Let $W$ be the ratio of independent normal random variables

\begin{equation} W=\frac{\mu_{1}+\sigma_{1}Z}{\mu_{2}+\sigma_{2}Z}=\frac{X_{1}}{X_{2}} \end{equation}

where, $Z\sim\mathcal{N}(0,1)$. Given that $\mathrm{Pr}(X_{2}>0)=1$, the cumulative distribution of $W$, $F(w)$, can be expressed as

\begin{equation} F(w) = \Phi\left(\frac{\mu_{2}w-\mu_{1}}{\sigma_{1}\sigma_{2}\sqrt{\frac{w^2}{\sigma_{1}^{2}}+\frac{1}{\sigma_{2}^{2}}}}\right) \end{equation}

where, $\Phi(x):=\int_{-\infty}^{x}\phi(t)\, \mathrm{d}t$ is the standard normal cdf, and $\phi(t):=\frac{1}{\sqrt{2\pi}}e^{-t^2/2}$ is the standard normal pdf. For more information on this distribution see "On the ratio of two correlated random variables" by D.V Hinkley JSTOR.org.

Since $X_{2}$ is assumed to have zero density at zero, the moments of the ratio density $f(w)$ exist. I want to find the skewness of $f(w)$ which requires knowing the first three raw moments, i.e. $\mathbf{E}[X]$, $\mathbf{E}[X^2]$, and $\mathbf{E}[X^3]$. To attempt this I first found the density of $W$ by differentiating $F(w)$. Let,

\begin{equation} a(w)=\frac{\mu_{2}w-\mu_{1}}{\sigma_{1}\sigma_{2}\sqrt{\frac{w^2}{\sigma_{1}^{2}}+\frac{1}{\sigma_{2}^{2}}}} \end{equation}

Then the density function is

\begin{equation} f(w)=a^{\prime}(w)\, \phi(a(w)) \end{equation}

The mean, $\mu_{W}$, is then

\begin{equation} \mu_{W}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}w\,a^{\prime}(w)\, \exp(-a^{2}(w)/2) \ \mathrm{d}w \end{equation}

Performing integration by parts where

\begin{equation} \begin{aligned} &u=w\\ &\mathrm{d}u=\mathrm{d}w\\ &\\ &\mathrm{d}v=a^{\prime}(w)\,\exp(-a^{2}(w)/2) \ \mathrm{d}w\\ &v= \sqrt{\frac{\pi}{2}}\, \mathrm{erf}\left(\frac{a(w)}{\sqrt{2}}\right) \end{aligned} \end{equation}


\begin{equation} \mu_{W}=\frac{w}{2}\, \mathrm{erf}\left(\frac{a(w)}{\sqrt{2}}\right)\bigg|_{w=-\infty}^{\infty}-\frac{1}{2}\int_{-\infty}^{\infty}\mathrm{erf}\left(\frac{a(w)}{\sqrt{2}}\right) \ \mathrm{d}w \end{equation}

$\mathrm{d}v$ was integrated here. The $uv$ term can be evaluated by writing it as

\begin{equation} \frac{1}{2}\,\lim_{w\to\infty}\left( w\, \mathrm{erf}\left(\frac{a(w)}{\sqrt{2}}\right) + w\, \mathrm{erf}\left(\frac{a(-w)}{\sqrt{2}}\right) \right) = -2\mu_{1}\,\phi\left(\frac{\mu_{2}}{\sigma_{2}}\right) \end{equation}

This result can be found here.

This gets me to

\begin{equation} \mu_{W}=-2\mu_{1}\,\phi\left(\frac{\mu_{2}}{\sigma_{2}}\right) -\frac{1}{2}\int_{-\infty}^{\infty}\mathrm{erf}\left(\frac{a(w)}{\sqrt{2}}\right) \ \mathrm{d}w \end{equation}

Consider only the remaining integral. Rewriting the $\mathrm{erf}$ using Taylor expansion gives

\begin{equation} {\frac{2}{\sqrt{\pi}}}\sum _{n=0}^{\infty}{\frac{(-1)^{n}}{n\,(2n+1)}} \int_{-\infty}^{\infty} \left(\frac{\mu_{2}w-\mu_{1}}{\sqrt{2}\,\sigma_{1}\sigma_{2}\sqrt{\frac{w^2}{\sigma_{1}^{2}}+\frac{1}{\sigma_{2}^{2}}}} \right)^{2n+1} \ \mathrm{d}w \end{equation}

I am able to get a solution to the indefinite integral for the above here...it's not of much use. After evaluating the antiderivative at the limits of integration might yield something more useful but I was not able to get that far. Does anyone see a way to the solution from here?

  • $\begingroup$ Have you tried re-writing $\text{erf}$ as an integral and convert the last integral into an integral over a half-plane? Maybe with some tricky change of variable and/or Fubini's theorem the last part can be made explicit or, at least, well-approximated by a simple function. $\endgroup$ – Jack D'Aurizio Dec 30 '16 at 21:00
  • $\begingroup$ @JackD'Aurizio I agree that rewriting $\mathrm{erf}$ is probably neccessary but its still not clear where to go from there. I originally thought I had evaluated the $w\times\mathrm{erf}(\cdots)$ as well but it looks like I may have been wrong. The $\mathrm{erf}$ is bounded while $w$ is not...resulting in infinities at both limits. $\endgroup$ – Aaron Hendrickson Dec 31 '16 at 19:20

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