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So the models are $H_0$: always get 0, and $H_1$: obtain 0 with probability 1/2 or 1 with probability 1/2. your question is equivalent to can we test whether a coin has heads on each side by tossing it many times. It’s pretty clear that we have pretty good power to reject $H_0$ when it’s false with only 5 or so flips.


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You may want to calculate the independence between two variables, A and B (test of independence) or if the distribution of A given B=B1 (first column) fits the distribution of A given B=B2 (second column). That is if P(A|B=B1)=P(A|B=B2). The two calculations are slightly different, the test of independence takes into account the differences of both ...


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The result is called Craig's theorem. Named after the 1943 note Note on the Independence of Certain Quadratic Forms by Craig. See this paper for details and history.


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To prove the independence and dependence of those. You may use the moment generating function (MGF). Hints: in page 10, of this document: https://ocw.mit.edu/courses/mathematics/18-443-statistics-for-applications-spring-2015/lecture-notes/MIT18_443S15_LEC1.pdf


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"Anti-lingering" Comment: Your stated contingency table is TBL as below: TBL = rbind(c(13,19,28,40), c(7,11,22,60)) TBL [,1] [,2] [,3] [,4] [1,] 13 19 28 40 [2,] 7 11 22 60 Verifying that both rows sum to $100,$ as stated: rowSums(TBL) [1] 100 100 A chi-squared test in R, shows significant departure from homogeneity at 5% ...


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To answer the question about correlation, we can note that $\frac{X_{(1)}}{X_{(n)}}=\frac{X_{(1)}/\theta}{X_{(n)}/\theta}=\frac{Y_{(1)}}{Y_{(n)}}$ where $Y_1,\ldots,Y_n$ are i.i.d uniform on $(0,1)$. As such the distribution of the ratio is free of $\theta$, hence it is an ancillary statistic. Moreover $X_{(n)}$ is a complete sufficient statistic for the $U(...


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You need the joint distribution of $X_{(1)}$ and $X_{(n)}$. see here for the derivation in the case of Unif(0,1). The joint density is $$f_{X_{1},X_{n}}(u,v)=n! \frac{(v-u)^{n-2}}{(n-2)!} \theta^{-n}$$ You can find the expected value of any function of the two random variables by integrating the function of $u$ and $v$ over the two-dimensional integral with ...


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You could try independent component analysis, see Making sense of independent component analysis. More information.


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So a game has been played 1554 times and you wonder what share of these uses happened by a good and how many by a bad. This can be done via a binomial test. Let's say we observed the game being played by good 1254 times and by bad 300 times. binom.test says > binom.test(c(1254, 300)) Exact binomial test data: c(1254, 300) number of successes = 1254,...


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For efficiency, use characteristic functions. Let $|\cdot|$ denote the determinant. The following demonstration repeatedly evaluates the multinormal ($n$ dimensional) integral in the form $$(2\pi)^{n/2}|T|^{n/2} = \int\cdots\int \exp\left(-\mathbf{x}\,T\,\mathbf{x}^\prime/2\right)\,\mathrm{d}^n\mathrm{x}$$ where $T$ is any symmetric matrix (with real or ...


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Let's first tackle dependence versus correlation and then we'll add causation into the mix. There are exceptions to almost every rule, some of them worth only a minimal footnote and you should start worrying about those exceptions only after you've been exposed to what is true as a rule. Correlation is one type of dependence. Dependence is a broader concept. ...


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Suppose you have $n = 1000$ observations from $\mathsf{Pois}(\lambda = 5).$ [Note: in some circumstances, one might estimate $\lambda$ directly from these data, as 4.968, but that estimate is not valid for our test whether the data fit a Poisson distribution.] set.seed(2021) x = rpois(1000, 5) table(x) x 0 1 2 3 4 5 6 7 8 9 10 11 12 ...


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This is an interesting question. You cannot check this assumption before analysis because the changepoints make the data non-stationary and so standard stationary methods for assessing independence/correlation are no longer valid. The best way is to perform a changepoint analysis and then look at the residuals (remove the time varying mean/variance) to see ...


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Your confusion may stem from the definitions of "outcome" vs "event", and from the difference between running a single random experiment vs several independent random experiments. A random experiment is an experiment whose outcome is uncertain, in the sense that if the experiment is repeated, the outcome can change. In your example, the ...


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Depends on what you mean by probability. If you are a frequentist, then conditional probability P(A|B) means something akin to how much B causes A. If you are Bayesian, P(A|B) only measures the logical connection between B and A. To sum up, for a Bayesian, the answer to your question is yes, for a frequentist, it is no. For ones, who are a mix of both, it is ...


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No, not in general. A covariance of zero between two random variables does not necessarily imply that they are independent. However, the statement is true if the variables are normally distributed.


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The following is a counterexample Say the following events (for binary values of X, Y and Z) have equal probabilities. X Y Z probability 0 1 1 1/4 1 0 1 1/4 0 0 0 1/4 1 1 0 1/4 Then $P(X=1) = P(X=0) = 0.5$ and $P(Y=1) = P(Y=0) = 0.5$ independent from $Z$. But $P(X| Y,Z) = \text{XOR}(Y,Z)$ and not independent from $Y,Z$ (where $\text{XOR}$ refers ...


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You argued with ... The question was how many students have a sister, not at least one sister. The term “at least one” means “1 or more”, so: If someone has a sister, it has indeed at least one sister. Reversely, if someone has at least one sister, he indeed has a sister. So these two statements are equivalent, and the result 120/300 is correct.


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Yes, 40 is included because those students have at least one sister; naturally, they're in the set of students that has at least one sister (i.e. 120 students). This means, there are 80 students with at least one sister and no brothers.


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