How do we prove that $g(x,y) = 1 + \theta[xe^{-x} - (1-\frac{2}{e})][ye^{-y} - (1-\frac{2}{e})]\geq 0$?

Question

Consider the following bivariate density function:

$$\displaystyle g(x,y) = 1 + \theta\left[xe^{-x} - \left(1-\frac{2}{e}\right)\right]\left[ye^{-y} - \left(1-\frac{2}{e}\right)\right]$$

where $(x,y)\in[0,1]^{2}$ and $\theta\in[-1,1]$.

How can we show that it is non-negative based on the analysis of its critical points? To be more precise, its critical points are given by $(1,1)$ and $(0.3904,0.3904)$.

Doing some analysis, we conclude that $g(1,1)$ is a maximum point and $(0.3904,0.3904)$ is a saddle point. Moreover, we also have to study its behavior on the border, from whence we find that $(1,0)$ and $(0,1)$ are critical points as well (both maximum points).

From now on I get a little lost. Based on such results, how do we conclude that $g(x,y) \geq 0$? Thanks in advance!

Answer 1

There is a simple way to prove this. First observe that the minimum w.r.t. $\theta$is attained at $\pm 1$ so what is stated is equivalent to $|xe^{-x} -(1-\frac 2 e)||ye^{-y} -(1-\frac 2 e)|\leq 1$. Let us prove that $|xe^{-x} -(1-\frac 2 e)|\leq 1$ and $|ye^{-y} -(1-\frac 2 e)|\leq 1$. For this just note that $xe^{-x} -(1-\frac 2 e)$ has positive derivative on $[0,1]$ so it is an increasing function. Its extrema are at the end points $x=0$ and $x=1$. At these two points it is trivial to verify the inequality, so we are done.

Answer 2

HINTS (in case this is homework)

• Define $f(z) = z e^{-z}$ with domain $z \in [0,1]$. Within this domain, if my own calculations are not buggy, you can show that $f$ is increasing.

• Find range of $f(z)$ for $z \in [0,1]$.

• Use it to find range of $[f(x) - a] [f(y) - a]$ for $(x,y) \in [0,1]^2$ where $a = (1 - 2/e)$.

• Finally you can prove that, for any $\theta, x, y$ in their respective domains, $g(x,y,\theta) = 1 + \theta [f(x) - a] [f(y) - a] \ge 0$.

Hopefully this recipe works for you?