# Tag Info

5

There's no unique solution I don't think that true discrete probability distribution can be recovered, unless you make some additional assumptions. Your situation is basically a problem of recovering the joint distribution from marginals. It is sometimes solved by using copulas in the industry, for example financial risk management, but usually for ...

4

Your question does not make this clear, but I'm going to assume that the bombs are initially distributed via simple-random-sampling without replacement over the cells (so a cell cannot contain more than one bomb). The question you have raised is essentially asking for the development of an estimation method for a probability distribution that can be ...

4

In a comment I indicated how the answer can be found from knowledge of the Non-central chi-squared distribution. Here I will sketch how to find it from the definition. Let $X$ be a random variable. Recall that the characteristic function of any random variable $g(X)$ is, by definition, a function of a real variable $t$ given by \phi_{g(X)}(t) = E\left[...

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They're not independent. Intuitively speaking, if $X^2=0$ , $XY$ must be $0$, so the two have a dependence. More formally, There are plenty of other ways to do it, but I'll focus on a simple contradiction. Let $Z=XY,W=X^2$, then we are asking if $Z$ and $X$ are independent. If they are, we should have $\operatorname{var}(Z|W=w)=\operatorname{var}(Z)=\... 3 The solution space (valid bomb configurations) can be viewed as the set of bipartite graphs with given degree sequence. (The grid is the biadjacency matrix.) Generating a uniform distribution on that space can be approached using Markov Chain Monte Carlo (MCMC) methods: every solution can be obtained from any other using a sequence of "switches," which in ... 2 The tickets-in-a-box model of random variables described at https://stats.stackexchange.com/a/54894/919 provides a helpful way to think about this. Imagine you have a box full of tickets on which are written various numbers in such a way that a blind draw of one ticket acts like observing$X.W$is a second number found on every ticket: half the ... 2 This is not a standard situation because it concerns unordered pairs. It needs a model and some analysis. The model describes the state pairs when there is no association between states. One plausible and flexible model supposes each state$s$is associated with a constant but unknown probability$\pi_s$. (We introduce these probabilities, and allow them ... 1 Imagine a plot of the joint distribution of$X,Y$. The condition$X+Y <a$can be depicted by a diagonal line$y=a-x$as a boundary. The points below it satisfy the condition. Second image Next, consider the marginal distribution by summing all the distributions of$X$for the various values of$Y$. This will be a sum of right truncated distributions$ ...

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"IID" means independent and identically distributed. Not all two-day returns are independent; but they are all identically distributed. They are identically distributed because the independence of the one-day returns $r_1(t)$ implies the bivariate random variables $(r_1(t-1), r_1(t))$ all have the same (2D) distributions: they have the same marginals (due ...

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