Tail probability bound on the expected value of measurable function of a random variable

probabilityprobability theory

I was reading Allan gut's graduate probability book and came across this problem,

Let $X$ be a non-negative random variable and $g$ be a non-negative, strictly increasing, differentiable function. Then,

$ Eg(X) < \infty \iff \sum_{n=1}^{\infty}g'(n)\textbf{P}(\textbf{X}>n) <\infty$ .

I am stuck and any help will be appreciated.

Best Answer

This result is false. As indicated by my comments, we can define a function $g$ that is strictly increasing, differentiable, and satisfies $g(n)=n$ and $g'(n)=0$ for all $n \in \{0, 1, 2, \ldots\}$ (see also this Math.SE post). Then $\sum_{n=1}^{\infty} g'(n)P[X>n]=0$ regardless of $X$, but if $X$ is non-negative and integer-valued then $g(X)=X$.

I found a PDF of the book here (see page 76).

Theorem 12.3 is wrong but can be fixed as follows: Let $X$ be a nonnegative random variable. Let $g:[0, \infty)\rightarrow\mathbb{R}$ be a function that is nonnegative, continuous, differentiable with a continuous derivative for all $x>0$, and satisfies $g'(x)\geq 0$ for all $x>0$. Then

a) $E[g(X)] = g(0) + \int_0^{\infty} g'(t) P[X>t]~\mathrm dt$

b) For $n \in \{1, 2, 3, \ldots\},$ define \begin{align} L_n &= \inf_{t \in [n, n+1]}g'(t)\\ H_n &= \sup_{t \in [n, n+1]}g'(t) \end{align} In addition to the conditions of part (a), suppose that $L_n/H_n\rightarrow 1$. Then $$E[g(X)]<\infty \iff \sum_{n=1}^{\infty} g'(n)P[X>n]<\infty.$$


The conditions for part (b) are satisfied by functions $g$ of the form $g(x) = x^c$ for any $c> 0$.


To prove part (b) I use the fact that if $\{a_n\}_{n=1}^{\infty}$ and $\{v_n\}_{n=1}^{\infty}$ are nonnegative sequences and $v_n\rightarrow 1,$ then $$ \sum_{n=1}^{\infty} a_n<\infty \iff \sum_{n=1}^{\infty} a_nv_n < \infty.$$


Proof of part (b): First suppose $\sum_{n=1}^{\infty} g'(n)P[X>n]<\infty$. Let $m$ be such that $L_n>0$ for all $n \geq m$. Then from part (a) \begin{align} E[g(X)] &= g(0) + \int_0^{\infty} g'(t)P[X>t]~\mathrm dt\\ &= g(0) + \int_0^m g'(t)P[X>t]~\mathrm dt + \int_m^{\infty} g'(t)P[X>t]~\mathrm dt\\ &\leq g(0) + \int_0^m g'(t)P[X>t]~\mathrm dt + \sum_{n=m}^{\infty} H_nP[X>n]\\ &\leq g(0) + \int_0^m g'(t)P[X>t]~\mathrm dt + \sum_{n=m}^{\infty} \frac{H_n}{L_n}g'(n)P[X>n]\\ \end{align}

where we have used the fact $g'(n)/L_n\geq 1$. The final expression is finite since we can utilize the previous fact with $v_n=\frac{H_n}{L_n}$ and $a_n=g'(n)P[X>n]$. Thus $E[g(X)]<\infty$. The other direction is similar.

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