Testing Numbers for Randomness Chi Square test
- 1. TESTING NUMBERS FOR RANDOMNESS CHI SQUARE TEST SUBMITTED TO: MS. PRIYANKA UIET, CSE PANJAB UNIVERSITY, CHANDIGARH PRESENTATION BY: UE213070 RACHIT SHARMA UE213071 RAGHAV GOEL UE213072 RAHUL RAJ UE213073 RIDHAM GUPTA
- 2. RECAP: Up until now what we have studied is:- ● What are random numbers? ● Importance of Random numbers ● Pseudo Random numbers ● Different methods to generate Random numbers:- ○ Mid-square random number generator ○ Residue method ○ Arithmetic Congruential Generator
- 3. INTRODUCTION In this module we will learn about how to test whether a given a sequence of numbers is random or not. For this we will firstly refer to the two important properties of random numbers which are: ● Uniformity- This means that the method from which random number is generated, every number has an equal chance of occurrence. ● Independence- This means that there is no correlation between the numbers present in the sequence. Eg. 0,2,4,6,8 The given above sequence is uniform but it is not independent because as we can see that the given sequence is of even numbers and we can predict the next number.
- 4. So for testing that whether a given sequence of numbers are random or not we have to check these two properties. So, test for uniformity is performed to validate a random number generator. These tests compare the generated random numbers with the theoretical uniform distribution. The two uniformity/frequency test we perform are:- ● Kolmogorv-Smirnov Test ● Chi-Square Test For testing independence we perform chi square test for serial autocorrelation.
- 17. CHI-SQUARED TEST • The Chi-Squared Test uses the sample statistic Where Oi is the observed number in the ith class, Ei is the expected number in the ith class and n is the number of classes. For the uniform distribution, Ei, the expected number in each class is given by For equally spaced classes, where N is the total number of observations. It can be shown that the sampling distribution of x2 is approximately the chi-square distribution with n-1 degree of freedom.
- 18. • This test involves the classification of 500 random numbers b/w 0 and 1 into 10 equal intervals, that is numbers less than or equal to 0.1 less than or equal to 0.2, 0.3, ...., 1.0. • A bar chart or histogram can than be plotted to illustrate the uniformity of distribution, If the 500 numbers were distributed among the 10 classes, with perfect uniformity, all bars will be of equal length of 50 numbers. • Chi-square is a characteristic of the distribution which is a measure of its randomness. • The statistic Chi-Square is computed by subtracting the number of random numbers in each class from the expected number(that is 50), squaring the difference, adding the squares for the ten classes, and dividing the sum by the expectation(50). • For example, if 47, 49, 55, 54, 46, 53, 51, 43, 46 and 54 are the numbers in 10 classes. Then their differences from 50 are 3, 1, 5, 4, 4, 3, 1, 7, 4, and 4; and sum of the squares of differences is 9 + 1 + 25 + 16 + 16 + 9 + 1 + 49 + 16 + 16 = 158 Chi- square = 158/50 = 3.16
- 19. • There are tables of Chi-square that will tell us about the goodness of the results. The acceptable value of Chi-square will depend upon the degrees of freedom which are one less than the number of classes, and the level of confidence we wish to place in our results. • In the present case, since there are 10 classes in which we have divided the random numbers, there are 9 degrees of freedom. At 95% confidence level the acceptable value of Chi-square for 9 degrees of freedom is 16.919 . • This means that when the value of Chi-square is less than or equal to 16.9, there are only 5/chances out of 100 that our results are wrong. Chi-Squared Test can be employed to compare different sets of random numbers or different random number generators. For each set of random numbers the statistics Chi- square is computed The one with smaller value of Chi-square is more uniformly distributed
- 20. CHI-SQUARED TEST - UNIFORMITY- RANDOM NUMBERS
- 21. CHI-SQUARED TEST - UNIFORMITY- RANDOM NUMBERS
- 22. CHI-SQUARED TEST - UNIFORMITY- RANDOM NUMBERS
- 23. CHI-SQUARED TEST – TESTING FOR AUTOCORRELATION • The uniformity test of random numbers is only a necessary test for randomness, not a sufficient one. A sequence of numbers may be perfectly uniform, and still not random. • For example, the sequence .1, .2, .3, .4., .1, .2, .3, ....., would give a perfectly uniform distribution with the chi-square value as zero. But the sequence can by no means be regarded as random. The numbers are not independent, as the occurrence of one number say .3 decides the next, which is to be .4 etc. This defect is called 'serial autocorrelation of adjacent pairs of indes the ne • The Chi-Squared test for serial autocorrelation makes use of a 10 x 10 matrix (checker board). The 10 classes described in the uniformity test are represented both along the rows and columns. If the classes are to be represented on a bar chart, 100 bars, one for each cell of the matrix, will be required. To reduce the number of groups, instead of 10, random numbers are divided into smaller number of classes as 3, or 4. Three classes will be: less than or equal to 0.33, less than or equal to 0.67 and less than or equal to 1.0. With three classes in rows and three in columns, there will be 9 cells
- 24. Let us consider the following random numbers : These 73 random numbers giving 72 pairs, are grouped into 9 classes with expectation of 8 in each group. Chi-square = 24/8 = 3.0
- 25. • The counts in different classes have been determined by taking the pairs of random numbers. • Pair .49 and .95 falls in class R1 <= 67 and R2 <= 1.0 Then the next pair is .95 and .82 which falls in class R1 <= 1.0 and R2 <= 1.0, Pair .82 and .19 falls in class R1 <= 1.0 and R2 <= 0.33 and so on. • Since the total number of pairs in 72, one less than the number of random numbers, the expectation is 8, that is 8 in each class. Then squares of differences (frequency - expectation) are determined and their sum is obtained, which on division by expectation gives the value of Chi-square as 3.0. • In this case there are two variables R1 and R2 and hence the degrees of freedom are nine minus two that is seven. The criterion value of x² (chi-square) for seven degrees of freedom at 95% confidence level is 14.067. The value of Chi-square obtained for the given set of random number is well within the acceptable limit, and hence, they are not serially autocorrelated.System
- 26. EXAMPLE. GIVEN BELOW IS A SEQUENCE OF RANDOM NUMBERS. PERFORM THE CHI- SQUARED TESTS TO CHECK THE NUMBERS FOR UNIFORM DISTRIBUTION AND SERIAL AUTOCORRELATION
- 27. • Uniformity Test i. For checking the random numbers for the uniformity, we will divide these in 10 class and take 90 random numbers to have the expected value in each class as 9. ii. There are ten classes and one variable, giving 9 degrees of freedom. For 9 degrees of freedom at 95% confidence level, the acceptable value of x^{2} is up to 16.9. Our value of x^{2} is well within the acceptable limit and hence, the random numbers in the given sequence are uniformly distributed
- 28. • Chi-Squared Test for Autocorrelation i. For this test, the overlapping pairs of random numbers are to be taken. Taking three classes; 33; 34 to 67 and 68 to 100 for each random number in the pair, the number of pairs in each is is obtained as follows. Since there are 9 classes, the expectation value is 10 The criterion value of x² for 7 degrees of freedom at 95% confidence level is 14.1, which is less than the value we have obtained for the given sequence. Thus, the given sequence of random numbers is serially autocorrelated at 95% confidence level.