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?
Best Answer
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.