Poisson Random Variable P [X = i ] = e - i / i! i.e. the probability that the number of events is i E [ X ] = for Poisson Random Variable  X 2 =

Signal to Noise Ratio I ∆I Object we are trying to detect Background Definitions:

To find  I, find the probability density function that describes the # photons / pixel ( ) Source Body Detector 1)X-ray emission is a Poisson process N 0 is the average number of emitted X-ray photons, or in the Poisson process.

2) Transmission -- Binomial Process transmitted p = e - ∫ u(z) dz interactingq = 1 - p 3) Cascade of a Poisson and Binary Process still has a Poisson Probability Density Function - Q(k) represents transmission process Still Poisson, with = p N 0 Average Transmission =pN 0 Variance = pN 0

Let the number of transmitted photons = N then describes the signal, Another way, Recall,

Units of Exposure (X) = Roentgen (R) is defined as a number of ion pairs created in air 1 Roentgen = 2.58 x coulombs/kg of air Ionizing energy creates energy in the body. Dose refers to energy deposition in the body. Units of Dose (D): ergs/ gram (CGS) or J/kg (SI) Section 4.6 of textbook

1 Rad - absorbed dose unit: expenditure of 100 ergs/gram 1 R produces 0.87 Rad in air How do Rads and Roentgen relate? -depends on tissue and energy -Rads/Roentgen > 1 for bone at lower energies -Rads/ Roentgen approximately 1 for soft tissue -independent of energy Section 4.6 of textbook

N =  AR exp[ - ∫  dz ] R = incident Roentgens A = pixel area (cm 2 ) photons / cm 2 / Roentgen Photon Energy

Dose Equivalent: H(dose equivalent) = D(dose) * Q(Quality Factor) Q is approximately 1 in medical imaging Q is approximately 10 for neutrons and protons Units of H = 1 siever (Sv)= 1 Gray.01 mSv=.001 Rad = 1 mrem Background radiation: mrem/yr Typical Exams: Chest X-ray = 10 mRad = 10 mrems CT Cardiac Exam = Several Rad Quantitative Feeling For Dose Fermilab Federal Limits : 5 Rads/year No one over 2.5

Let t = exp [ - ∫  dz ] Add a recorder with quantum efficiency  Example chest x-ray: 50 mRad  = 0.25 Res = 1 mm t = 0.05 What is the SNR as a function of C?

Have we made an image yet? Consider the detector M  X light photons / capture  Y light photons Transmitted And captured Photons Poisson What are the zeroth order statistics on Y? M Y =  X m m=1 Y depends on the number of x-ray photons M that hit the screen, a Poisson process. Every photon that hits the screen creates a random number of light photons, also a Poisson process.

What is the mean of Y? ( This will give us the signal level in terms of light photons) Mean Expectation of a Sum is Sum of Expectations (Always) Each Random Variable X has same mean. There will be M terms in sum. There will be M terms in the sum E [Y] = E [M] E [X] Sum of random variables E [M] =  Ncaptured x-ray photons / element E [X] = g 1 mean # light photons / single x-ray capture so the mean number of light photons is E[Y] =  N g 1.

What is the variance of Y? ( This will give us the std deviation) We will not prove this but we will consider the variance in Y as a sum of two variances. The first will be due to the uncertainty in the number of light photons generated per each X-ray photon, X m. The second will be an uncertainty in M, the number of incident X-ray photons. To prove this, we would have to look at E[Y 2 ]. The square of the summation would be complicated, but all the cross terms would greatly simplify since each process X in the summation is independent of each other.

What is the variance of Y? ( This will give us the std deviation) If M was the only random variable and X was a constant, then the summation would simply be Y = MX. The variance of Y,   y =X 2   m Recall multiplying a random variable by a constant increases its variance by the square of the constant. X is actually a Random variable, so we will write X as E[X] and the uncertainy due to M as   y =[E[X]] 2   m If M were considered fixed and each X in the sum was considered a random variable, then the variance of the sum of M random variables would simply be M *   x. We can make this simplification since each process that makes light photons upon being hit by a x-ray photon is independent of each other.

 M 2 =  N Recall M is a Poisson Process  X 2 = g 1 Generating light photons is also Poisson  Y 2 =  Ng 1 +  Ng 1 2 Uncertainty due to X Uncertainty due to M Dividing numerator and denominator by g 1

Actually, half of photons escape and energy efficiency rate of screen is only 5%. This gives us a g 1 = 500 Since g 1 >> 1, What can we expect for the limit of g 1, the generation rate of light photons?

We still must generate pixel grains Y W = ∑ Z m where W is the number of silver grains developed m=1 Y  Z  W grains / pixel Light Photons / pixel Z = developed Silver grains / light photons Let E[Z] = g 2, the number of light photons to develop one grain of film. Then,  z 2 = g 2 also since this is a Poisson process, i.e. the mean is the variance. E[W] = E[Y] E[Z]  W 2 = E[Y]  z 2 +  Y 2 E 2 [Z] uncertainty in gain factor zuncertainty in light photons

Let E [ Z ] = g 2,, the mean number of light photons needed to develop a grain of film

Recall g 1 = 500( light photons per X-ray) g 2 = 1/200 light photon to develop a grain of film That is one grain of film requires 200 light photons. Is 1/g 1 g 2 <<1? Is 1/g 1 << 1? What is the lesson of cascaded gains we have learned?