# why isn't the variance of Fano noise proportional to the number of photon arrivals?

I've been reading "Photon Transfer" by James Janesick and in Chapter 3 (http://spie.org/samples/PM170.pdf) he describes the various sources of noise in a CCD. Basically, incoming photons interact with the CCD and cause some electrons to jump the bandgap. Both of these processes have some inherent randomness and must be described statistically - the photon arrival rate's variability gives rise to the shot noise, while the variability in electrons produced from a given photon impact gives rise to the Fano noise.

The photon arrival is a Poisson process, so if m is the expected number of photon arrivals during a given exposure, the actual number of arrivals is a Poisson distributed random variable N(m), and var(N) = m.

Now here's the part I don't understand. I quote from Janesick, editing slightly for clarity (p. 26):

If all the energy of an interacting photon was spent in the production of electron-hole (e-h) pairs, then there would be no variation in the number of e-h pairs produced. On the other hand, if the energy was partitioned between breaking covalent bonds and lattice vibrations, or if phonon production was uncorrelated, Poisson statistics would apply. But neither is the case in nature. The variance in multiple electron-hole charge generation, called Fano noise, is empirically described by

$$\sigma_{FN}^2 = F_F \eta_i = F_F \frac{h_v}{E_{e-h}}$$

where [$\sigma_{FN}^2$ is the variance of the Fano noise], and $F_F$ is referred to as the Fano factor, which is defined by the variance in the number of electrons generated divided by the average number of electrons generated per interacting photon. [notation: $h_v$ and $E_{e-h}$ were defined to be the photon energy in eV and the energy required to create an electron-hole pair, and $\eta_i$ is the ratio of the two. So $\eta_i$ is the expected number of electrons per incident photon, if all photon energy goes into production of electrons.]

Based on this description of the process, I would assume that each incident photon liberates an uncertain number of electrons, and therefore each incident photon should contribute a constant amount of variance. So I think the Fano noise variance should be proportional to the incident number of photons. But according to Janesick, the Fano noise contributes a constant variance independent of the number of photon arrivals.

To be more mathematically precise, I think that the whole process should be modeled as a compound Poisson process, since both the number of arrivals and the number of electrons liberated per arrival are stochastic quantities. The wikipedia article on the compound Poisson process shows that the variance of the compound process has two components (one for the randomness of arrivals and one for the random size of the arrivals), and both components are proportional to the mean number of arrivals.

So can anyone explain the discrepancy between my expectations and the book's equations? I feel I must have misunderstood what Fano noise is -- because if it's what I think it is, the natural model of it contradicts Janesick's equation.

By the way, I know that $\sigma_{FN}^2$ is not referring to the variance per incident photon. If it were, then Janesick would multiply it by the mean photon arrivals when calculating the total image noise, but he doesn't. Here is his calculation of the total noise variance (equation 3.17, p. 34, irrelevant flat field term removed): $$\sigma_{TOTAL}^2 = \sigma^2_{READ} + \eta_i F_F + \eta_i S$$

where $S$ is the expected number of electrons from the exposure. The first term is from shot noise, the second from Fano noise, the third is from shot noise. Note that the shot noise term scales linearly with $S$ but the Fano noise term is independent of $S$.

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Could you elaborate a bit more on the equations you refer to in your last paragraph? I've usually heard "Fano noise" as reffering to (as you suspect) the stochastic nature of pair production in the detector, with the "Fano factor" describing the degree to which that process has sub-Poissonian variance. In other words, Fano noise is not another noise source in addition to pair production variance, is simply descriptive of the fact that this noise is sub-Poissonian because that process is not truly a Poisson process. Because the F factor simply modifies the shot noise from the expected value... –  Colin K Oct 19 '11 at 15:24
...it would not need to be computed per-photon; but I"m not sure exactly how the author presents this calculation, so I want to see the equation before I commit to an answer. –  Colin K Oct 19 '11 at 15:25
I edited the OP to respond to your question. The total noise includes a shot noise term whose variance scales linearly with incident intensity, but the Fano noise term doesn't. –  Paul Oct 19 '11 at 15:33
Ok I think I'm on the right track with this, unfortunately now I've got to go eat my lunch or I won't have any other time in the work day! In hunting down an answer I found a PDF of the book chapter you refer to, you may want to edit it in:spie.org/samples/PM170.pdf –  Colin K Oct 19 '11 at 16:51
I think I got it. The basic point is that the equation you quoted is in fact the result for a single photon interaction. The way the author stated it, and the particular symbolism he uses, make things confusing, but I'm pretty sure about this. I'll explain in an answer when I get the chance. –  Colin K Oct 19 '11 at 17:16

It seems that the confusion is due to some unfortunate notation. As the OP states, Fano noise is due to the variance in photoelectron production per incident photon, and this should indeed be signal-dependent. However, the author also states that the total noise is given by:

$$\tag{1} \sigma^2_\mathrm{TOTAL} = \sigma^2_\mathrm{READ} + \eta_i F_F + \eta_i S$$

This equation is confusing, because the first and third term appear to be the familiar terms for read noise and Photon noise, but the second term, representing Fano noise, appears to have no dependance on the signal level. Matters are further confused because the notation used by the author is subtly different from what I (and I assume most others?) are familiar with.

First we will figure out this notation, then we will clarify the confusion about Fano noise, and finally we will see if we can make sense of equation 1. I'll be referring to the chapter of the text in question, "Photon Transfer" by James Janesick, which can be found for free here. I will explicitly state when I am referring to an equation from Janesick, while references to my own equations will simply be by number.

## Notation

To be clear, I'm not saying this notation is bad. It's just different from what most people are used to, but similar enough to be confusing if you aren't careful.

$$\begin{array} {l l} \sigma^2 & \quad \textrm{Variance} \\ \eta_i & \quad \textrm{The expectation value for the numper of photoelectrons per incident photon} \\ F_F & \quad \textrm{Fano coefficient} \\ S & \quad \textrm{Expected Number of photoelectrons in an exposure} \\ N_P & \quad \textrm{Number of photons incident in a given exposure} \\ P_I & \quad \textrm{Expected Number of photons incident on the detector}\\ \end{array}$$

Note the following relations as well:

$$\begin{array} {r c l} P_I &=& \left<N_P\right> \\ S &=& \left< S_\textrm{PEAK} \right> \\ S_\textrm{PEAK} &=& \eta_i N_P \\ S &=& \eta_i P_I \end{array}$$

where $\left<X\right>$ represents the expectation value of the random variable $X$. So remember that $S$ and $P_I$ are the mean values of their respective distributions, while $N_P$ and $S_\textrm{PEAK}$ are the specific values measured in a particular trial.

Another subtlety of this author's notation is that $\eta_i$ is not simply the quantum efficiency, and is not necessarily limited to [0, 1). In this author's notation, $\eta_i$ includes both of the quantities that I would refer to separately as the quantum efficiency $\eta$, and the gain $G$. So I may substitute $\eta_i \rightarrow \eta G$ where $\eta$ is a unitless and $0 \leq \eta \leq 1$.

As an example, the third term in equation 1 is supposed to represent the contribution to the variance of the photoelectron number due to the Poisson distributed statistics of the photon signal (i.e. photon noise). If we make all of my preferred substitutions we get:

$$\begin{array}{r c l c r c l} \sigma^2_\textrm{photon} &=& \eta_i S &\quad& S &\rightarrow& \eta_i P_I \\ \sigma^2_\textrm{photon} &=& \eta^2_i P_I &\quad& \eta_i &\rightarrow& \eta G \\ \sigma^2_\textrm{photon} &=& \eta^2 G^2 P_I \end{array}$$

...which is the familiar expression for the variance contribution from photon noise is a system with gain. This also agrees with Janesick equations 3.5 and 3.6. So far, so good.

## Fano Noise

For a moment let's ignore equation 1, and follow Janesick in section 3.3, where he introduces Fano noise. Janesick uses the example of a CCD or some other detector array, so we may, interchangeably, speak of the signal from a single "trial," or from one pixel, or from a single exposure. Furthermore, each pixel in the array will see a photon source with the same $P_I$, while the number of photons that hit each pixel, $N_P$, will generally not be exactly $P_I$, but will follow a Poisson distribution with $P_I$ as its mean.

In Janesick equation 3.7, he gives the variance in electron-hole generation as:

$$\tag{2} \sigma^2_\textrm{FN} = F_F \eta_i$$

where $F_F$ is a constant, $0\leq F_F \leq 1$ describing the strength of the noise. Looking at this equation, we can be confident that we are considering the electron-hole production from a single photon interaction. Why? Well, consider the case where $F_F = 0$. If this expression is for a single photon interaction, then the $F_F = 0$ case corresponds to the situation where each photon produces exactly the same number of photoelectrons, and there is no extra noise term associated with the electron-hole production process. Now consider the $F_F = 1$ case. This, according to the text preceeding the equation in Janesick, corresponds to the electron-hole production process being purely Poissonian and indeed the equation would reduce to

$$\sigma^2_\textrm{FN} = \eta_i$$

which simply tells us that the variance from the electron-hole generation process would be equal to the expected number of photoelectrons from one photon($\eta_i N_P$ where $N_P = 1$). In other words the variance would equal the mean, as expected for a Poisson process.

So now we understand Fano noise for a single photon interaction, and we see that $F_F$ describes the degree to which this noise is less noisy than a simple Poisson process.

Moving on, Janesick goes back to the CCD array example, and plots a histogram of the charge measured on each pixel after exposing the array to an average incident signal of 3 photons per pixel, with $\eta_i = 10e^- \textrm{/photon}$. So we are looking at the result of many trials with $P_I=3 \textrm{photons}$ and $S = 30e^-$, but there are peaks around $(10, 20, 30, \ldots) e^-$ per pixel because $N_P$ is Poisson distributed. Each peak is broadened by read noise, but we also see that the peaks at larger $N_P$ are wider. Why? Janesick equation 3.9 tells us:

$$\sigma^2_\textrm{FN,PEAK} = N_P F_F \eta_i$$

In other words, Fano noise varies as a function of signal, but not the mean signal $P_I$. Rather, it varies with the number of incident photons incident in a single trial.

Now things become confusing. Janesick tells us that to combine shot noise and Fano noise, we add the variances in the normal way:

$$\sigma^2_\textrm{SHOT+FN} = \sigma^2_\textrm{SHOT} + \sigma^2_\textrm{FN}$$

No surprise there, but we are wondering what expression will give us $\sigma^2_\textrm{FN}$? Shot noise, as we know, varies with the signal $P_I$ or $S$, but we have not yet seen how to extend the $N_P$ dependance of Fano noise to an expression containing the mean signal level. At this point we understand Fano noise as a contribution to the variance in photoelectron number from pixels with a given number of incident photons, but this number itself is Poisson distributed. Janesick substitutes equation 2 (his equation 3.7) to state that

$$\tag{3} \sigma^2_\textrm{SHOT+FN} = \eta_i \left(S + F_F \right)$$

..but he does not explicitly justify this substitution.

## The Confusion, and Resolution

It is this final substitution which leads to the form we see in equation 1, and the confusion can be traced to a lack of clarity on how that substitution is justified. Possibly there is some theorem which states that variance of a process in which the variances of each trial are themselves distributed with a given variance can be added in such a way as to arrive at equation 3, but I am unaware of such a theorem.

I suspect that the correct resolution to this confusion is two-fold: First, the noise term described in Janesick equation 3.7 is purely an empirical description, so one could assume that the form of equation 1 is simply an empirical estimate of noise in a detector where Fano effects are non-negligible.

Second, it is only for small signals where Fano noise will dominate, so when considering Fano noise we can take $N_P \approx 1$. Thus we can justify the substitution made by Janesick. This assumption would seem appropriate, as the signal to noise ratio (SNR) due to photon noise will scale with $\sqrt{P_I}$, while the SNR for Fano noise will scale like $\sqrt{\frac{N_P \eta_i}{F_F}}$, so for Fano noise to dominate over photon noise would imply high gain and low photon flux. In other words, Fano noise is associated with single photon counting detectors.

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If this answer is indeed what Janesick is doing, then I'm kind of bothered by his approach. It sounds like he only cares about getting his equation right in the case where Fano noise dominates. This strikes me as a bit silly. The equations can be made exact by modeling the photoelectron generation as a compound Poisson process, as I mentioned previously. The photons arrive via a Poisson process, and for each photon that enters, it liberates a random number of electrons. As described in compound Poisson process, you get two terms... –  Paul Oct 22 '11 at 17:36
... in the calculation of the variance, via the law of total variance. The first is the variance of the number of photon arrivals, and the second is the conditional variance of the electrons liberated, given the number of photons that did arrive. In formulas, let $N_e$ be the number of photoelectrons liberated in a given exposure. Then $N_e = \sum_p^{N_P} f_i$, where $f_p$ is the number of electrons liberated by photon $p$. $f_p$ has a mean of $\eta_i$ as you said at the top, and a variance of $F_F \eta_i$ by Janesick's eq. (3.7)... –  Paul Oct 22 '11 at 17:58
... conditioning on the value of $N_P$ in the law of total variance we obtain the expression $$Var(N_e) = \mathbb{E} Var(N_e|N_P) + Var(\mathbb{E}(N_e|N_P)) = \mathbb{E} (F_F \eta_i N_P) + Var(\eta_i N_P) = F_F \eta_i P_I + \eta_i^2 P_I$$. Substituting $S = \eta_i P_I$ we get $$\sigma_{SHOT+FN}^2 = Var(N_e) = (F_F + \eta_i)S$$, which is identical to Janesick's expression **except that $S$ and $\eta_i$ are switched!!** [note: in the previous comment, $f_i$ should be $f_p$, and the bottom of the sigma should be $p = 1$ not $p$] –  Paul Oct 22 '11 at 18:08
All of this simply follows mathematically if my compound Poisson model is correct, and a compound Poisson model is strongly suggested by Janesick's verbiage when he introduces Fano noise. Which brings me back to my original problem -- either Janesick messed up his math or I have misunderstood what Fano noise is. –  Paul Oct 22 '11 at 18:21
@Paul: I think you are correct, but I wouldn't say janesick messed up his math. I'd definitely agree that I don't like the way he writes, but there are a lot of authors who have a bad habit of making assumptions or approximations that they don't tell us about. –  Colin K Oct 22 '11 at 19:16

The standard deviation of a system of n measurements is typically proportional to sqrt(n). So the variance is naturally proportional to n. So the proportionality is not surprising.

For the situation where one makes n binomial tests with probability p of success, the expected number of successes is np. The standard deviation for an estimate of $p$ is $\sqrt(p(1-p)/n$. (Note, I'm assuming that $n$ is large and ignoring differences between $n$ and $n-1$ which arise depending on whether one has prior knowledge of the actual probability $p$.) Thus the standard deviation for an estimate of $np$ is $\sqrt{np(1-p)}$, and the variance is the square of this, i.e. $np(1-p)$ which is proportional to $n$.

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I understand all this as you'll see in the comments below. My question is more subtle than this. –  Paul Oct 22 '11 at 18:06