IF probability of occurance of A , infinitely often is 0, does it mean probability of occurance of A complement , infinitely often is 1???
The opposite of A occurring infinitely often is not
'not A' occurring infinitely often
but
'A' not occurring infinitely often
i.e.
'A' occurring only a finite number of times.
As Mark pointed out, the meaning of this for not A depends on whether you have a finite number of trials ('occurrences') or not:
finite number of trials: also not A can only occur a finite number of times
infinite number of trials: A only occurs for a finite number of trials, so not A occurs for the rest of the trials which are infinitely many (assuming only A and not A can occur in each trial).
No it doesn't.
If you have a finite number of trials then neither A nor complement A will happen an infinite number of times.
If you have an infinite number of trials then one or both of A and complement A must occur an infinite number of times. Furthermore if the probability of A remains constant then the only possibilities are:
A never occurs, complement A always occurs.
A always occurs, complement A never occurs.
Both A and complement A occur an infinite number of times.
This is interesting.
Try instead a finite number of trials.
We do 100 trials where the outcomes are A or A'. If the the probability of getting any As is zero then it follows that the probability of getting 100 A' is 1.
Extend to 1000 trials ... 1000(A) probability 0, 1000(a') probability 1.
Extend to infinity ...
How surprising, yes it appears that the answer is "yes".
Related
It is possible to print to several hundred decimal places a square root in bc, as it is in C. However in C it is only accurate to 15. I have checked the square root of 2 to 50 decimal places and it is accurate but what is the limit in bc? I can't find any reference to this.
To how many decimal places is bc accurate?
bc is an arbitrary precision calculator. Arbitrary precision just tells us how many digits it can represent (as many as will fit in memory), but doesn't tell us anything about accuracy.
However in C it is only accurate to 15
C uses your processor's built-in floating point hardware. This is fast, but has a fixed number of bits to represent each number, so is obviously fixed rather than arbitrary precision.
Any arbitrary precision system will have more ... precision than this, but could of course still be inaccurate. Knowing how many digits can be stored doesn't tell us whether they're correct.
However, the GNU implementation of bc is open source, so we can just see what it does.
The bc_sqrt function uses an iterative approximation (Newton's method, although the same technique was apparently known by the Babylonians in at least 1,000BC).
This approximation is just run, improving each time, until two consecutive guesses differ by less than the precision requested. That is, if you ask for 1,000 digits, it'll keep going until the difference is at most in the 1,001st digit.
The only exception is when you ask for an N-digit result and the original number has more than N digits. It'll use the larger of the two as its target precision.
Since the convergence rate of this algorithm is faster than one digit per iteration, there seems little risk of two consecutive iterations agreeing to some N digits without also being correct to N digits.
I am reading the Rabin-Karb algorithm from Sedgewick. The book says:
We use a random prime Q taking as large a value as possible while
avoiding overflow
At first reading I didn't notice the significance of random and when I saw that in the code a long is used my first thoughts were:
a) Use Eratosthene's sieve to find a big prime that fits a long
or
b) look up from a list of primes any prime large enough that is greater than int and use it as a constant.
But then the rest of the explanation says:
We will use a long value greater than 10^20 making the probability
that a collision happens less than 10^-20
This part got me confused since a long can not fit 10^20 let alone a value greater than that.
Then when I checked the calculation for the prime the book defers to an exercise that has just the following hint:
A random n-digit number is prime with probability proportional to 1/n
What does that mean?
So basically what I don't get is:
a) what is the meaning of using a random prime? Why can't we just pre-calculate it and use it as a constant?
b) why is the 10^20 mentioned since it is out of range for long?
c) How is that hint helpful? What does it mean exactly?
Once again, Sedgewick has tried to simplify an algorithm and gotten the details slightly wrong. First, as you observe, 1020 cannot be represented in 64 bits. Even taking a prime close to 263 − 1, however, you probably would want a bit of room to multiply the normal way without overflowing so that the subsequent modulo is correct. The answer uses a 31-bit prime, which makes this easy but only offers collision probabilities in the 10−9 range.
The original version uses Rabin fingerprints and a random irreducible polynomial over 𝔽2[x], which from the perspective of algebraic number theory behaves a lot like a random prime over the integers. If we choose the polynomial to be degree 32 or 64, then the fingerprints fit perfectly into a computer word of the appropriate length, and polynomial addition and subtraction both work out to bitwise XOR, so there is no overflow.
Now, Sedgewick presumably didn't want to explain how polynomial rings work. Fine. If I had to implement this approach in practice, I'd choose a prime p close to the max that was easy to mod by with cheap instructions (I'm partial to 231 − 227 + 1; EDIT actually 231 − 1 works even better since we don't need a smooth prime here) and then choose a random number in [1, p−1] to evaluate the polynomials at (this is how Wikipedia explains it). The reason that we need some randomness is that otherwise the oblivious adversary could choose an input that would be guaranteed to have a lot of hash collisions, which would severely degrade the running time.
Sedgewick wanted to follow the original a little more closely than that, however, which in essence evaluates the polynomials at a fixed value of x (literally x in the original version that uses polynomial rings). He needs a random prime so that the oblivious adversary can't engineer collisions. Sieving numbers big enough is quite inefficient, so he turns to the Prime Number Theorem (which is the math behind his hint, but it holds only asymptotically, which makes a big mess theoretically) and a fast primality test (which can be probabilistic; the cases where it fails won't influence the correctness of the algorithm, and they are rare enough that they won't affect the expected running time).
I'm not sure how he proves a formal bound on the collision probability. My rough idea is basically, show that there are enough primes in the window of interest, use the Chinese Remainder Theorem to show that it's impossible for there to be a collision for too many primes at once, conclude that the collision probability is bounded by the probability of picking a bad prime, which is low. But the Prime Number Theorem holds only asymptotically, so we have to rely on computer experiments regarding the density of primes in machine word ranges. Not great.
I have two related question on population statistics. I'm not a statistician, but would appreciate pointers to learn more.
I have a process that results from flipping a three sided coin (results: A, B, C) and I compute the statistic t=(A-C)/(A+B+C). In my problem, I have a set that randomly divides itself into sets X and Y, maybe uniformly, maybe not. I compute t for X and Y. I want to know whether the difference I observe in those two t values is likely due to chance or not.
Now if this were a simple binomial distribution (i.e., I'm just counting who ends up in X or Y), I'd know what to do: I compute n=|X|+|Y|, σ=sqrt(np(1-p)) (and I assume my p=.5), and then I compare to the normal distribution. So, for example, if I observed |X|=45 and |Y|=55, I'd say σ=5 and so I expect to have this variation from the mean μ=50 by chance 68.27% of the time. Alternately, I expect greater deviation from the mean 31.73% of the time.
There's an intermediate problem, which also interests me and which I think may help me understand the main problem, where I measure some property of members of A and B. Let's say 25% in A measure positive and 66% in B measure positive. (A and B aren't the same cardinality -- the selection process isn't uniform.) I would like to know if I expect this difference by chance.
As a first draft, I computed t as though it were measuring coin flips, but I'm pretty sure that's not actually right.
Any pointers on what the correct way to model this is?
First problem
For the three-sided coin problem, have a look at the multinomial distribution. It's the distribution to use for a "binomial" problem with more then 2 outcomes.
Here is the example from Wikipedia (https://en.wikipedia.org/wiki/Multinomial_distribution):
Suppose that in a three-way election for a large country, candidate A received 20% of the votes, candidate B received 30% of the votes, and candidate C received 50% of the votes. If six voters are selected randomly, what is the probability that there will be exactly one supporter for candidate A, two supporters for candidate B and three supporters for candidate C in the sample?
Note: Since we’re assuming that the voting population is large, it is reasonable and permissible to think of the probabilities as unchanging once a voter is selected for the sample. Technically speaking this is sampling without replacement, so the correct distribution is the multivariate hypergeometric distribution, but the distributions converge as the population grows large.
Second problem
The second problem seems to be a problem for cross-tabs. Then use the "Chi-squared test for association" to test whether there is a significant association between your variables. And use the "standardized residuals" of your cross-tab to identify which of the assiciations is more likely to occur and which is less likely.
I am confused about the concept of p-value. In general, if the p-value is greater than alpha which is generally 0.05, we are fail to reject null hypothesis and if the p-value is less than alpha, we reject null hypothesis. As I understand, if the p-value is greater than alpha, difference between two group is just coming from sampling error or by chance.So far everything is okay. However, if the p-value is less than alpha, the result is statistically significant, I was supposing it to be statistically nonsignificant ( because, in case p-value is less than alpha we reject null hypothesis).
Basically, if result statistically significant, reject null hypothesis. But, how a hypothesis can be rejected, if it is statistically significant? From the word of "statistically significant", I am understanding that the result is good.
You are mistaking what the significance means in terms of the p-value.
I will try to explain below:
Let's assume a test about the means of two populations being equal. We will perform a t-test to test that by drawing one sample from each population and calculating the p-value.
The null hypothesis and the alternative:
H0: m1 - m2 = 0
H1: m1 - m2 != 0
Which is a two-tailed test (although not important for this).
Let's assume that you get a p-value of 0.01 and your alpha is 0.05. The p-value is the probability of the means being equal when sampling from the two populations (m1 and m2). This means that there is a 1% probability that the means will be equal or in other words only 1 out of 100 sample pairs will have a mean difference of 0.
Such a low probability of the two means being equal makes us confident (makes us certain) that the means of the populations are not equal and thus we consider the result to be statistically significant.
What is the threshold that makes us think that a result is significant? That is determined by the significance level (a) which in this case is 5%.
The p-value being less than the significance level is what makes us think that the result is significant and therefore we are certain that we can reject the null hypothesis since the probability of the NULL hypothesis being true is very low.
I hope that makes sense now!
Let me make an example that I often use with my pupils, in order to explan the concepts of null hypothesis, alpha, & significance.
Let's say we're playing a round of Poker. I deal the cards & we make our bets. Hey, lucky me! I got a flush on my first hand. You curse your luck and we deal again. I get another flush and win. Another round, and again, I get 4 aces: at this point you kick the table and call me a cheater: "this is bs! You're trying to rob me!"
Let's explain this in terms of probability: There is a possibility associated with getting a flush on the first hand: anyone can get lucky. There's a smaller probability of getting too lucky twice in a row. There is finally a probability of getting really lucky three times in a row. But for the third shot, you are stating: "the probability that you get SO LUCKY is TOO SMALL. I REJECT the idea that you're just lucky. I'm calling you a cheater". That is, you rejected the null hypothesis (the hypothesis that nothing is going on!)
The null hypothesis is, in all cases: "This thing we are observing is an effect of randomness". In our example, the null hypothesis states: "I'm just getting all these good hands one after the other, because i'm lucky"
p-value is the value associated with an event, given that it happens randomly. You can calculate the odds of getting good hands in poker after properly shuffling the deck. Or for example: if I toss a fair coin 20 times, the odss that I get 20 heads in a row is 1/(2^20) = 0.000000953 (really small). That's the p-value for 20 heads in a row, tossing 20 times.
"Statistically significant", means "This event seems to be weird. It has a really tiny probability of happening by chance. So, i'll reject the null hypothesis."
Alpha, or critical p-value, is the magic point where you "kick the table", and reject the null hypothesis. In experimental applications, you define this in advance (alpha=0.05, e.g.) In our poker example, you can call me a cheater after three lucky hands, or after 10 out of 12, and so on. It's a threshold of probability.
okay for p-value you should at least know about null hypothesis and alternate hypothesis
null hypothesis means take an example we have 2 flowers and it is saying there is no significant difference between them
and alternate hypothesis is saying that there is significant difference between them
and yes what is the significant value for p- value most of the data scientist take as 0.05 but it is based on researches(value of level of significant)
0.5
0.05
0.01
0.001
can be taken as p-value
okay now p-value is taken by you but what to do next
if your model p-value is 0.03 and significant value you have taken 0.05 so you have to reject null hypothesis means there is significant difference between 2 flowers or simple as stated
p-value of your model < level of significant than reject it
and your model p-value is >level of significant than null hypothesis is going to accept.
Generally speaking when you are numerically evaluating and integral, say in MATLAB do I just pick a large number for the bounds or is there a way to tell MATLAB to "take the limit?"
I am assuming that you just use the large number because different machines would be able to handle numbers of different magnitudes.
I am just wondering if their is a way to improve my code. I am doing lots of expected value calculations via Monte Carlo and often use the trapezoid method to check my self of my degrees of freedom are small enough.
Strictly speaking, it's impossible to evaluate a numerical integral out to infinity. In most cases, if the integral in question is finite, you can simply integrate over a reasonably large range. To converge at a stable value, the integral of the normal error has to be less than 10 sigma -- this value is, for better or worse, as equal as you are going to get to evaluating the same integral all the way out to infinity.
It depends very much on what type of function you want to integrate. If it is "smooth" (no jumps - preferably not in any derivatives either, but that becomes progressively less important) and finite, that you have two main choices (limiting myself to the simplest approach):
1. if it is periodic, here meaning: could you put the left and right ends together and the also there have no jumps in value (and derivatives...): distribute your points evenly over the interval and just sample the functionvalues to get the estimated average, and than multiply by the length of the interval to get your integral.
2. if not periodic: use Legendre-integration.
Monte-carlo is almost invariably a poor method: it progresses very slow towards (machine-)precision: for any additional significant digit you need to apply 100 times more points!
The two methods above, for periodic and non-periodic "nice" (smooth etcetera) functions gives fair results already with a very small number of sample-points and then progresses very rapidly towards more precision: 1 of 2 points more usually adds several digits to your precision! This far outweighs the burden that you have to throw away all parts of the previous result when you want to apply a next effort with more sample points: you REPLACE the previous set of points with a fresh new one, while in Monte-Carlo you can just simply add points to the existing set and so refine the outcome.