Consider the following equation:
$p\left(1-\int _{\mu}^{x} f(y)dy\right) \left[p\left(1-\int _{\mu}^{x} f(y)dy\right)+(1-p)q \right]-xf(x)p(1-p)q=0$,
where $p,q \in [0,1]$, $f(\cdot)$ is the probability density of the normal distribution and $\mu$ is the mean of the variable $x$.
My question is: is there any root $x^{\ast}$ in the above equation? How can I prove the existence?
What I have already known:
(i) If $\mu = \frac{\sigma \sqrt{2\pi}(p+(1-p)q)}{(1-p)q}$, then $x^{\ast}=\mu$ is a root, where $\sigma$ is the variance of $x$. But I want to find a $x^{\ast}$ for an arbitrary $\mu >0$.
(ii) For $x$ large or lower enough, the expression is positive. However, I cannot prove that it is negative for some other $x$. So it is impossible for me to use some result like the Mean Value Theorem.
(iii) A necessary condition for the existence is $x>0$, independent of the value of the mean.
What I expect: in the general case, there are at least two roots, one higher than $\mu$ and another lower. It would be great if I could prove that!
Thank you!