Signal to Noise

Hypothesis Testing For the purposes of hypothesis testing in the context of the sciences, there are two buckets of hypotheses. Here, the main goal of any good scientific experiment is to discern the likelihood measurements of the experiment are drawn from a statistical model generated by either of the two hypotheses.

These two hypotheses are 1. Null Hypothesis 2. Alternate Hypothesis

The Null Hypothesis is the supposition that the interrogated phenomenon does not occur or that the measurements resulting from the experiment are consistent with random fluctuations in the measurement apparatus.

The Alternate Hypothesis is the supposition that the interrogated phenomenon does occur.

The goal of any good science experiment is to discern the likelihood that measurements generated by the experiment are generated by one hypothesis or the other. In most cases, we can never truly *know*, but we can be *damn sure*.

In the context of photometry and other measures derived from photometric methods (e.g., variation in light curves), the Null Hypothesis is the supposition that the observer only observed noise and not any astrophysical source. The Alternate Hypothesis is therefore the supposition that the astrophysical source was observed. We must endeavor to generate statistical models of each if we are to have any hope of testing them with data. Let's discuss the provenance of the sources of detector signals and noise in the context of photometry all while remembering them as random processes. Then we'll do some arm-wavy math to justify some nice and *normal*-looking statistical models for each hypothesis.

As discussed in Data Reduction I:Top of Telescope: For any given CCD-like detector, the light frame reported by the detector is the direct sum of a few disparate sources of counts.

$$L_{ij}(\lambda, t) = \biggl(\epsilon_{ij}(\lambda)I_{ij}(\lambda) + T_{ij}\biggr)t + B_{ij}$$

The Light frame ($L_{ij}(\lambda)$; measured in counts) is the sum of:

1. The counts resulting from the counts generated by photoelectrons across a heterogeneous detector and optical system and over some integration time ($t$) $\epsilon_{ij}(\lambda)I_{ij}(\lambda)t$
2. The counts resulting from the detector and optical system even when no light is incident on the optical system. With no light, the optical system is said to be in-the-dark and so any counts generated from such a system constitute the so-called Dark counts. $D_{ij}(t) = T_{ij}t + B_{ij}$. The Dark counts is the result of thermally agitated electrons finding themselves in the read-out circuitry as well as the relatively constant base-level of electronic noise owing to a variety of sources.

What's worse is that there is no guarantee that each pixel must abide the same random distribution as its neighbors. There may be pixel-level or optical system-level defects ensuring this. So each pixel in each frame is a sampling of some unknown probability distribution.

The measurement of counts in the pixels of each light frame are clearly owing to a few separate phenomenon occurring in unison at the time of observation. The generation of photoelectrons from incident light, the liberation of electrons owing to their temperature, and the liberation of electrons due to electronic effects are all inherently random, Possionian processes.

These processes can be isolated and sampled repeatedly, allowing us to build-out a rough distribution of measured counts. For each the Light and the Dark, we can take (separately) an ensemble average and recover what is called the “signal” from each of these sources of counts.

Therefore, a sequence of $N$ Light frames, $\{L_{ij}^0(\lambda), ~L_{ij}^1(\lambda), ~L_{ij}^2(\lambda), \cdots , ~L_{ij}^N(\lambda)\}$ can be averaged to yield the average $\langle L_{ij}^0(\lambda)\rangle$.

Being more clear, consider a the small array of pixels constituting a single frame read-out from a detector with pixels labeled $i$ & $j$.

We can imagine rotating this array onto its side …

And finally, considering that each pixel of each frame captured is a sampling of a probability distribution, we should sample that distribution $N$ times to generate a stack of frames.

If each pixel value (the counts) from the $i,j$-th pixel is a random sampling of a Poisson Distribution, then $L^n_{ij}\thicksim\mathcal{Pois(\lambda)} \rightarrow \mathcal{Norm(\lambda, \sqrt{\lambda})}$ if $\lambda$ is large enough (and it is for these purposes!).