20

The problem is due to the fact that the radius is not uniformly distributed. Namely, if $(X,Y)$ is uniformly distributed over $$\left\{ (x,y);\ x^2+y^2\le 1\right\}$$ then the (polar coordinates) change of variables $$R=(X^2+Y^2)^{1/2}\qquad A=\text{sign}(Y)\arccos(X/R)$$ has the density $$\frac{1}{\pi} \mathbb{I}_{(0,1)}(r)\left|\frac{\text{d}(X,Y)}{\text{d}...


14

The simplest and least error-prone approach would be rejection sampling: generate uniformly distributed points in the square around your circle, and only keep those that are in the circle. nn <- 1e4 radius <- 1 set.seed(1) # for reproducibility foo <- cbind(runif(nn,-radius,radius),runif(nn,-radius,radius)) plot(foo[rowSums(foo^2)<radius^2,],pch=...


6

You can find the mathematics of this situation in a related question here. The method is set out in Xi'an's excellent answer, and it can be summarised by the following requirements: $$\begin{matrix} R^2 \sim \text{U}(0,1) \quad \ \ & & & X = R \cos (\theta), \\[6pt] \theta \sim \text{U}(0, 2\pi) & & & Y = R \sin(\theta). \\[6pt] \end{...


3

The figure (below) answers the question. It reveals a complication: when the distribution is not continuous at $X=a,$ we have to take care not to include the chance of the event $X=a\lt Y$ in our calculation. This is done by "sneaking up" to the answer as a limit. Therein lies the interest in this question. By definition, the joint probability ...


1

Yes. Maybe start with $n=9$ to test $H_0=0$ vs. $H_1=1,$ rejecting if $\bar X_9≥0.5.$ Under $H_0,\bar X_9\sim\mathsf{Norm}(0,σ=1/3).$ Find, $α=P(\bar X_9≥.5|H_0).$ Under $H_1,\; \bar X_9\sim\mathsf{Norm}(1,1/3).$ Find $β=P(\bar X_9<.5|H_1).$ You'll get $α=β=0.0668.$ Draw a sketch with two densities of $\bar X_9,$ shading in areas that represent $α$ and ...


1

The sentence "7% of men ... are colorblind." means "Among men, 7% are colorblind." or in symbols, $P(C|M) = 0.07.$ By contrast, $P(M\cap C) = 3.43\%$ ought to go into English as "In the US population 3.43% are colorblind men." However, too many imprecise and/or innumerate writers are not careful to distinguish between $|$ and $\...


1

In short, you haven't been careful with the bounds of integration in your integral for $p(y)$. Let $$I_A (x) = \begin{cases} 1, & \text{ if $x \in A$} \\ 0, & \text{ if $x \notin A$} \\ \end{cases} $$ denote the indicator function of the set $A$. We're given in the problem that $X \sim U([0, 1])$ and $Y|X=x \sim U([0, x])$, which means that $p(x) = ...


1

You have overlooked the fact that $x \geq y$, implicit in the constraint on $P(Y|X) = 1/x$ that $0 \leq y \leq x$. As a consequence of this constraint, the lower bounds of the final two integrals in your next equation should be: $$\int_{0}^{1} P(Y|X)P(X)dx = \int_{Y}^{1}P(Y|X)\cdot 1 dx = \int_{Y}^{1}\dfrac{1}{x} dx $$ which of course equals $-\ln(Y)$. ...


1

The Do you know part makes it a bit hard to guess. Some other approximations: The expected Fisher information (commonly used in generalised linear models, where it's equal to the observed Fisher information under the canonical link but not in general) The variance of the influence functions (either the empirical influence functions, giving the jackknife, or ...


1

For a complete sufficient statistic "all information contained in the complete sufficient statistic is useful", whereas for a regular sufficient statistic this does not have to be the case. An example is helpful. Let $\textbf{X}=(X_1,\dots,X_n)$ where $X_i\sim \mathcal{N}(\mu,1)$. After some basic manipulations we can write the likelihood function ...


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