ = $\displaystyle {S \over \sqrt{S + A B}}$

Breaking out the dependence on exposure time and telescope area, this is:

$\displaystyle {S\over N}$ = $\displaystyle {S^\prime \over \sqrt{S^\prime + A B^\prime }}$$\displaystyle \sqrt{{T}}$$\displaystyle \sqrt{{t}}$

In the signal-limited case, S > > BA, we get

$\displaystyle {S\over N}$ = $\displaystyle \sqrt{{S}}$ = $\displaystyle \sqrt{{S^\prime t T}}$

In the background limited case, BA > > S, and

$\displaystyle {S\over N}$ = $\displaystyle {S\over \sqrt{AB}}$ = $\displaystyle {S^\prime\over \sqrt{B^\prime A}}$$\displaystyle \sqrt{{t T}}$

As one goes to fainter objects, the S/N drops, and it drops faster when you're background limited. This illustrates the importance of dark-sky sites, and also the importance of good image quality.

Consider two telescopes of collecting area, T1 and T2. If we observe for the same exposure time on each and want to know how much fainter we can see with the larger telescope at a given S/N, we find:

S2 = $\displaystyle {T_1\over T_2}$S1

for the signal-limited case, but

S2 = $\displaystyle \sqrt{{T_1\over T_2}}$S1

for the background-limited case.

Understand how Poisson uncertainties in the background c...
...lity because
the amount of background included in the measurement.}

Instrumental noise

In addition to the uncertainties from Poisson statistics (statistical noise), there may be additional terms from instrumental uncertainties. A common example of this that is applicable for CCD detectors is readout noise, which is additive noise (with zero mean!) that comes from the detector and is independent of signal level. For a detector whose readout noise is characterized by σrn,

$\displaystyle {S\over N}$ = $\displaystyle {S \over \sqrt{S + B A_{pix} + \sigma_{rn}^2}}$

if a measurement is made in a single pixel. If an object is spread over Npix pixels, then

$\displaystyle {S\over N}$ = $\displaystyle {S \over \sqrt{S + B A + N_{pix}\sigma_{rn}^2}}$

For large σrn, the behavior is the same as the background limited case. This makes it clear that if you have readout noise, image quality (and/or proper optics to keep an object from covering too many pixels) is important for maximizing S/N. It is also clear that it is critical to have minimum read-noise for low background applications (e.g., spectroscopy).

There are other possible additional terms in the noise equation, arising from things like dark current, digitization noise, uncertainties in sky determination, uncertainties from photometric technique, etc. (we'll discuss some of these later on), but in most applications, the three sources discussed so far – signal noise, background noise, and readout noise – are the dominant noise sources.

Note the applications where one is likely to be signal dominated, background dominated, and readout noise dominated.

Understand readout noise and how it represented by a nor...
...nces readout noise is an
important contributor to the total noise.}

Error propagation

Why are the three uncertainty terms in the noise equation added in quadrature? The measured quantity (S) is a sum of S + B - < B > + < R >, where < R > = 0 since readout noise has zero mean. The uncertainty in a summed series is computed by addding the individual uncertainties in quadature; in the equation above, we have neglected the uncertainty in < B >. To understand why they add in quadrature, let's consider general error propagation.

More reasons to consider error propagation: let's say we want to make some calculations (e.g., calibration, unit conversion, averaging, conversion to magnitudes, calculation of colors, etc.) using these observations: we need to be able to estimate the uncertainties in the calculated quantities that depend on our measured quantities.

Consider what happens if you have several known quantities with known error distributions and you combine these into some new quantity: we wish to know what the uncertainty is in the new quantity.

x = f (u, v,....)

The question is what is σx if we know σu, σv, etc.?

As long as uncertainties are small:

xi - < x > ∼(ui - < u > )$\displaystyle \left(\vphantom{{{\partial{x}}\over{\partial {u}}}}\right.$$\displaystyle {{\partial{x}}\over{\partial {u}}}$$\displaystyle \left.\vphantom{{{\partial{x}}\over{\partial {u}}}}\right)$ + (vi - < v > )$\displaystyle \left(\vphantom{{{\partial{x}}\over{\partial {v}}}}\right.$$\displaystyle {{\partial{x}}\over{\partial {v}}}$$\displaystyle \left.\vphantom{{{\partial{x}}\over{\partial {v}}}}\right)$ + ...

σx2 = lim(N - > ∞)$\displaystyle {1\over N}$$\displaystyle \sum$(xi - < x > )2

= σu2$\displaystyle \left(\vphantom{{{\partial{x}}\over{\partial {u}}}}\right.$$\displaystyle {{\partial{x}}\over{\partial {u}}}$$\displaystyle \left.\vphantom{{{\partial{x}}\over{\partial {u}}}}\right)^{2}_{}$ + σv2$\displaystyle \left(\vphantom{{{\partial{x}}\over{\partial {v}}}}\right.$$\displaystyle {{\partial{x}}\over{\partial {v}}}$$\displaystyle \left.\vphantom{{{\partial{x}}\over{\partial {v}}}}\right)^{2}_{}$ +2σuv2$\displaystyle {{\partial{x}}\over{\partial {u}}}$$\displaystyle {{\partial{x}}\over{\partial {v}}}$ + ...

The last term is the covariance, which relates to whether uncertainties are correlated.

σuv2 = lim(n - > ∞)$\displaystyle {1\over N}$$\displaystyle \sum$(ui - < u > )(vi - < v > )

If uncertainties are uncorrelated, then σuv = 0 because there's equal chance of getting opposite signs on vi for any given ui. When working out uncertainties, make sure to consider whether there are correlated errors! If there are, you may be able to reformulate quantities so that they have independent errors: this can be very useful!

Examples for uncorrelated errors:

Distribution of resultant uncertainties

When propagating errors, even though you can calculate the variances in the new variables, the distribution of uncertainties in the new variables is not, in general, the same as the distribution of uncertainties in the original variables, e.g. if uncertainties in individual variables are normally distributed, uncertainties in output variable are not necessarily.

When two variables are added, however, the output is normally distributed.

\textit{Know the error propagation formula and how to apply it.}

Determining sample parameters: averaging measurements

We've covered uncertainties in single measurements. Next we turn to averaging measurements. Say we have multiple observations and want the best estimate of the mean and variance of the population, e.g. multiple measurements of stellar brightness. Here we'll define the best estimate of the mean as the value which maximizes the likelihood that our estimate equals the true parent population mean.

For equal uncertainties, this estimate just gives our normal expression for the sample mean:

$\displaystyle \bar{x}$ = $\displaystyle {\sum x_i \over N}$

Using error propagation, the estimate of the uncertainty in the sample mean is given by:

σ$\scriptstyle \bar{x}$2 = $\displaystyle \sum$$\displaystyle {\sigma_i^2\over N^2}$ = $\displaystyle {\sigma^2\over N}$

But what if uncertainties on each observation aren't equal, say for example we have observations made with several different exposure times? Then the optimal determination of the mean is using a:

weightedmean = $\displaystyle {\sum {x_i\over\sigma_i^2} \over \sum {1\over\sigma_i^2}}$

and the estimated uncertainty in this value is given by:

σμ2 = $\displaystyle \sum$$\displaystyle {{\sigma_i^2\over \sigma_i^4} \over ({\sum {1\over \sigma_i^2}})^2}$ = $\displaystyle {1 \over \sum{1\over\sigma_i^2}}$

where the σi's are individual weights/errors (people often talk about the weight of an observation, i.e. ${1\over\sigma_i^2}$: large weight means small uncertainty.

This is a standard result for determining sample means from a set of observations with different weights.

However, there can sometimes be a subtlety in applying this formula, which has to do with the question: how do we go about choosing the weights/errors, σi? We know we can estimate σ using Poisson statistics for a given count rate, but remember that this is a sample variance (which may be based on a single observation!) not the true population variance. This can lead to biases.

Consider observations of a star made on three nights, with measurements of 40, 50, and 60 photons. It's clear that the mean observation is 50 photons. However, beware of the being trapped by your undertainty estimates. From each observation alone, you would estimate uncertainties of $\sqrt{{40}}$, $\sqrt{{50}}$, and $\sqrt{{60}}$. If you plug these uncertainty estimates into a computation of the weighted mean, you'll get a mean rate of 48.64!

Using the individual estimates of the variances, we'll bias values to lower rates, since these will have estimated higher S/N.

Note that it's pretty obvious from this example that you should just weight all observations equally. However, note that this certainly isn't always the right thing to do. For example, consider the situation in which you have three exposures of different exposure times, and you are calculating the photon rate (counts/s). Here you probably want to give the longer exposures higher weight (at least, if they are signal or background limited). In this case, you again don't want to use the individual uncertainty estimates or you'll introduce a bias. There is a simple solution here also: just weight the observations by the exposure time. However, while this works fine for Poisson uncertainties (variances proportional to count rate), it isn't strictly correct if there are instrumental uncertainties as well which don't scale with exposure time. For example, the presence of readout noise can have this effect; if all exposures are readout noise dominated, then one would want to weight them equally, if readout noise dominates the shorter but not the longer exposures, once would want to weight the longer exposures even higher than expected for the exposure time ratios! The only way to properly average measurements in this case is to estimate a sample mean, then use this value scaled to the appropriate exposure times as the input for the Poisson uncertainties.

Another subtlety: averaging counts and converting to magnitudes is not the same as averaging magnitudes!

\textit{Understand how the distinction between sample variance
...biases when calculating a weighted
mean, and how to overcome this.}

Can you split exposures?

Although from S/N considerations, one can determine the required number of counts you need (exposure time) to do your science, when observing, one must also consider the question of whether this time should be collected in single or in multiple exposures, i.e. how long individual exposures should be. There are several reasons why one might imagine that it is nicer to have a sequence of shorter exposures rather than one single longer exposure (e.g., tracking, monitoring of photometric conditions, cosmic ray rejection, saturation issues), so we need to consider under what circumstances doing this results in poorer S/N.

Consider the object with photon flux S, background surface brightness B, and detector with readout noise σrn. A single short exposure of time t has a variance:

σS2 = STt + BATt + Npixσrn2

If N exposures are summed, the resulting variance will be

σN2 = S2

If a single long exposure of length Nt is taken, we get

σL2 = STNt + BATNt + Npixσrn2.

The ratio of the noises, or the inverse ratio of the S/N (since the total signal measured is the same in both cases), is

$\displaystyle {\sigma_N\over \sigma_L}$ = $\displaystyle \sqrt{{
S^\prime T Nt + B^\prime A T Nt + NN_{pix} \sigma_{rn}^2 \over
S^\prime T Nt + B^\prime A T Nt + N_{pix} \sigma_{rn}^2 }}$

The only difference is in the readout noise term! In the signal- or background-limited regimes, exposures can be added with no loss of S/N. However, if readout noise is significant, then splitting exposures leads to reduced S/N.

\textit{Understand under what circumstances you can split exposur...
...asons why you might want to split an exposure into shorter

Random errors vs systematic errors

So far, we've been discussing random errors. There is an additional, usually more troublesome, type of errors known as systematic errors. These don't occur randomly but rather are correlated with some, possibly unknown, variable relating to your observations, and can have the effect of not just adding spread around the true value that you are trying to measure, but actually measuring the wrong mean.

EXAMPLE : flat fielding


Note also that in some cases, systematic errors can masquerade as random errors in your test observations (or be missing altogether if you don't take data in exactly the same way), but actually be systematic in your science observations.

EXAMPLE: flat fielding, subpixel QE variations.

Note that error analysis from expected random errors may be the only clue you get to discovering systematic errors. To discover systematic errors, plot residuals vs. everything!

\textit{Understand the distinction between random and systematic errors.}

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