0

Let $T$ be a real valued absolutely continous r.v. such that $\lim\limits_{t\to +\infty} t^a \mathbb{P}(T>t)=1$ for some $a<1$, is it true that its expected value must be infinite?

I found the related statement in some notes I am reading, and being a bit rusty on these things I might really use some help.

It is straightforward to see this if $T$ is non-negative, but is it true also if $T$ possibly takes on negative values? If that is not the case, which would be a counterexample?

Many thanks in advance!

Bubu
  • 17
  • 2
  • 3
    I think we get $E(T^{+}) = \infty$, from which, based on whether $E(T^{-})$ is finite or not, we conclude either $E(T) = \infty$ or $E(T)$ does not exist. – Mason Aug 08 '24 at 16:39

1 Answers1

2

I'll just sum up everything in this answer.

Ther exists some $M>0$ such that for all $t\geq M$, you have $t^{a}P(T>t)\geq \frac{1}{2}$ .

For any positive random variable $X$, you have $E(X)=\int_{0}^{\infty}P(X>t)\,dt$

So here, $$E(T^{+})=\int_{0}^{\infty}P(T>t)\,dt= \int_{0}^{M}P(T>t)+\int_{M}^{\infty}P(T>t)\,dt$$

And $$\int_{M}^{\infty}P(T>t)\,dt=\int_{M}^{\infty}P(T>t)t^{a}\cdot\frac{1}{t^{a}}\,dt\geq \int_{M}^{\infty}\frac{1}{2t^{a}}\,dt=+\infty$$ for $a<1$.

Thus $E(T^{+})$ is necessarily $+\infty$. If for some reason $T$ is not a positive random variable and $E(T^{-})$ is also $+\infty$, then the expectation does not exist. If $E(T^{-})<\infty$, then too $E(T^{+})$ is $+\infty$.

Mr. Gandalf Sauron
  • 17,114