Search Results

Let's generate 20 independent and identically distributed (iid) standard Normal variates and estimate their mean and standard deviation with the usual formulas (mean = sum/count, etc.). … To test goodness of fit, create four bins with cutpoints at the quartiles of a standard normal: -0.675, 0, +0.657, and use the bin counts to generate a Chi-squared statistic. …

That's fine: you can easily generate a random orthogonal matrix. Just wrap $n^2$ iid standard Normal values into a square matrix and then orthogonalize it. … The QR decomposition will do that, as in this code n <- 5 p <- qr.Q(qr(matrix(rnorm(n^2), n))) This works because the $n$-variate multinormal distribution so generated is "elliptical": it is invariant …

:$$X=\sqrt{-2\log U_1}\,\cos(2\pi U_2)$$(and even two since $Y=\sqrt{-2\log U_1}\,\sin(2\pi U_2)$ is another normal variate independent from $X$). … By comparison, if I exploit the very same 3000 die outcomes $D_i$ to create enough digits of a pseudo-uniform as$$U=\sum_{i=1}^k \dfrac{D_i-1}{6^i}$$with $k=15$ (note that 6¹⁵>10¹¹, hence I generate more …

Heavy tails While the Cauchy is symmetric and roughly bell shaped, somewhat like the normal distribution, it has much heavier tails (and less of a "shoulder"). … For example, there's a small but distinct probability that a Cauchy random variable will lay more than 1000 interquartile ranges from the median -- roughly of the same order as a normal random variable …

This approach seems circular, though: how do we generate a uniform variate with a process that needs a uniform variate to begin with? … For fixed $k$, the asymptotic computational cost is therefore $O(n\log(k))$ = $O(n)$, needing $2n(1+1/k)$ Normal variates to generate $n$ uniform values.) …

iterations for each permutation test (n.iter), a reference to the function test to compute the test statistic (you will see momentarily why we might not want to hard-code this), and two functions to generate … <- replicate(n.reps, f(n.x, n.y, n.per.rep, dist.y=rnorm)) hist(normal, breaks=20) plot(normal) exponential <- replicate(n.reps, f(n.x, n.y, n.per.rep, dist.y=function(n) rgamma(n, 1))) hist(exponential …

Finally, for each pseudorandom variate, $x_i$, that you have generated, generate a pseudorandom error variate, $e_i$, with the appropriate error variance $v_e$, and compute the correlated pseudorandom … variate, $y_i$, by multiplying and adding. …

For example, to generate a set of $N$ data from a normal distribution that will have a given sample mean, $\bar x$, and variance, $s^2$, you will need to fix the values of two points: $y$ and $z$. … Alternatively, you could generate a small number of samples and use the one with the smallest (say) Kolmogorov-Smirnov statistic. …

Gauss copula The Gauss copula is defined implicitely from the multivariate normal distribution, that is, the Gauss copula is the copula associated with a multivariate normal distribution. … Generate a vector $Z = (Z_1, \ldots, Z_d)'$ of independent standard normal variates. Set $X = AZ$ Return $U = (\Phi(X_1), \ldots, \Phi(X_d))'$. …

Therefore, $$ {\rm cor}(X,Y) = \frac{ {\rm cov}(X,Y) }{ \sqrt{ {\rm var}(X)^{2} } } = \rho $$ So if you can generate data from $F_{X}$, you can generate a variate, $Y$, that has a specified correlation … This is due to the fact that sums of normally distributed variables are normal; that is not a general property of distributions. …

The relationship between the mean and variance of a lognormal variate and the mean and variance of the corresponding normal variate is: $\mathbb E(X/Y) = \mathbb E e^Z = \exp \{\mu_Z + \frac{1}{2}\sigma … }$ $\mathrm{Var}(X/Y) = \mathrm{Var}(e^Z) = \exp \{2\mu_Z + 2\sigma^2_Z\} - \exp \{2\mu_Z + \sigma^2_Z\} \>.$ This can be rather easily derived by considering the moment-generating function of the normal …

When $X$ and $Y$ are standard normal, $$XY = ((X+Y)/2)^2 - ((X-Y)/2)^2$$ is a difference of two independent scaled Chi-squared variates. … Because the mgf of a chi-squared variate is $1/\sqrt{1 - 2\omega}$, the mgf of $((X+Y)/2)^2$ is $1/\sqrt{1-\omega}$ and the mgf of $-((X-Y)/2)^2$ is $1/\sqrt{1+\omega}$. …

So it may make perfect sense to model apple weights by a normal distribution, but that's not the real thing; the normal distribution is a convenient abstraction, a tool. … If you test normality, you may well fail to reject normality, but you would also fail to reject an infinite number of non-normal distributions at the same time, and at least some of those would have a …

Note that the collection of dependent variables ($Y$'s) is not assumed to be normal. … of independent variable values, that might be very non-normal. …

If you want to avoid sorting, instead generate $n+1$ independent exponentially-distributed variates. Normalize their cumulative sum to the range $(0,1)$ by dividing by the sum. … For many more (amusing) ways to simulate independent uniform variates, see Simulating draws from a Uniform Distribution using draws from a Normal Distribution. …