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The image below illustrates intuitively why the transformed variable has a different distribution: I have drawn two parallel lines. On the lowest line I have plotted evenly spaced points at $0.1, 0.2, ..., 1.1, 1.2$ which represent the velocity $v$. On the upper line I have draw points according to the formula $t=0.1/v$ (note I reversed the axis it has 1.2 ...

14

Height, for instance, is often modelled as being normal. Maybe the height of men is something like 5 foot 10 with a standard deviation of 2 inches. We know negative height is unphysical, but under this model, the probability of observing a negative height is essentially zero. We use the model anyway because it is a good enough approximation. All models ...

10

Doesn't the normal distribution allow for negative values? Correct. It also has no upper bound. In one part of my textbook, it says that a normal distribution could be good for modeling exam scores. In spite of the previous statements, nevertheless this is sometimes the case. If you have many components to the test, not too strongly related (e.g. so ...

4

Yes, this is an instance of inverse Gaussian. It has been observed that there is an inverse relationship between the cumulant generating function of the time to cover a unit distance and the cumulant generating function of the distance covered in a unit time. Because the distance covered in a unit time (in this case, walking speed) is approximately normal, ...

3

Based on the answer provided by the OP, I believe the issue is not with the transformation into the copula space (i.e. applying the inverse CDF of a standard normal), but rather the transformation into uniform random variables from the raw data. Recall that with copula models, there are two parts. First is modeling the marginal distributions for each ...

3

There is no good single summary statistic for the type of distribution you have plotted, or, really, for any multimodal distribution. That is, you can calculate anything you'd like: Mean, median, mode, interquartile range .... whatever. But none of these are good representations of data that has multiple modes. Even for data that is perfectly normally ...

3

Since you have $\mu = X\beta$ this is the identity link. To fit an exponential GLM, you fit a gamma GLM but instead of estimating the dispersion parameter $\phi$ from the data, you specify that it's 1. This doesn't affect the parameter estimation, only the standard errors. [Consequently in R you simply specify the dispersion parameter when calling summary ...

3

I don't think there is a c++ library, but you can easily write your code: 1) on wikipedia there is a method to simulate a two parameters inverse gaussian http://en.wikipedia.org/wiki/Inverse_Gaussian_distribution. You can use this and then shift the distributon to have a three parameters distribution 2) you can simulate a normal variable with the same two ...

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Prior is the distribution for the parameter that you assume "before seeing the data", you seem to be talking about likelihood, i.e. the distribution to describe the data itself. If your outcomes are relative changes, that can be either positive, or negative, choosing inverse Gaussian is a bad idea, because it is a distribution for non-negative outcomes. You ...

2

For simplicity write $x=x_1$ and $y=x_2,$ so that $$2\bar x = x+y,\quad s = \frac{1}{x} + \frac{1}{y} - \frac{4}{x+y}.\tag{*}$$ Their Jacobian $J(\bar x, s)$ can be computed by comparing the differential elements in the two coordinate systems \mathrm{d} \bar x \wedge \mathrm{d} s = \frac{1}{2}(\mathrm{d} x + \mathrm{d} y) \wedge \left(-\frac{\mathrm{d} ... 2 With some more searching I found a library by Richard Saucier which contained the three parameter lognormal among other distributions. The documentation for the library is here. Thank you niandra82 for your reply. When I looked at the library code you were quite right in that it would have been easy enough to write myself. The wikipedia definitions of the ... 1 the inverse transform of the standard deviation is wrong. Mean and standard variation have to be transformed differently. Here's a brief explanation: Transform Let's assume random variable Y with mean \mu_Y and variance \sigma^2_Y. The "Scaler" subtracts some constant a and divides the result by a factor b. The transformed variable equals Z = \... 1 \begin{align*} \left(\frac{\lambda}{2\pi x^3}\right)^{\frac{1}{2}} \exp{\left(-\lambda \frac{(x-\mu)^2}{2\mu^2x}\right)} &= \exp{\left(\ln{\left(\frac{\lambda}{2\pi x^3}\right)^{\frac{1}{2}}}\right)} \exp{\left(-\lambda \frac{(x-\mu)^2}{2\mu^2x}\right)} \\ &= \exp{\left( -\lambda \frac{x^2-2x\mu +\mu^2}{2\,\mu^2x} - \frac{\ln{2\pi x^3}-\ln{\lambda}}{... 1 Try: G. A. Whitmore & V. Seshadri (1987) A Heuristic Derivation of the Inverse Gaussian Distribution, The American Statistician, 41:4, 280-281 1 It turns out that the answer lies in looking at the pdf of the Inverse Gaussian. The full conditional for each \tau^2_i is\dfrac{1}{\tau^2_i}| \beta, \sigma^2, y \sim \text{Inverse Gaussian}\left(\sqrt{\dfrac{\lambda^2 \sigma^2}{\beta_i^2}}, \lambda^2 \right).  \begin{align*} f(\tau^2_i | \beta, \sigma^2, y) & = \sqrt{\dfrac{\lambda^2}{2 \pi}} \...

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