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Consider the following random quadratic equation, $$ x^2 + Z x + Y = 0, $$ where, $$ \begin{gathered} Z \sim \mathcal{N}(\mu_Z,\sigma_Z), \qquad Y \sim \mathcal{N}(\mu_Y,\sigma_Y). \end{gathered} $$ What is the distribution (PDF) of the solution(s) $x$, and what are the moments?

I am interested in any solvable case or approximation to the solution; approximating just the first two moments would be sufficient. Assume $Z$ and $Y$ are uncorrelated; however, a solution to the correlated case would be interesting.

Emmy B
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    Which branch are you interested in and how do you define it? Note that you have to permit $x$ to be Complex. – whuber Dec 23 '23 at 19:40
  • @whuber By branch, I mean pick either a + or - in the quadratic formula. Maybe a different definition is needed, but I'm not sure what that might be yet. The probability that $x$ is complex is $\operatorname{P}(Z^2 > 4Y)$ and generically not 1. In my particular problem, parameters are chosen so that $x$ is rarely complex. Therefore, in the solution, it is acceptable to me (and maybe even preferred) that $x$ is assumed to be real. – Emmy B Dec 23 '23 at 19:52
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    Your comment contradicts your question: if "parameters are chosen," or if "$x$ is assumed to be real," then the parameters $(Z,Y)$ cannot possibly be Normally distributed, can they? (Or if by "parameters" you mean the means and SDs of the distributions of $Z$ and $Y,$ I doubt such choices exist unless one is very generous about what "rarely complex" means -- and that calls into question whether any solution for the PDF would be meaningful.) Please tell us what your problem really is. – whuber Dec 23 '23 at 21:23
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    @whuber By parameters, I do mean the means and variances of $Z$ and $Y$. If $Z^2 \gg 4 Y$, then the solution will often be real. By computing the moments of $Z^2 - 4 Y$ and applying Chebyshev's inequality, one can show that the probability that $x$ is complex approaches zero asymptotically when $\sigma_Y^2 \ll (\mu_Z^2 - 4 \mu_Y)^2 - 2 (\mu_Z^2 + 4 \mu_Y) \sigma_Z^2 - \sigma_Z^4$. I am able to confirm this with numerical experiments. For the sake of generality of the question, a complex answer is preferred. I was merely trying to indicate that I am interested in any possibly tractable cases. – Emmy B Dec 23 '23 at 21:42
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    That is all correct. But since it is (very) unlikely one can find an analytical solution, your numerical experiments are perhaps the most effective way to answer your question. That is, perhaps the most productive approach is to consider how to perform an efficient numerical computation of the distribution. – whuber Dec 24 '23 at 12:38
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    BTW, You can approximate the PDF in terms of Owen's T function. – whuber Dec 24 '23 at 13:57
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    @whuber I sat and thought about this for a little bit, and I came up with an approximation that works in a special, relevant case. I'll answer my own question to start things off in the hope that someone might eventually come along and figure out a nice way to expand it. – Emmy B Dec 24 '23 at 21:36

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Here is a special case to get things started:

Consider the typically real case where $Z^2 \gg 4Y$. By the quadratic formula: $$ x_\pm = \frac{1}{2} \left( -Z \pm \sqrt{ Z^2 - 4Y } \right). $$ Write the radicand as a zero-mean fluctuating term and a deterministic part: $$ \sqrt{ (Z^2 - 4Y - \operatorname{E}[Z^2 - 4Y]) + \operatorname{E}[Z^2 - 4Y] } $$ The solution is typically real when the probability the radicand is less than zero is rare. By Chebyshev's inequality, this is equivalent to, $$ |Z^2 - 4Y - \operatorname{E}[Z^2 - 4Y]| \ll \operatorname{E}[Z^2 - 4Y]. $$ Using this, perform a Taylor expansion in the fluctuating part about the mean: $$ \sqrt{ (Z^2 - 4Y - \operatorname{E}[Z^2 - 4Y]) + \operatorname{E}[Z^2 - 4Y] } = \sqrt{\operatorname{E}[Z^2 - 4Y]} + \frac{Z^2 - 4Y - \operatorname{E}[Z^2 - 4Y]}{2\sqrt{\operatorname{E}[Z^2 - 4Y]}} + O((Z^2 - 4Y - \operatorname{E}[Z^2 - 4Y])^2). $$ In this limit, the solutions are approximately, $$ x_\pm \approx \frac{1}{2} \left( -Z \pm \left[ \sqrt{\operatorname{E}[Z^2 - 4Y]} + \frac{Z^2 - 4Y - \operatorname{E}[Z^2 - 4Y]}{2\sqrt{\operatorname{E}[Z^2 - 4Y]}} \right] \right). $$ One can then compute the moment generating function or any of the moments using a computer algebra package like Mathematica: Mathematica output

Emmy B
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