# Conditional probability  (Page 6/6)

 Page 6 / 6

The calculations, as in [link] , are simple but can be tedious. We have an m-procedure called bayes to perform the calculations easily. The probabilities $P\left({A}_{i}\right)$ are put into a matrix PA and the conditional probabilities $P\left(E|{A}_{i}\right)$ are put into matrix PEA. The desired probabilities $P\left({A}_{i}|E\right)$ and $P\left({A}_{i}|{E}^{c}\right)$ are calculated and displayed

## Matlab calculations for [link]

>>PEA = [0.10 0.02 0.06];>>PA = [0.2 0.5 0.3];>>bayes Requires input PEA = [P(E|A1) P(E|A2) ... P(E|An)]and PA = [P(A1) P(A2) ... P(An)] Determines PAE = [P(A1|E) P(A2|E) ... P(An|E)]and PAEc = [P(A1|Ec) P(A2|Ec) ... P(An|Ec)] Enter matrix PEA of conditional probabilities PEAEnter matrix PA of probabilities PA P(E) = 0.048P(E|Ai) P(Ai) P(Ai|E) P(Ai|Ec) 0.1000 0.2000 0.4167 0.18910.0200 0.5000 0.2083 0.5147 0.0600 0.3000 0.3750 0.2962Various quantities are in the matrices PEA, PA, PAE, PAEc, named above

The procedure displays the results in tabular form, as shown. In addition, the various quantities are in the workspace in the matrices named, so that theymay be used in further calculations without recopying.

The following variation of Bayes' rule is applicable in many practical situations.

(CP3*) Ratio form of Bayes' rule $\frac{P\left(A|C\right)}{P\left(B|C\right)}=\frac{P\left(AC\right)}{P\left(BC\right)}=\frac{P\left(C|A\right)}{P\left(C|B\right)}\cdot \frac{P\left(A\right)}{P\left(B\right)}$

The left hand member is called the posterior odds , which is the odds after knowledge of the occurrence of the conditioning event. The second fraction in the right hand member is the prior odds , which is the odds before knowledge of the occurrence of the conditioning event C . The first fraction in the right hand member is known as the likelihood ratio . It is the ratio of the probabilities (or likelihoods) of C for the two different probability measures $P\left(\phantom{\rule{0.166667em}{0ex}}\cdot \phantom{\rule{0.166667em}{0ex}}|A\right)$ and $P\left(\phantom{\rule{0.166667em}{0ex}}\cdot \phantom{\rule{0.166667em}{0ex}}|B\right)$ .

## A performance test

As a part of a routine maintenance procedure, a computer is given a performance test. The machine seems to be operating so well that the prior odds it is satisfactory are takento be ten to one. The test has probability 0.05 of a false positive and 0.01 of a false negative. A test is performed. The result is positive. What are the posterior odds the device isoperating properly?

SOLUTION

Let S be the event the computer is operating satisfactorily and let T be the event the test is favorable. The data are $P\left(S\right)/P\left({S}^{c}\right)=10$ , $P\left(T|{S}^{c}\right)=0.05$ , and $P\left({T}^{c}|S\right)=0.01$ . Then by the ratio form of Bayes' rule

$\frac{P\left(S|T\right)}{P\left({S}^{c}|T\right)}=\frac{P\left(T|S\right)}{P\left(T|{S}^{c}\right)}\cdot \frac{P\left(S\right)}{P\left({S}^{c}\right)}=\frac{0.99}{0.05}\cdot 10=198\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{so}\phantom{\rule{4.pt}{0ex}}\text{that}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(S|T\right)=\frac{198}{199}=0.9950$

The following property serves to establish in the chapters on "Independence of Events" and "Conditional Independence" a number of important properties for the concept of independence and of conditional independence of events.

(CP4) Some equivalent conditions If $0 and $0 , then

$P\left(A|B\right)*P\left(A\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{4pt}{0ex}}\mathrm{iff}\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(B|A\right)*P\left(B\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{4pt}{0ex}}\mathrm{iff}\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(AB\right)*P\left(A\right)P\left(B\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}$
$P\left(AB\right)*P\left(A\right)P\left(B\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{4pt}{0ex}}\mathrm{iff}\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({A}^{c}{B}^{c}\right)*P\left({A}^{c}\right)P\left({B}^{c}\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{4pt}{0ex}}\mathrm{iff}\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(A{B}^{c}\right)\diamond P\left(A\right)P\left({B}^{c}\right)$

where $*\text{is}<,\le ,=,\ge ,\text{or}>$ and $\diamond \text{is}>,\ge ,=,\le ,\text{or}<$ , respectively.

Because of the role of this property in the theory of independence and conditional independence, we examine the derivation of these results.

VERIFICATION of (CP4)

1. $P\left(AB\right)*P\left(A\right)P\left(B\right)$ iff $P\left(A|B\right)*P\left(A\right)$ (divide by $P\left(B\right)$ — may exchange A and A c )
2. $P\left(AB\right)*P\left(A\right)P\left(B\right)$ iff $P\left(B|A\right)*P\left(B\right)$ (divide by $P\left(A\right)$ — may exchange B and B c )
3. $P\left(AB\right)*P\left(A\right)P\left(B\right)$ iff $\left[P\left(A\right)-P\left(A{B}^{c}\right)\right]*P\left(A\right)\left[1-P\left({B}^{c}\right)$ iff $-P\left(A{B}^{c}\right)*-P\left(A\right)P\left({B}^{c}\right)$ iff $P\left(A{B}^{c}\right)\diamond P\left(A\right)P\left({B}^{c}\right)$
4. We may use c to get $P\left(AB\right)*P\left(A\right)P\left(B\right)$ iff $P\left(A{B}^{c}\right)\diamond P\left(A\right)P\left({B}^{c}\right)$ iff $P\left({A}^{c}{B}^{c}\right)*P\left({A}^{c}\right)P\left({B}^{c}\right)$

$\square$

A number of important and useful propositons may be derived from these.

1. $P\left(A|B\right)+P\left({A}^{c}|B\right)=1$ , but, in general, $P\left(A|B\right)+P\left(A|{B}^{c}\right)\ne 1$ .
2. $P\left(A|B\right)>P\left(A\right)$ iff $P\left(A|{B}^{c}\right) .
3. $P\left({A}^{c}|B\right)>P\left({A}^{c}\right)$ iff $P\left(A|B\right) .
4. $P\left(A|B\right)>P\left(A\right)$ iff $P\left({A}^{c}|{B}^{c}\right)>P\left({A}^{c}\right)$ .

VERIFICATION — Exercises (see problem set)

$\square$

## Repeated conditioning

Suppose conditioning by the event C has occurred. Additional information is then received that event D has occurred. We have a new conditioning event $CD$ . There are two possibilities:

1. Reassign the conditional probabilities. ${P}_{C}\left(A\right)$ becomes
${P}_{C}\left(A|D\right)=\frac{{P}_{C}\left(AD\right)}{{P}_{C}\left(D\right)}=\frac{P\left(ACD\right)}{P\left(CD\right)}$
2. Reassign the total probabilities: $P\left(A\right)$ becomes
${P}_{CD}\left(A\right)=P\left(A|CD\right)=\frac{P\left(ACD\right)}{P\left(CD\right)}$

Basic result : ${P}_{C}\left(A|D\right)=P\left(A|CD\right)={P}_{D}\left(A|C\right)$ . Thus repeated conditioning by two events may be done in any order, or may be done in one step. This result extends easilyto repeated conditioning by any finite number of events. This result is important in extending the concept of "Independence of Events" to "Conditional Independence" . These conditions are important for many problems of probable inference.

#### Questions & Answers

a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
how did I we'll learn this
Noor Reply
f(x)= 2|x+5| find f(-6)
Prince Reply
f(n)= 2n + 1
Samantha Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
can nanotechnology change the direction of the face of the world
Prasenjit Reply
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.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive
Samson Reply

### Read also:

#### Get the best Algebra and trigonometry course in your pocket!

Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Applied probability' conversation and receive update notifications?

 By Anonymous User By Jordon Humphreys By