# Tag Info

85

First of all, you must understand that there is no such thing as a perfectly fair coin, because there is nothing in the real world that conforms perfectly to some theoretical model. So a useful definition of a "fair coin" is one, that for practical purposes behaves like fair - that is, no human flipping it for even a very long time, would be able to tell a ...

31

If you take a coin you have modified so that it always lands in heads and you get 1000 heads then the probability of it being unfair is 100%. If you take a coin you have crafted yourself and carefully made sure that it is a fair coin and then you get 1000 heads then the probability of it being unfair is 0%. Next, you fill a box with coins of both types, ...

14

In order to assign a probability to this event you need to start with a prior probability which can then be updated based on the data. Probabilities can't really be derived from experience alone, you need to start with some sort of "inductive bias" in order to draw conclusions from evidence. The heuristic approach of hypothesis testing gives a framework ...

9

From the Langford&Langford paper, suppose you have heard $m$ songs, of which $i$ were different, and $i<m$. You wish to find the size of the playlist, whose most likely value (in the maximum likelihood sense) is $\hat{N}$, which is given by $$\hat{N}=\left\lfloor \frac{1}{1-y^\star} \right\rfloor$$ where $\lfloor \cdot \rfloor$ denotes the floor ...

6

In my comment, posted this link to a more thorough treatment of the general question being asked. However, I will directly answer the question using the attained result on posterior density functions, as discussed in the linked article. In particular, we note that if $r$ is the actual probability that the coin will land on "heads" rather than "tails", then ...

4

For "almost all" priors, the answer is 100%: you have to specify an interval of "fairness" in order to a nonzero value. Why? Because the bias $p$ can be any real number between $0$ and $1$, so the probability that it is a very specific real number is $1/\infty = 0$; hence the probability that it is $1/2$ is 0. However, what I can tell you is that after your ...

4

As @dsaxton suggests, one method is 'capture-recapture' (also called 'mark-recapture'). See the Wikipedia article for some technical details (if you can deal with the needlessly confusing notation). Here is the basic method with an intuitive argument for it. Method. Listen to $c = 20$ distinct songs, noting their titles. Then listen to another $r = 20$ ...

4

I'm not sure I understand what you are trying to say. There is no medium in stats. A medium might be a size of your soft drink at a fast food place or possibly someone who can communicate with the spirit world, but I can't think of any standard meaning it would have in statistics. Median on the other hand does have a definition but seems to have little to ...

3

You seem to be quite confused about the use of the median. Typically, you find the median of a set of numbers, not a single number. Your initial example has you finding the median of four apples, which doesn't really make sense. Imagine, instead, you have a set of buckets containing apples. Suppose you have four buckets, containing 1, 2, 3, and 4 apples. The ...

3

Comment on simulation results for method based on time to first repeat. Method: Let $Y$ be the wait (number of songs) until the first repeat. Then estimate the size $N$ of the playlist as $\hat N = Y(Y-1)/2.$ This certainly satisfies OP's request for computational simplicity. Some simulation results: Based on 100,000 iterations for each of ten values of ...

3

If they are numbered $0$ to $n$, that's $n+1$ numbers so there's one missing. To make the problem nicer, I'll suppose they are numbered $1$ to $n$. A statistic $L$ is an upper limit for $n$ with confidence level $p$ if for every possible $n$, $P(L \ge n) \ge p$. In this case if $M$ is the maximum number observed and $L = c M$ where $c > 1$ is a ...

3

$X_n$ is an $(\mathcal{F}_n)$-martingale: $$\mathbb{E}[X_k(X_m-X_l)]=\mathbb{E}[X_k\mathbb{E}[(X_m-X_l)|\mathcal{F}_k]]$$ $$=\mathbb{E}[X_k(X_k-X_k)]=0$$

2

In every answer, you have $\frac15$ chance to guess the correct answer. You need that chance to occur all the $50$ time, and since all of your answers are independent events, you need to multiply these probabilties, therefore your chance: $(\frac{1}{5})^{50}=\frac{1}{5^{50}}$, which is not too big, you should learn for your exams. :)

2

A A A A ^ | At that midpoint, how many complete A's are on the left? 2. On the right 2. You do not have 0.5 A's at the midpoint. ie | A | A | A | A | +---+---+---+---+ 0 1 2 3 4 The line is now measuring (the width of) the A's. The midpoint is at 2. | A | A | A | A | --+---+---+---+ 1 2 3 4 This is NOT how you should draw the ...

2

Consider any $\pi\in (0,1)^I$ with $\sum_{i\in I}\pi_i=1$. Then $$\sum_{i\in I} p_i\ln p_i-\sum_{i\in I} p_i\ln \pi_i=-\sum_{i\in I} p_i\ln \frac{\pi_i}{p_i}$$ $$\ge -\sum_{i\in I} p_i\ln \frac{\pi_i}{p_i}\ge -\ln\left[\sum_{i\in I} p_i\frac{\pi_i}{p_i}\right]=-\ln\left[\sum_{i\in I}\pi_i\right]=0$$ This inequality shows that $$\max_{\pi} \sum_{i\in ... 2 You can only get integer counts and cannot always get counts that will be exactly 5% above and 5% below the mean. So Because P(27.61 ≤ X ≤38.6) = P(27.61 ≤ X < 28) + P(28 ≤ X ≤ 38) + P(38 < X ≤ 38.61) = 0 + P(28 ≤ X ≤ 38) + 0 = P(28 ≤ X ≤ 38), you just add up the probabilities of getting 28 through 38. That is the probability of being within 5 percent ... 2 Z=(X,Y)' is not necessary bivariate normal (it works if X and Y are independent). There are lots of counterexamples (e.g. link). Another example: \psi_1 and \psi_2 are independent standard normal r.v.s and$$(X, Y)=(\psi_1,|\psi_2|)1\{\psi_1\ge 0\}+(\psi_1,-|\psi_2|)1\{\psi_1< 0\}$$which is not bivariate normal. However, marginals X and Y ... 2 I think the confusion comes from translating the hypotenuse. Let's be explicit about it. Suppose your original triangle is made by intersecting the lines y = 0, x = 0, and y = -x + 10. Two of those lines are easy to translate, so the "good" triangle has borders along y = 2 and x = 2. the hypotenuse is more subtle. The translated hypotenuse has slope -1 ... 2 There are indeed 24 ways to return to (0,0) if we move in all four directions, but we can also get there moving only north and south or only east and west. Each possibility gives us \binom{4}{2}=6 options. Hence there are 24+12=36 ways to return to (0,0) in 4 moves. On the other hand there are 4^4 possible movement sequences. Final answer is ... 1 If I understand correctly, the question is about sample variance. Here is a one-dimensional example. We have two samples, 1,0,0,0 and the subsample 1,0. The first sample variance is$$\frac{1}{4}\left(\frac{9}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}\right).$$This is \frac{3}{16}. The variance of the subsample is ... 1 See to answer this one you need to consider (p,q) and (q,p) as different results but (p,p) are considered as same results Then we need to think how many possibilities are there if we throw red and green die simultaneously.There are 36 possibilities as follows: (1,1),(1,2),(1,3)....(1,6)  (2,1)....(2,6) ... (6,1)....(6,6) then for getting ... 1 Tabulating the difference is not too onerous, as your sample space is 36. Assuming the columns are the result of the Red die throw ordered 1,2,3,4,5,6, and the rows the results of the Green die, also ordered 1,2,3,4,5,6 (or vice versa), one obtains$$ \left( \begin{array}{cccccc} 0 & 1 & 2 & 3 & 4 & 5 \\ 1 & 0 & 1 & 2 ...

1

Going north at the first step, there is 9 possible paths NSNS, NSEW, NSSN, NSWE, NEWS, NESW, NWES, NWSE, NNSS So in total there is 4*9 = 36 possible paths to get back at the origin after 4 hop

1

Consider the multisets $A=\{\!\{0,2,2,3\}\!\}$ and $B=\{\!\{0,2,2,3,5\}\!\}$, both with mode $2$. Then $$A+B=\{\!\{0,2,2,2,2,3,3,4,4,4,4,5,5,5,5,5,6,7,7,8\}\!\}\;,$$ whose mode is $5$.

1

Your idea is substantially correct. I suppose the text intended you to consider the binomial distribution of the number $X$ of successes that would give $\hat p = X/100$ in the desired range. Then using the approximate normal distribution $Norm(\mu = 24, \sigma = \sqrt{100(.24)(1-.24)}),$ with the continuity correction, gives a slightly more precise ...

1

The issue is that you confuse two problems. You have $4$ apples, but can you choose to have $0$ apple or not? In this case, the value you can choose from are $\left\{0,1,2,3,4\right\}$. In the other case, you can choose from $\left\{1,2,3,4\right\}$...

1

the integral's lower limit is $1$ not $-\infty$. With that you will get the correct cdf.

1

There are $\frac{12!}{2!^6}$ anagrams of the string 112233445566 and $6^{12}$ possible outcomes. What else do you need?

1

If you want to estimate some functional of the original distribution, then you can always apply the same functional to an estimate of that distribution (the "plug-in principle"). In the case of moments, this would just amount to integrating the EDF, or treating the jumps as point masses and calculating the moments of this discrete distribution.

1

You asked if this specific string of flips happens, what is the probability the coin is unfair? And it seems to me that most answer here are answering the question if the coin was fair, what is the probability of this specific string of flips happening? If I try to literally answer your question, I get stuck unless we make additional ...

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