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Diffusion of a single particle is a random walk in two dimensions. As a function of time, $t$, the probability density for its location will therefore be Gaussian, centered at the particle's original …

Therefore, this Normal approximation works for narrow slices not too far into the tails of the bivariate distribution. …

As explained at https://stats.stackexchange.com/a/71303/919, this level set is the image of an ellipse that has been skewed upwards. Here is part of the original ellipse, with vertical arrows indicat …

By definition, the ECDF $F$ at any location $(x,y)$ counts the data points that lie to the left and beneath $(x,y)$. Specifically, writing $(x_i,y_i), i=1, 2, \ldots, n$ for the data points (which ma …

By subtraction, you would like to show that when $(\varepsilon_1, \varepsilon_2)$ has a bivariate Normal distribution with covariance $\Sigma$, then $u = \varepsilon_1 - \frac{\sigma_{12}}{\sigma_{22}} …

Notice that $$\mathbb{E}(\min\{\sigma\tau Z^2, c\}) \to 0$$ as $\sigma\tau\to 0.$ However, provided both $\sigma$ and $\tau$ are nonzero and $n=2$ it is possible for the outcomes to be $$\{(X_i,Y_i), …

By definition, $F$ is the distribution function of a bivariate random variable $(X,Y)$ when $${\Pr}(a\lt X \le b,\, c\lt Y \le d) = F(b,d) - F(a,d) - F(b,c) + F(a,c)$$ for all real numbers $a\le b,\, …

Another interpretation of the question is that it requests a procedure to test for inclusion within the ellipses created by a bivariate normal approximation to the data. … <- cov(p) The formula requires inversion of the variance-covariance matrix: sigma.inv = solve(sigma, matrix(c(1,0,0,1),2,2)) The ellipse "height" function is the negative of the logarithm of the bivariate …

Also, positional errors are bivariate normal with 95% probability of ship's actual position being within 1 mile of the expected position. … A bivariate normal distribution with no correlation and variances of $\sigma^2$ for each of the coordinates has a total probability of $1 - \exp(-x^2/(2\sigma^2))$ within a distance $x$ of its mean. …

This is a common theme in many bivariate probability problems: probabilities are specified or constructed by fixing one variable and changing the other (here, fixing $X$ and changing $Y$) but the question …

Often, the most straightforward way to find the distribution of a variable defined in terms of other random variables is to compute its cumulative distribution function. For any number $y$ this funct …

As Stephen Kolassa suggests, it helps to draw a picture of the transformation $g$. Here are two: a contour plot of its values and a 3D perspective plot showing $z = g(x,y)$. Use basic definitions …

When $(X,Y)$ is a bivariate normal distribution, $c\to -\infty$ as $|b-a|\to 0$. … (The apparent exception $\rho=0$ can be handled by starting with, say, a bivariate distribution with Normal marginals whose support is confined to the lines $y=\pm x$.) …

.$ Colors denote values of the bivariate density with $\rho=2/5.$ Its regression line $y=\rho x$ is shown in white. … .$ Reversing the roles of $X$ and $Y$ changes nothing in the reasoning and only swaps $x$ and $y$ in the result: from the symmetric expression for the bivariate Normal density, the situation is identical …

Find the distribution function of $Y_1$ with a picture, then differentiate it. Observe that because $(X_1,X_2)$ lies in the first quadrant and $Y_1$ is the angle subtended by this point, $0\le Y_1 \le …