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I have a given gamma distribution as: $f(x;k,\theta) = \frac{1}{\Gamma(k)\theta^{k}}x^{k-1}e^{\frac{-x}{\theta}}$ and a non-centrality parameter $\delta$. Now, I need to find the pdf of this non-central gamma distribution $f(x;k,\theta,\delta)$?

I have found an expression of this in a paper by Oliveira and Ferreira. However, the pdf expression is in terms of shape parameter and non-centrality parameter only, which is given as $f(x;k,\delta) = \displaystyle\sum_{i=0}^{\infty}e^{\frac{-\delta}{2}}\left(\frac{\delta}{2}\right)^i \left[ \frac{1}{\Gamma(k+i)}e^{-x}x^{k+i-1}\right]$.

Is there an expression for pdf that incorporates x,$\theta$,k, and $\delta$? Or, any approximations to make the non-central distribution to the central distributions?

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up vote 1 down vote accepted

Let $y = g(x)$ be a 1 to 1 transformation, so $x = g^{-1}(y)$

In your case $Y = \frac{1}{X+\delta}$, so $X = \frac{1}{Y}-\delta=g^{-1}(Y)$

$f_Y(y) = |\frac{d}{dy}(g^{-1}(y))|f_X(g^{-1}(y))$

Why is this the case? $f_X(g^{-1}(y))$ is just $f_X(x)$ with $y$ plugged in, $|\frac{d}{dy}(g^{-1}(y))|$ is the change of variable term - same as doing a substitution in an integral

For your example, we get

$|\frac{d}{dy}(\frac{1}{y}-\delta)| = \frac{1}{y^2}$

$\dfrac{(\frac{1}{y}-\delta)^{k-1} e^{-\frac{1}{\theta}(\frac{1}{y}-\delta)}}{\theta^k \Gamma (k)} \times \frac{1}{y^2} = \dfrac{(\frac{1}{y}-\delta)^{k-1} e^{-\frac{1}{\theta}(\frac{1}{y}-\delta)}}{y^2\theta^k \Gamma (k)} $

so $f_Y(y) = \dfrac{(\frac{1}{y}-\delta)^{k-1} e^{-\frac{1}{\theta}(\frac{1}{y}-\delta)}}{y^2\theta^k \Gamma (k)}$ for $y<\frac{1}{\delta}$

Now your job:

make sure you understand this and try to do it for normal distribution.

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thanks. was of great help. and could the mean and variance be generalized too? :) – Sam Jan 21 '14 at 22:02
@Sam please be more specific. if you mean, can we calculate the mean of this thing. I believe the answer is no. for $\delta = 0$, this is known. see but $1/(y+\delta)$ is pretty nasty, so i do not think so. – Lost1 Jan 21 '14 at 22:09
I could actually get to this, I know $X$ is gamma distributed so its easy for mean(X) and var(X). Now $(X+\delta)$ has mean=mean(X)+$\delta$ and variance = var(X). So, I am looking for an alternative to calculate the mean of $\dfrac{1}{X+\delta}$. I agree it might get too nasty. – Sam Jan 21 '14 at 22:15
@Sam yes, unfortunately $E(1/(X+\delta))$ is almost never a nice object. – Lost1 Jan 21 '14 at 22:17

As far as my monte-carlo simulation and closed form expression match, the non-central gamma could be well approximated by Amoroso distribution i.e., $f(x;k,\theta,\delta) = \frac{1}{\theta^k\Gamma(\theta)}(x-\delta)^{k-1}e^{\left(\frac{-(x-\delta)}{\theta} \right)}$ where $\delta$ is the location parameter.

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What is non central gamma? If you just want to shift the distribution to the right by delta. I.e. Do $\delta+X$ where $X$ is a gamma distribution. This is pretty clear via change of variable. – Lost1 Jan 21 '14 at 17:49
Well I was talking about the distribution when the gamma distribution is shifted by some known parameters. – Sam Jan 21 '14 at 17:52
Yeah so $X+\delta$ where $X$ is a standard gamma, this answer is correct. Have you rad my answer to your other question. – Lost1 Jan 21 '14 at 18:02
so, you mean say $X$ is gamma distributed with pdf $f_{X}(x)$ and $X+\delta$ has some distribution with pdf $\delta+f_{X}(x)$ or $f_{X}(x+\delta)$. and neither is true. – Sam Jan 21 '14 at 18:18
Nope, i am or saying that. I am saying $X+\delta$ has pdf $f_X(x-\delta)$, isn't this obvious from your answer? This is via a change of variable argument, rather than all this simulation you did, which is not a proof. This is actually for any distribution with a density function/mass function. – Lost1 Jan 21 '14 at 18:37

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