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Simple question, but every time I think I have the answer, I get stuck.

Let $ F(x)=\frac{1}{x^2}*(2*\Phi(x)-1) +(2/3)*\frac{1}{x}*(\phi(x)) +(2/3)*\frac{4}{x^3}*(\phi(x)-\phi(0))$

where $\phi$ is the standard normal PDF and $\Phi$ is the standard normal CDF.

Show that $F(x)>0$ for all $x>0$

So far I have that the limit as x->0 = $\infty$ and that as x->$\infty$, $ F(x)->0$

But, then showing the first derivative is negative leads to the same problems as showing the equation above is positive and I'm left where I started.

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up vote 0 down vote accepted

Here is one possible approach.

Assuming that you mean you want to prove that $$F(x) = \left.\left.\frac{1}{x^2}\right[2\Phi(x)-1\right] + \frac{2}{3x}\phi(x) + \left.\left.\frac{8}{3x^3}\right[\phi(x)-\phi(0)\right] > 0 ~~\text{for}~ x > 0,$$ which holds if $$(2\Phi(x)-1) + \frac{2}{3}x\phi(x) + \left.\left.\frac{8}{3x}\right[\phi(x)-\phi(0)\right] > 0 ~~\text{for}~ x > 0,$$ notice that $\Phi(x)-\Phi(0) = \Phi(x)-\frac{1}{2}$ is the probability that a standard normal random variable has value in $[0,x]$ and that $\frac{\mathrm d}{\mathrm dx}\phi(x) = \phi^{\prime}(x) = -x\phi(x)$. Thus, we can write that for $x >0$, $$\begin{align*} x^2F(x) &= (2\Phi(x)-1) + \frac{2}{3}x\phi(x) + \left.\left.\frac{8}{3x}\right[\phi(x)-\phi(0)\right]\\ &= 2\int_0^x \phi(t)\,\mathrm dt + \frac{2}{3}\phi(x)\int_0^x 1\,\mathrm dt + \frac{8}{3x} \int_0^x \phi^{\prime}(t)\,\mathrm dt\\ &= \int_0^x \left[2 - \frac{8t}{3x}\right]\phi(t) + \frac{2}{3}\phi(x)\,\mathrm dt \end{align*}$$ The integrand has value $2\phi(0) + \frac{2}{3}\phi(x) > 0$ at $t = 0$, $\frac{2}{3}\phi(x)$ at $t = \frac{3}{4}x$, and value $0$ at $t = x$. So if the integrand can be proved to be nonnegative on $(0,x)$, you are done.

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