# 4.25 Type-based detection

 Page 1 / 1

Perhaps the ultimate non-parametric detector makes no assumptions about the observations' probability distribution under either model. Here, we assumethat data representative of each model are available to train the detection algorithm. One approach uses artificial neuralnetworks, which are difficult to analyze in terms of both optimality and performance. When the observations arediscrete-valued, a provable optimal detection algorithm ( Gutman ) can be derived using the theory of types .

For a two-model evaluation problem, let $\stackrel{}{r}$ (length $\stackrel{}{L}$ ) denote training data representative of some unknown probability distribution $P$ . We assume that the data have statistically independentcomponents. To derive a non-parametric detector, form a generalized likelihood ratio to distinguish whether a set ofobservations $r$ (length $L$ ) has the same distribution as the training data or a different one $Q$ . $\lg (r)=\lg \left(\frac{\max\{P, , Q , P(r)Q(\stackrel{}{r})\}}{\max\{P , P(r)P(\stackrel{}{r})\}}\right)$ Because a type is the maximum likelihood estimate of the probability distribution (see Histogram Estimators ), we simply substitute types for the training data and observationsprobability distributions into the likelihood ratio. The probability of a set of observations having a probabilitydistribution identical to its type equals $e^{-(L((P)))}$ . Thus, the log likelihood ratio becomes $\lg (r)=\lg \left(\frac{e^{-(L(({P}_{r})))}e^{-(\stackrel{}{L}(({P}_{\stackrel{}{r}})))}}{e^{-((L+\stackrel{}{L})(({P}_{r,\stackrel{}{r}})))}}\right)$ The denominator term means that the training and observed data are lumped together to form a type. This type equals the linearcombination of the types for the training and observed data weighted by their relative lengths. $({P}_{r,\stackrel{}{r}})=\frac{L({P}_{r})+\stackrel{}{L}({P}_{\stackrel{}{r}})}{L+\stackrel{}{L}}$ Returning to the log likelihood ratio, we have that $\lg (r)=-(L(({P}_{r})))-\stackrel{}{L}(({P}_{\stackrel{}{r}}))+(L+\stackrel{}{L})(\frac{L({P}_{r})+\stackrel{}{L}({P}_{\stackrel{}{r}})}{L+\stackrel{}{L}})$ Note that the last term equals $(L+\stackrel{}{L})(\frac{L({P}_{r})+\stackrel{}{L}({P}_{\stackrel{}{r}})}{L+\stackrel{}{L}})=-\sum (L({P}_{r}(a))+\stackrel{}{L}({P}_{\stackrel{}{r}}(a)))\lg \left(\frac{L({P}_{r}(a))+\stackrel{}{L}({P}_{\stackrel{}{r}}(a))}{L+\stackrel{}{L}}\right)$ which means it can be combined with the other terms to yield the simple expression for the log likelihood ratio.

$\lg (r)=L(({P}_{r}), ({P}_{r,\stackrel{}{r}}))+\stackrel{}{L}(({P}_{\stackrel{}{r}}), ({P}_{r,\stackrel{}{r}}))$

When the training data and the observed data are drawn from the same distribution, the Kullback-Leibler distances will besmall. When the distributions differ, the distances will be larger. Defining ${}_{0}$ to be the model that the training data and observations have the same distribution and ${}_{1}$ that they don't, Gutman showed that when we use the decision rule $\frac{1}{L}\lg (r)\underset{{}_{0}}{\overset{{}_{1}}{}}$ its false-alarm probability has an exponential rate at least as large as the threshold and the miss probability is thesmallest among all decision rules based on training data. $\lim_{L\to }L\to$ 1 L P F and ${P}_{M}$ minimum.

We can extend these results to the $K$ -model case if we have training data ${\stackrel{}{r}}_{i}$ (each of length $\stackrel{}{L}$ ) that represent model ${}_{i}$ , $i\in \{0, , K-1\}$ . Given observed data $r$ (length $L$ ), we calculate the log likelihood function given above for each model todetermine whether the observations closely resemble the tested training data or not. More precisely, define the sufficientstatistics ${}_{i}$ according to ${}_{i}=(({P}_{r}), ({P}_{r,{\stackrel{}{r}}_{i}}))+\frac{\stackrel{}{L}}{L}(({P}_{\stackrel{}{r}}), ({P}_{r,{\stackrel{}{r}}_{i}}))-$ Ideally, this statistic would be negative for one of the training sets (matching it) and positive for all of the others(not matching them). However, we could also have the observation matching more than one training set. In all such cases, wedefine a rejection region ${}_{?}$ similar to what we defined in sequential model evaluation . Thus, we define the ${i}^{\mathrm{th}}$ decision region ${}_{i}$ according to ${}_{i}< 0$ and ${}_{j}> 0$ , $j\neq i$ and the rejection region as the complement of ${}_{i0}^{K1}{}_{i}$ . Note that all decision regions depend on the value of  , a number we must choose. Regardless of the value chosen, the probability ofconfusing models - choosing some model other than the true one - has an exponential rate that is at least  for all models. Because of the presence of a rejection region, another kind of "error" isto not choose any model. This decision rule is optimal in the sense that no other training-data-based decision rule has asmaller rejection region than the type-based one.

Because it controls the exponential rate of confusing models, we would like  to be as large as possible. However, the rejection region grows as  increases; choosing too large a value could make virtually all decisionsrejections. What we want to ensure is that $\lim_{L\to }L\to$ i ? 0 . Obtaining this behavior requires that $\lim_{L\to }L\to$ L L 0 : As the length of the observations increases, so must the size of the training set. In summary, $\text{for some}i:(\lim_{L\to }L\to )\implies$ L L 0 0 L i ? 1 $\forall i\colon ((\lim_{L\to }L\to ))\implies$ L L 0 0 L 1 L i ? 0 The critical value ${}_{0}$ depends on the true distributions underlying the models. The exponential rate of the rejection probability  also depends on the true distributions. These results mean that if sufficient trainingdata are available and the decision threshold is not too large, then we can perform optimal detection based entirely on data! Asthe number of observations increases (and the amount of training data as well), the critical threshold ${}_{0}$ becomes the Kullback-Liebler distance between the unknown models. In other words, the type-based detector becomesoptimal!

#### Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!