Behavior of normal distribution when close to infinity

Question

Let $X$ be a normal distribution with mean $0$ and variance $\sigma^2$. For $c>0$, define a function $f(c)=c\mathbb{P}(X>c)$, then what's the behavior of $f$? For example, what is $\lim(f)_{c\to\infty}$? Is $f$ montone?

Answer 1

Another way to see this is to apply Chebyshev's Inequality, which states that for a random variable $X$ with mean $\mu$ and variance $\sigma^2$, the probability that $X$ deviates from $\mu$ by at least $k \sigma$ is no more than $1/k^2$. Taking $X$ to be normally distributed with mean $0$ and variance $\sigma^2$, we have $$\mathbb{P}(X > c) \leq \mathbb{P}(|X| \geq c) \leq \sigma^2/c^2 \implies f(c) = c \mathbb{P}(X > c) \leq \sigma^2 / c$$ which tends to $0$ as $c \to \infty$, since $\sigma$ is fixed. This shows that you did not need the assumption of normality to show that $c \mathbb{P}(X > c)$ tends to zero: indeed this statement holds for any random variable $X$ with finite mean and finite variance.

In fact $\mathbb{P}(X > c)$ tends much more quickly to zero if $X$ is normally distributed (it decreases exponentially in $c$, as opposed to the polynomial bound offered by Chebyshev).

Answer 2

$\lim_{c \to \infty} \frac{P(X>c)}{1/c}=\lim_{c \to \infty} \frac{\int_{c}^{\infty} \frac{1}{\sqrt{2\pi} }e^{-\frac{x^2}{2\sigma^2}}dx}{1/c}\overset{\text{Hopital}}{=}\lim_{c \to \infty} \frac{ \frac{1}{\sqrt{2\pi} }e^{-\frac{c^2}{2\sigma^2}} }{\frac{-1}{c^2}}= 0$