4
$\begingroup$

I am asked to prove the following lower bound for the normal tail. Let $X\sim N(0,1)$

$$P(X\geq a)\geq c\exp\left(-a-\frac{a^{2}}{2}\right)$$ for some $c>0$ .

To do this , as a hint I am asked to find the density of $N(0,1)$ with respect to $N(a,1)$ which I have computed as $\exp\left(-xa+\frac{a^{2}}{2}\right)$ and then asked to show that

$$ \bigg(F(t)-F(0)\bigg)\exp\left(-a(t+a)+\frac{a^{2}}{2}\right)\leq P(X\geq a)$$ for all $a,t>0$ . Where $F$ is the the cdf of a standard Gaussian.

So I would have to make $t=1$ .

But I am unable to show the above inequality. How do I use the density?

For example if $\mu\sim N(a,1)$ then the LHS is $$\displaystyle\int_{0}^{t}\exp\left(-xa+\frac{a^{2}}{2}-a^{2}-at-\frac{a^{2}}{2}\right)\,d\mu(x) = \int_{0}^{t}\exp(-a(x+t))\,d\mu(x)$$ . How do I now go from here to $\int_{a}^{\infty}\exp\left(-xa+\frac{a^{2}}{2}\right)\,d\mu(x)=P(X\geq a)$ ?

$\endgroup$
2
  • $\begingroup$ By ''the density of $N(0,1)$ with respect to $N(a,1)$'' you mean the conditional density function, right ? $\endgroup$
    – Hamdiken
    Jan 18, 2023 at 22:53
  • $\begingroup$ I mean the Radon-Nikodym derivative . If $\mu\sim N(a,1)$ and $\nu\sim N(0,1)$ then for any Borel set $\nu(A)=\int_{A} \exp(-xa+\frac{a^{2}}{2})\,d\mu$ . Note that I am integrating wrt $\mu$ and not the Lebesgue measure. $\endgroup$
    – Dovahkiin
    Jan 19, 2023 at 9:58

1 Answer 1

1
$\begingroup$

I don't know why this is becoming a recurrent theme in my questions but again I am posting an answer to my own question .

$$P(X\geq a)=\int_{a}^{\infty}\exp(-ax+\frac{a^{2}}{2})\,d\mu(x)=\exp(-a(a+t)+\frac{a^{2}}{2})\int_{a}^{\infty}\exp(-a(x-t)+a^{2})\,d\mu(x)\\=\exp(-a(a+t)+\frac{a^{2}}{2})\bigg(\int_{a}^{a+t}\exp(-a(x-t)+a^{2})\,d\mu(x)+\int_{a+t}^{\infty}\exp(-a(x-t)+a^{2})\,d\mu(x)\bigg)$$

Thus

$$P(X\geq a)\geq \exp(-a(a+t)+\frac{a^{2}}{2})\int_{a}^{a+t}\exp(-a(x-t)+a^{2})\,d\mu(x)\,$$

Now for $x\in(a,a+t]$ we have $\exp(-a(x-t)+a^{2})\geq 1$

Thus $$P(X\geq a)\geq \exp(-a(a+t)+\frac{a^{2}}{2})\int_{a}^{a+t}\,1 \,d\mu=\exp(-a(a+t)+\frac{a^{2}}{2})\mu(a,a+t]$$

Now as $\mu$ has mean $a$ , $\mu(a,a+t]= F(t)-F(0)$ where $F$ is the cdf of a standard normal distribution.

Thus $P(X\geq a)\geq \exp(-a(a+t)+\frac{a^{2}}{2})\bigg(F(t)-F(0)\bigg)$ and we get our bound when $t=1$.

I'd appreciate other answers showing other methods though.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .