Why are these two upper bounds the same?

Question

I'm trying to upper bound the quantity $\int_t^{\infty} e^{-x^2} dx$ as a function of $t$. I came up with two ways of doing this:

1. We have that $(x-t)^2 = x^2 -2tx + t^2 \ge 0 \implies -x^2 \le -2tx + t^2$, so that $\int_t^{\infty} e^{-x^2} dx \le \int_t^{\infty} e^{-2tx}e^{t^2} dx= e^{t^2} \int_t^{\infty} e^{-2tx} dx = e^{t^2} \left(-\frac{1}{2t} e^{-2tx}\bigg\rvert_{x=t}^{x=\infty}\right) = \frac{1}{2t}e^{-t^2}$.

2. We have, on the domain $x \in [t, \infty)$, that $x \ge t \implies 1 \le \frac{x}{t}$, so that $\int_t^{\infty}e^{-x^2} dx \le \int_t^{\infty} \frac{x}{t} e^{-x^2} dx = \frac{1}{t} \left(-\frac{1}{2}e^{-x^2} \bigg \rvert_{x=t}^{x=\infty}\right) = \frac{1}{2t} e^{-t^2}$.

Is it a coincidence that these two totally different methods give the same upper bounds? Or is there a connection between these two estimates that I'm not realizing?

Answer 1

You could also try and use the gaussian integral and taylor series to get that: $$\int_t^\infty e^{-x^2}dx=\frac{\sqrt{\pi}}{2}+\sum_{n=0}^\infty\frac{(-1)^{n+1}t^{2n+1}}{(2n+1)n!}$$ or if you take the first few terms you get: $$I\approx\frac{\sqrt{\pi}}{2}+\frac{t^3}{3}-t+O(-t^5)$$

Answer 2

The coincedence is that from the point of view of an integral your estimations are the same. Why?

Let's recall the substitution rule for integrals: For $\varphi \in C^1([a,b])$ it holds $$\int_{a}^{b} f(\varphi(x)) \cdot \varphi'(x)\,\mathrm{d}x = \int_{\varphi(a)}^{\varphi(b)} f(x)\,\mathrm{d}x$$

If we now have a closer look we can see that both of your estimations are of the form of the LHS:

For $$f(x) = \frac{1}{2}\exp(-tx)$$ and $$\varphi_1(x) = 2x-t$$ we get $$\int_{t}^{\infty} f(\varphi_1(x)) \cdot \varphi_1'(x)\,\mathrm{d}x = \int_t^\infty \exp(-2tx+t^2) \,\mathrm{d}x$$ for your first estimator while taking

$$\varphi_2(x) = \frac{x^2}{t}$$ it gives us $$\int_{t}^{\infty} f(\varphi_2(x)) \cdot \varphi_2'(x)\,\mathrm{d}x = \int_t^\infty \frac{x}{t}\exp(-x^2) \,\mathrm{d}x$$ so your second estimator. And because it holds $$\varphi_1(t) = \varphi_2(t) = t$$ as well as $$\varphi_1(\infty) = \varphi_2(\infty) = \infty$$ both functions give the same result by using the substitution rule namely $$\int_{\varphi(a)}^{\varphi(b)} f(x)\,\mathrm{d}x = \int_t^\infty \frac{1}{2}\exp(-tx) \,\mathrm{d}x$$

Moreover: Any suitable $\varphi \in C^1([t,\infty))$ with $\varphi(t) = t$ and $\varphi(\infty) = \infty$ will lead to the same result.

So what you did in both cases was actually estimating $\exp(-x^2)$ by $f(\varphi(x)) \cdot \varphi'(x)$ for a suitable $\varphi$