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If a random variable $x$ follows a Gamma distribution $\Gamma(k,\theta)$, is it possible to transform the variable and make it follows $\Gamma(k=0,\theta=1)$, like we transfer the variable which follows a Gaussian distribution to Normal distribution through $(x-\mu)/\sigma$ ($\mu$ is the mean and $\sigma$ is the standard deviation)?

Many thanks.

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No.

(a) All normal distributions have basically the same shape: $\mu$ shifts position left or right, $\sigma$ shrinks or squeezes. So one can 'standardize' by transforming to get $\mu=0$ and $\sigma=1.$

[This makes it possible to work most problems involving the normal distribution using printed tables of one normal distribution, the standard normal distribution.]

(b) However, for the gamma family of distributions, $k$ is a shape parameter. Look at different shapes in Wikipedia article on 'gamma distribution'. Also, both parameters of gamma must be positive, so $k = 0$ isn't possible (unless you're using some kind of nonstandard parameterization).

[Specific information about a subfamily of gamma with integer and half-integer shape parameters (chi-squared distributions) is widely tabled, but most problems involving gamma distributions are solved using software. Exponential distributions are also in the gamma family (shape parameter 1), and its CDF is available in closed form so computations are relatively easy without specialized software or printed tables.]

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  • $\begingroup$ Thanks a lot, actually, I'm studying pearson's III, which can be considered as gamma distribution but with a transformation. The solution in wikipedia seems not right, as I draw the pdf using the function of "pearspdf" gets different result from using the function gampdf and the transformations. Maybe it's my mistake... $\endgroup$ – Fay Feb 3 '17 at 13:22

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