Critical value for the separation of two normal distributions.

Question

Consider the equation:

$$x \mathrm{e}^{\frac1{2} ((x-z)^2 - x^2)} + y(x - z) = 0$$

And limiting $z > 0$ and $y \ge 1$, for what values of $z$ and $y$ does the function in $x$ have exactly 2 real solutions? Looking for a function $z(y)$, but I can work with any implicit closed form or even approximate solution, as I've spent some time playing around with it, and I do not believe that there is an explicit closed form solution to this problem.

Motivation: Consider two normally distributed random variables $\mathcal{N}(0, 1)$ and $\mathcal{N}(z, 1)$ which are drawn from with probabilities $\frac{1}{y+1}$ and $\frac{y}{y+1}$. The function $z(y)$ are the $z$ are exactly the critical value for the separation of the two modes. For values of $z$ less than $z(y)$ there are not two separate modes, for values greater than this there are.

A few known values of this function are $z(1) = 2$, and $z(2) \approx 2.627509131962$.

Answer 1

From @OmnipotentEntity’s answer, let $x_0^2=e^w\implies z=\frac1{x_0}+x_0=2\cosh(\frac w2)$

$$y=x_0^2\exp\left(\frac{1}{2}\left(\frac{1}{x_0^2}-x_0^2\right)\right)=e^{\sinh(w)-w}$$

This post inverts $\sinh(x)+ax$ via an inverse Laplace transform; define its inverse as $f_a(x)$. Thus, $z(y)=2\cosh(\frac12f_{-1}(\ln(y))$:

$$\boxed{y(z) = \frac{\sqrt{z^4-4 z^2+2 \sqrt{z^2-4} z-\sqrt{z^2-4} z^3+2}}{ \sqrt{2 e^{-z \sqrt{z^2-4}}}}\implies z(y)=2\cos\left(\frac14\int_{c-i\infty}^{c+i\infty}\frac{y^s}s\operatorname A_{-s}(s)ds\right)}$$

shown here:

is $z(2)$ which matches the question. $\operatorname A_v(z)$ is the associated Anger Weber function. Perhaps this explicit expression will help.

Answer 2

If we let $f(x) = x \mathrm{e}^{\frac1{2} ((x-z)^2 - x^2)} + y(x - z)$ then at our critical points we know that $f(x_0) = 0$ and $f'(x_0) = 0$. We can consider this a system of two simultaneous equations and solve for $y$ and $z$ in terms of $x_0$, then eliminate $x_0$ to find $y(z)$. However, this function is not invertible.

We have:

$$ z = \frac1{x_0} + x_0 $$ $$ y = y=x_0^2\exp\frac{1}{2}\left(\frac{1}{x_0^2}-x_0^2\right) $$

We can express $y(z)$ as

$$ y(z) = \frac{\sqrt{z^4-4 z^2+2 \sqrt{z^2-4} z-\sqrt{z^2-4} z^3+2}}{ \sqrt{2 e^{-z \sqrt{z^2-4}}}} $$

Answer 3

Another approach is as follows. Conceptually, I think it is pretty much what you have shown in your solution. I just feel this is a bit more intepretable. Let $f(x) := x\exp\left(\frac{1}{2} ((x-z)^2 - x^2))\right)$ and $g(x) = -y(x - z)$. What you are essentially searching for is that the equation $f(x) = g(x)$ has exactly two 2 real solutions.

Given $g(x)$ is a line, the only way the above relation can hold (for the given $f(x)$) is when this said line is a tangent to $f(x)$ and it cuts it again exactly once more at another point. Furthermore, it has to be a tangent at a point $x > 1/z$. This follows from the fact that $f(x)$ is increasing for $x \leq 1/z$ and any tangent will cut it exactly once. In fact, this will give you the region for exactly one solution. Let $x_0 > 1/z$ and the tangent at $x_0$ is given as $f'(x_0) x + (f(x_0) - x_0 f'(x_0))$. You want this to be the same as $g(x)$. On comparing the equations, we obtain, \begin{align*} y & = -f'(x_0) \\ yz & = f(x_0) - x_0 f'(x_0). \end{align*} This implies $x_0$ satisfies the following relation, \begin{align*} f'(x_0)(x_0 - z) & = f(x_0) \\ \implies \exp\left(\frac{z^2}{2} - zx_0\right)(1 - x_0z)(x_0 - z) & = x_0\exp\left(\frac{z^2}{2} - zx_0\right) \\ \implies x_0^2 - x_0 z + 1 = 0. \end{align*}

On solving this, we obtain, $x_0 = \frac{1}{2}\left(z - \sqrt{z^2 - 4}\right)$ (assuming $z \geq 2$). We choose this root since $x_0 > 1/z$.

The required value of $y = -f'(x_0)$ for the above value (which I think matches what you obtained). The condition $z \geq 2$ is a consequence of $y \geq 1$, implying that this solution covers the entire range.