Let $X\sim N(0,1)$. I want to prove the next innequality holds for $x\geq 0$:
$$\left(\dfrac{x}{x^2+1}\right)f(x)\ \leq \ \mathbb{P}[X\geq x]$$
where $f(x)$ is the pdf of $X$. I've already read a proof here, but I'm having trouble with the calculation.
The idea is to proof $g(x) = \mathbb{P}[X\geq x] – \left(\frac{x}{x^2+1}\right)f(x) > 0$ for $x\geq 0$. The proof for $x=0$ it's easy, but the problem comes when assuming $x>0$. At one step, the author takes the derivative of $g(x)$ to conclude $g(x)$ is positive, which results:
$$g^{\prime}(x) = -2 \dfrac{e^{-x^2/2}}{(x^2 + 1)^2}.$$
My problem comes when I compute the derivative trying to reach that result: computing directly the derivative my problem comes in the term
$$\dfrac{d}{dx} \mathbb{P}[X\geq x] = \dfrac{d}{dx}\int_{x}^{\infty}\dfrac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}dt,$$
I try to apply Leibniz integral rule but the upper limit of integral was a problem, and sincerely, although that derivative wasn't a problem, I have no idea how the author reach that expression since
$$\dfrac{d}{dx}\left[\left(\frac{x}{x^2+1}\right)f(x)\right] = -\dfrac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \dfrac{x^4 + 2 x^2 – 1}{(x^2 + 1)^2}.$$
Having this trouble with this approach to the proof, I found that another way to prove the bound was answered here where the next innequality holds since $x\leq t$:
$$\dfrac{1}{x^2}\int_{x}^{\infty} e^{-\frac{t^2}{2}}dt \geq \int_{x}^{\infty}\dfrac{1}{t^2}e^{-\frac{t^2}{2}}dt,$$
the suggestion here was applying integration by parts to the right side of the innequality, so, I consider:
$$u = e^{-\frac{t^2}{2}}, \quad dv = \dfrac{1}{t^2}dt$$
but then again I'm having trouble trying to compute the integral since:
$$\int \dfrac{1}{t^2}e^{-\frac{t^2}{2}}dt = -\dfrac{e^{-\frac{t^2}{2}}}{t} – \int e^{-\frac{t^2}{2}}dt$$
and I'm stuck trying to understand how can I reach that this equality hold
$$-\dfrac{e^{-\frac{t^2}{2}}}{t}\bigg\vert_{t=x}^{\infty} – \int_{t=x}^{\infty} e^{-\frac{t^2}{2}}dt = \dfrac{1}{x(x^2 + 1)}e^{-\frac{x^2}{2}}$$
since that will finish the alternative proof. Any help to conclude any of the calculations here would be appreciated.
Best Answer
$\frac d {dt} [\int_t^{\infty} e^{-x^{2}/2}-\frac t {1+t^{2}} e^{-t^{2}/2}]=-e^{-t^{2}/2}+\frac {t^{2}} {1+t^{2}} e^{-t^{2}/2}-e^{-t^{2}/2} \frac {(1+t^{2})-t(2t)} {(1+t^{2})^{2}}$ by product rule and quotient rule. A little simplification gives exactly the derivative quoted in the link.