Transformation the inverse Gaussian probability distribution

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 $V$?

• Your last equation is a quadratic in $x$ (multiply by $x$ on both sides), so solve for $x$ using the well-known formula. – StubbornAtom Jul 17 '18 at 7:17
• @StubbornAtom, Do I need to solve the quadratic equation in $x$ to find V? Thanks. – score324 Jul 17 '18 at 22:13
• 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. – StubbornAtom Jul 18 '18 at 5:59