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I am looking to upper bound the expression:

$$\frac{x \phi(x)}{2 \Phi(x) - 1}$$ for $x \geq 0$. Here $\phi(x), \Phi(x)$ denote the PDF/CDF (respectively) of a $\mathcal{N}(0, 1)$ random variable.

Based on numerical plots it appears to be upper bounded by $\frac{1}{2}$, with that bound achieved at $x = 0$. Could anyone please show how to obtain this using elementary methods and properties of the standard normal random variable?

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One remark first: technically $\frac{x\phi(x)}{2\Phi(x)-1}$ is undefined at $x=0$, because plugging in zero leads to an expression of the form $\frac{0}{0}$. To resolve this problem, we note that by l'Hopital's Rule, $$\lim_{x\to 0}\frac{x\phi(x)}{2\Phi(x)-1}=\lim_{x\to 0}\frac{\phi(x)+x\phi'(x)}{2\phi(x)}=\frac{1}{2},$$ (the last equality by direct evaluation) and thus $\frac{x\phi(x)}{2\Phi(x)-1}$ has a continuous extension to all of $\mathbb{R}$. So when $x=0$ we interpret $\frac{x\phi(x)}{2\Phi(x)-1}=\frac{1}{2}$. Okay, with that technicality out of the way, to show that for $x\geq 0$ $$\frac{x\phi(x)}{2\Phi(x)-1}\leq \frac{1}{2}$$ is equivalent to showing that $$x\phi(x)\leq \Phi(x)-\frac{1}{2},$$ and it suffices to show that if $f(x):=x\phi(x)$ and $g(x):=\Phi(x)-\frac{1}{2}$, then $f(0)=g(0)$ and that $f'(x)\leq g'(x)$ for all $x\geq 0$. We have $$f'(x)=\phi(x)+x\phi'(x)\text{ and }g'(x)=\phi(x),$$ so showing $f'(x)\leq g'(x)$ reduces to showing $x\phi'(x)\leq 0$, but this clearly holds because by direct calculation $x\phi'(x)=-x^2\phi(x)\leq 0$. This finishes the proof. Please feel free to comment below if you aren't sure about any of the above; I omitted some details which I thought are easy to verify.

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$$\frac{x \,\phi(x)}{2 \Phi(x) - 1}=\frac{x\,e^{-\frac{x^2}{2}} }{\sqrt{2 \pi }\, \text{erf}\left(\frac{x}{\sqrt{2}}\right)}$$

Using Padé approximants $$\frac{x \,\phi(x)}{2 \Phi(x) - 1} \leq \frac 12 \,\frac{1-\frac{2 }{9}x^2+\frac{1}{63}x^4 } {1+\frac{1}{9}x^2+\frac{8 }{945}x^4 }$$

Expanded as series $$\frac 12 \,\frac{1-\frac{2 }{9}x^2+\frac{1}{63}x^4 } {1+\frac{1}{9}x^2+\frac{8 }{945}x^4 }-\frac{x \,\phi(x)}{2 \Phi(x) - 1}=\frac{4 x^{10}}{654885}+O\left(x^{12}\right)$$

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