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Is thisthe function $e^{x^2/2}\Phi(x)$ monotone increasing?

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Hello,

Here is an interesting problem. It looks elementary, but it has taken me some efforts without solving it. Let

$$h(x) = e^{x^2/2} \Phi(x),\qquad \text{with}\quad \Phi(x):=\int_{-\infty}^x \frac{e^{-y^2/2}}{\sqrt{2\pi}} dy.$$

The question is whether the function $h(x)$ is monotone increasing over $R$? Are there some work dealing with such function?

It seems a quite easy problem. By taking the first derivative, we need to prove that

$$h(x)' = h(x) x + \frac{1}{\sqrt{2\pi}} \ge 0.$$ which again, not obvious (for $x<0$). Some facts, that might be useful, are:

$$\lim_{x\rightarrow -\infty} h(x) =0, \quad \lim_{x\rightarrow -\infty} h(x)' =0.$$

Thank you very much for any hints!

Anand

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Is this function monotone increasing?

Hello,

Let

$$h(x) = e^{x^2/2} \Phi(x),\qquad \text{with}\quad \Phi(x):=\int_{-\infty}^x \frac{e^{-y^2/2}}{\sqrt{2\pi}} dy.$$

The question is whether the function $h(x)$ is monotone increasing over $R$? Are there some work dealing with such function?

It seems a quite easy problem. By taking the first derivative, we need to prove that

$$h(x)' = h(x) x + \frac{1}{\sqrt{2\pi}} \ge 0.$$ which again, not obvious (for $x<0$). Some facts, that might be useful, are:

$$\lim_{x\rightarrow -\infty} h(x) =0, \quad \lim_{x\rightarrow -\infty} h(x)' =0.$$

Thank you very much for any hints!

Anand