Exponential bound for tail of standard normal distributed random variable

Question

Let $X\sim N(0,1)$ and $a\geq 0$. I have to show that $$\mathbb{P}(X\geq a)\leq\frac{\exp(\frac{-a^2}{2})}{1+a}$$

I have no problem showing that $\mathbb{P}(X\geq a)\leq \frac{\exp(\frac{-a^2}{2})}{a\cdot\sqrt{2\pi}}$ which can be done by computation of the integral, but I haven't found a way to prove the inequality above. I tried to use the fact that $$\mathbb{P}(X\geq a) = \mathbb{P}(h(X)\geq h(a))\leq \frac{\mathbb{E}[h(X)]}{h(a)}$$ but I haven't found a suitable function $h$, yet. Can anybody give me an advice or an idea how to go forward?

Answer 1

We have $$ \mathbb{P}(X \ge a) = \int_a^\infty \frac{1}{\sqrt{2\pi}}\mathrm{e}^{-x^2/2}\mathrm{d} x = \int_0^\infty \frac{1}{\sqrt{2\pi}}\mathrm{e}^{-a^2/2 - ay - y^2/2}\mathrm{d} y. $$

If $0 \le a < 1$, we have \begin{align} \mathbb{P}(X \ge a) &\le \int_0^\infty \frac{1}{\sqrt{2\pi}}\mathrm{e}^{-a^2/2 - y^2/2}\mathrm{d} y\\ &= \frac{1}{2}\mathrm{e}^{-a^2/2}\\ &\le \frac{1}{1 + a}\mathrm{e}^{-a^2/2}. \end{align}

If $a \ge 1$, we have \begin{align} \mathbb{P}(X \ge a) &\le \int_0^\infty \frac{1}{\sqrt{2\pi}}\mathrm{e}^{-a^2/2 - ay}\mathrm{d} y\\ &= \frac{1}{a\sqrt{2\pi}}\mathrm{e}^{-a^2/2}\\ &\le \frac{1}{1 + a}\mathrm{e}^{-a^2/2} \end{align} where we have used $a\sqrt{2\pi} > 1 + a$.

We are done.