Bounding Probability with Large Variance

Question

Dear amazing community,

Suppose I have a random variable $X\sim N(0,d)$ where $d \gg 1$ (i.e., normal distribution with a big variance). I want an upper bound for the following probability (as a function of $d$):

$$\Pr(|X|\leq 1)<\quad ?$$

Does anyone have a suggestion? You can assume $d$ to be as big as you want. Many thanks!

Answer 1

It's equivalent to give a lower bound for $P(X>1)$, and thus for $P(Z>\sqrt{d})$ for standard normal $Z$. This can be done using integration by parts. The first step gives

$$\int_a^\infty e^{-x^2/2} dx = e^{-a^2/2}/a-\int_a^\infty e^{-x^2/2}/x^2 dx.$$

This gives an upper bound, by replacing the second integral with $0$. You get a lower bound by integrating by parts again:

$$\int_a^\infty e^{-x^2/2} dx = e^{-a^2/2}/a-e^{-a^2/2}/a^3+\dots$$

where the dots are an integral of a positive function. Thus $\int_a^\infty e^{-x^2/2} dx \geq e^{-a^2/2} \left ( 1/a-1/a^3 \right )$. You can use this to get an upper bound on $P(|Z| \leq \sqrt{d})=1-2P(Z \geq \sqrt{d})$.

Repeating this procedure gives a whole sequence of upper and lower tail bounds, of the form $\sum_{k=1}^n (-1)^{k-1} c_k e^{-a^2/2} a^{1-2k}$, where $c_k$ are positive constants, odd values of $n$ give upper tail bounds, and even values of $n$ give lower tail bounds. However, since the $c_k$ grow rapidly, the errors eventually grow instead of shrinking.

Answer 2

Let $\mathrm{CDF(x)}$ be the cumulative distribution function of the normal function.

$\mathrm{CDF(x)} = \frac{1}{2} \left( 1+ \mathrm{erf} \left( \frac{x - \mu}{\sigma \sqrt{2}} \right) \right)$

$\mathrm{Pr(|X|)} \le 1 = \mathrm{CDF(1)} - \mathrm{CDF(-1)} = \frac{1}{2} \left( \mathrm{erf} \left( \frac{1}{\sqrt{2d}} \right) - \mathrm{erf} \left( -\frac{1}{\sqrt{2d}} \right) \right) = \mathrm{erf} \left( \frac{1}{\sqrt{2d}} \right) $

Then you can read arXiv:math/0607694 [math.CA] to choose your favourite inequality.

Answer 3

The PDF has $$ f(x) = \frac{1}{\sqrt{2\pi d}} e^{-\frac{1}{2d}x^2} \le \frac{1}{\sqrt{2\pi d}}$$ and thus we have $$ P(|X|<1) =\int_{-1}^1 f(x)dx < \frac{2}{\sqrt{2\pi d}}.$$

If $d\gg1,$ we have $f(x) \approx \frac{1}{\sqrt{2\pi d}}$ over the range (since the exponential barely decays at all), so this becomes pretty accurate.