8

No, this is not possible: as whuber notes, you are trying to determine two parameters (the mean and the standard deviation), having only one data point. For instance, a 25th percentile at $1$ can come from an uncountably infinite one-parameter family of normal distributions, because for any mean $\mu>1$ (do you see why $\mu$ must be larger than $1$?), you ...


5

The half normal distribution is usually parametrised so that the $\sigma$ from the corresponding normal distribution is its scale parameter. However, this is not the standard deviation of it, which is $\sigma\sqrt{1-\frac{2}{\pi}}$, see https://en.wikipedia.org/wiki/Half-normal_distribution. It should be intuitive that the sd is smaller than $\sigma$; there ...


5

If you know the mean, you can determine the variance. Let's assume that $q_{0.25} = -0.67$ and $\mu = \mu_0$ with unknown $\sigma^2$. Then $$f_X(x) = \dfrac{1}{\sigma\sqrt{2\pi}}\exp\bigg(-\dfrac{1}{2}\dfrac{(x - \mu_0)^2}{\sigma}\bigg)$$ $$ F_X(x) = \int_{-\infty}^xf_X(u)du \implies F_X(0.25) = \int_{-\infty}^{0.25}f_X(u)du = 0.25 $$ This solution will be ...


4

If you randomly assign a sign (+ or -) to each member of the sample generated from a half-normal distribution, the result will be indistinguishable from a sample generated from the corresponding unfolded normal distribution. Your method wouldn't work - for example, if your sample size is 2, it would be very surprising to draw the sample 4, -4 from a normal ...


3

A very simple, general-purpose solution: First, write a function that takes parameters as an input, and returns the different between the predicted PDF for those parameters and the actual PDF (I've used the sum of squared differences here). Then, use optim() to find the parameters than minimise this function. x = seq(-3,3,0.1) pdf = dnorm(x, mean = -.5, sd = ...


3

For a normal density function $f,$ if you have a grid of points X and corresponding density values $y = f(x),$ then you can use numerical integration to find $\mu$ and $\sigma.$ [See Note (2) at the end.] If you have many realizations $X_i$ from the distribution, you can estimate the population mean $\mu$ by the sample mean $\bar X$ and the population SD $\...


3

With data $x_1, x_2, \ldots, x_n,$ you propose synthesizing a dataset of $2n$ values $|x_1|, -|x_1|, |x_2|, -|x_2|, \ldots, |x_n|, -|x_n|.$ Because each absolute value $|x_i|$ balances its negative $-|x_i|,$ the mean is zero. The usual standard deviation estimator therefore reduces to the square root of $$\frac{1}{2n-1}\left((|x_1|-0)^2 + (-|x_1|-0)^2 + \...


2

I don't see where you've described any correlation structure (despite the title using the term "correlated"). So maybe starting small with $n=2$ and assuming independence will be a start. Using Mathematica (or probably just basic algebra) one can find the distribution of $$W_2=X_1^2/(X_1^2+X_2^2)$$ where $X_i \sim N(0,\sigma_i^2)$. dist = ...


2

The setup seems to imply that $(\theta_1,\ldots\theta_M)$ are only sampled once. $\theta_x$ for given $x$ will always be the same regardless of $n$. There is no $\theta^*_x$ different from the original $\theta_x$. Note that there is a certain confusion of notation between random variables and their realisations (which is often found in Bayesian notation). I ...


2

I assume you want to say multi-gaussian assumption. It is quite improbable that will be true in so high dimension, so here is a way to approach your problem. Try independent component analysis, see Making sense of independent component analysis. This will in practice search for linear combinations that are as far away from a normal distribution as possible. ...


2

This largely depends on the robustness of your inferences to errors in the distribution When you are dealing with quantities that are either directly observable, or for which there is some close estimator (e.g., residuals for error terms), you can use the data to make non-parametric estimates of the distribution. Assuming you have sufficient data to do this,...


1

$P(|Y| > 1.96\, |\, X = 1) < 0.06,$ but $P(|Y| > 1.96,|\, X = 2) > .32.$ This can be shown using printed normal tables. Illustration by simulation in R below: set.seed(722) x = sample(1:2, 10^6, rep=T) y = rnorm(10^6, 0, x) mean(abs(y[x==1]) > 1.96) [1] 0.05046732 mean(abs(y[x==2]) > 1.96) [1] 0.3280174


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