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Suppose $X$ has pdf $I(\mu, \tau)$ with density,

$$\sqrt{\frac{\tau}{2\pi x^{3}}}\exp\{-\frac{\tau}{2x\mu^{2}}(x-\mu)^2\}\quad; x>0, \quad \tau,\mu>0$$

I want to find the distribution of $V = \dfrac{\tau(x-\mu)^{2}}{x\mu^{2}}$.

My work: The distribution of $V$ can be written as follows.

$$f(v) = f_{X}(g^{-1}(v))\bigg|\frac{dg^{-1}(v)}{dv}\bigg|$$

Solving $x$ using $V$ we can find that $x - \dfrac{\mu^{2}}{x} = 2\mu + \dfrac{\mu^{2}v}{\tau} $.

So, my question is how to get x on one side of the equation (I have $\mu$ on my left hand side)? Are there any methods to derive the distribution of $V$?

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    $\begingroup$ Your last equation is a quadratic in $x$ (multiply by $x$ on both sides), so solve for $x$ using the well-known formula. $\endgroup$
    – StubbornAtom
    Commented Jul 17, 2018 at 7:17
  • $\begingroup$ @StubbornAtom, Do I need to solve the quadratic equation in $x$ to find V? Thanks. $\endgroup$
    – score324
    Commented Jul 17, 2018 at 22:13
  • $\begingroup$ Yes. If $g(x)=v$, then $x=g^{-1}(v)$ (if the inverse exists). You have to find $x$ in terms of $v$ to apply the transformation formula. $\endgroup$
    – StubbornAtom
    Commented Jul 18, 2018 at 5:59
  • $\begingroup$ I am afraid I missed this question and asked the same. With the derivation that $V$ is a $\chi^1_1$ variate provided as the answer. $\endgroup$
    – Xi'an
    Commented Nov 10, 2020 at 13:45
  • $\begingroup$ Note that you cannot use the Jacobian formula here as $V$ is not a bijective transform of $X$. The trick consists in using first $Y=\min\{X,\mu^2/X\}$ and showing that$$W = \dfrac{\tau(Y-\mu)^{2}}{Y\mu^{2}}$$ has the same distribution as $V$ but is a bijective transform of $Y$. $\endgroup$
    – Xi'an
    Commented Nov 10, 2020 at 13:48

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