Upper bound on integrand that converges to zero?

Question

Let $H(x)$ be the cumulative distribution function for a continuous nonnegative random variable $X$ that has a finite mean. Let $\tilde{H}(x) = 1 - H(x)$ (the tail probability).

For $r \geq 1$, suppose that the integral $\int_{0}^{\infty} x^r \tilde{H}(x) \, dx$ is convergent. (Remark: This condition arises when the $(r+1)$th moment of $X$ is finite. While the integral is indeed convergent for $r=0$ by assumption of $X$ having a finite mean, this case is not being considered here.)

Does there necessarily exist a function $B(x)$ such that $\tilde{H}(x) \leq \displaystyle\frac{B(x)}{x^{r+1}}$ where $B(x) \to 0$ (monotonically decreasing) as $x \to \infty$?

Remarks

As this is my third attempt to formulate a proposition that can do what I’m seeking, here is what I have tried:

• I am aware that I cannot necessarily claim that $x^r \tilde{H}(x) \to 0$ as this does not necessarily hold for improper integrals, even if they converge. I can make that conclusion if it is established that $x^r \tilde{H}(x) \to \ell$ for some finite limit $\ell$, but the existence of $\ell$ does not necessarily follow solely from $x^r$ being monotone increasing to infinity and $\tilde{H}(x)$ being monotone decreasing to zero.

• But if $\tilde{H}(x) = \displaystyle\frac{B(x)}{x^{r+1}}$ with $B(x) = c>0$ constant then the integral $\int_{0}^{\infty} x^r \tilde{H}(x) \, dx$ is divergent.

That $\tilde{H}(x)$ decreases monotonically to zero (from being a tail probability) seems to make all the difference. I tried constructing a counterexample for $r = 1$ in which $$ \displaystyle \int_{0}^{\infty} x \tilde{H}(x) \, dx = \int_{0}^{\infty} D(x) \, dx $$ is convergent but $D(x)$ is nondecreasing. I thought to construct $$ \tilde{H}(x) = \frac{D(x)}{x} $$ where $D$ is mostly zero except that at each integer $n$ there is a triangle of height 1 and area $1/n^2$ (so $\int_{0}^{\infty} D(x) \, dx$ converges but $D(x)$ does not decrease to zero). However $$ \tilde{H}'\!(x) = \frac{x D'\!(x) - D(x)}{x^2} = \frac{1}{x}\left( D'\!(x) - \tilde{H}(x) \right) $$ whenever $\tilde{H}(x)$ is differentiable. Thus as $x$ increases and hence $\tilde{H}(x)$ decreases, the gradient that I can introduce via $D'\!(x)$ must also decrease to preserve $\tilde{H}'\!(x) \leq 0$. So in particular, if I tried to specify that $D(n) = 1$ for each positive integer $n$ then I would be restricted to $D'\!(x) \leq 1/n$ for the triangle at $n$, which leads to a triangle of area $\geq 1/2n$. Doing so causes $\int_{0}^{\infty} D(x) \, dx$ to diverge.

(Many thanks again.)

Answer 1

This answer requires that $\tilde{H}(x)$ is differentiable on an interval $(a,\infty)$. In the following analysis, $\tilde{H}(x)$ has been replaced with $f(x)$ to make the assumptions more explicit and traceable. I have refactored the logic into small units to increase reusability.

Lemma 1. Let function $f(x)$ be real-valued and nonnegative on $(a,\infty)$, and suppose that for $r > -1$ the integral $\int_a^\infty x^r f(x) \, dx$ is convergent. Write $$\displaystyle f(x) = \frac{g(x)}{x^{r+1}}$$ Suppose further that $g(x) \to c$ as $x \to \infty$ where $c$ is finite. Then $c = 0$.

Proof of Lemma 1. For $x > a$, $f(x)$ is nonnegative so $g(x)$ is nonnegative hence $c \geq 0$. Now suppose that $c > 0$ and let $\epsilon = \tfrac{1}{2}c$. Then $\epsilon > 0$, so from $g(x) \to c$: $$ \text{there exists}\ x' > a \ \text{such that if}\ x > x' \ \text{then}\ |g(x) - c| < \epsilon $$ or equivalently $$ \text{there exists}\ x' > a \ \text{such that if}\ x > x' \ \text{then}\ c - \epsilon < g(x) < c + \epsilon $$ so in particular $$ \text{there exists}\ x' > a \ \text{such that if}\ x > x' \ \text{then}\ g(x) > c - \epsilon = \tfrac{1}{2}c $$ Therefore $$\begin{align*} \int_a^\infty x^r f(x) \, dx &= \int_a^\infty \frac{g(x)}{x} \, dx \\&= \int_a^{x'} \frac{g(x)}{x} \, dx + \int_{x'}^\infty \frac{g(x)}{x} \, dx \\&\geq \int_a^{x'} \frac{g(x)}{x} \, dx + \int_{x'}^\infty \frac{c}{2x} \, dx \end{align*}$$ Now the second term on the righthand side diverges to infinity but the lefthand side is a convergent integral by assumption. Contradiction, therefore $c = 0$. $\blacksquare$

Lemma 2. Let function $f(x)$ be real-valued and nonnegative on $(a,\infty)$, and suppose that for $r > -1$ the integral $\int_a^\infty x^r f(x) \, dx$ is convergent. Write $$\displaystyle f(x) = \frac{g(x)}{x^{r+1}}$$ Suppose further that $f(x)$ is differentiable and nonincreasing on its domain. Then there exists $c \geq 0$ (finite) such that $g(x) \to c$.

Proof of Lemma 2.

Let $\left\{ x_n \right\}_n$ be any strictly increasing sequence of numbers drawn from $(a,\infty)$. Construct $\left\{ g_n \right\}_n$ by putting $g_n = g(x_n)$ for each $n$. Proceeding in four steps:

1. It is sufficient to show that there exists $c$ finite such that $g_n \to c$ as $n \to \infty$. The function $f(x)$ is nonincreasing so its limit as $x \to \infty$ is either finite or $-\infty$. Furthermore $f(x)$ is nonnegative so the option of $-\infty$ is excluded and the limit must be nonnegative. Meanwhile the function $d(x) = x^{r+1} \to \infty$ as $x \to \infty$. Now $g(x) = f(x)d(x)$ therefore $g(x)$ either has a finite, nonnegative limit or it diverges to infinity. So suppose there exist sequences $\left\{ x_n \right\}_n$ and $\left\{ x'_m \right\}_m$ such that if $g_n = g(x_n)$ for each $n$ and $g'_m = g(x'_m)$ for each $m$ then $g_n \to c$ as $n \to \infty$ and $g'_m \to c'$ as $m \to \infty$ but $c \neq c'$. Construct sequence $\left\{ x''_k \right\}_k$ by merging $\left\{ x_n \right\}_n$ and $\left\{ x'_m \right\}_m$ in increasing order. Then as $k \to \infty$, $g(x''_k)$ does not converge. Contradiction hence $c = c'$. Therefore if $g_n \to c$ finite as $n \to \infty$ for any given sequence $\left\{ x_n \right\}_n$ then $g(x) \to c$ as $x \to \infty$.

2. $\displaystyle g'(x) \leq (r+1)\frac{g(x)}{x}$ for all $x > a$. The function $f(x)$ is differentiable on $(a,\infty)$. Indeed $$\begin{align*} f'(x) &= -\frac{r+1}{x^{r+2}}g(x) + \frac{1}{x^{r+1}}g'(x) \\&= \frac{1}{x^{r+1}}\left( g'(x) - (r+1)\frac{g(x)}{x} \right) \end{align*}$$ on that interval. Now for $x > a$, $f(x)$ is nonincreasing so $f'(x) \leq 0$. Meanwhile $1/x^{r+1}$ is nonnegative. Therefore $\displaystyle g'(x) - (r+1)\frac{g(x)}{x} \leq 0$ for $x > a$ and result follows immediately.

3. For all $\epsilon > 0$ there exists $t > 0$ such that if $t \leq x_i \leq x_j$ then $g_j - g_i < \epsilon$. Let $\epsilon > 0$. The integral $\int_a^\infty x^r f(x) \, dx$ is convergent so there exists $t$ such that $$ \int_{t}^\infty x^r f(x) \, dx < \frac{\epsilon}{r+1} $$ noting that $r+1 > 0$ by assumption. Now if $t \leq x_i \leq x_j$ then $$\begin{align*} \int_{t}^\infty x^r f(x) \, dx &\geq \int_{x_i}^{x_j} x^r f(x) \, dx & \text{as $f(x)$ is nonnegative} \\&= \int_{x_i}^{x_j} \frac{g(x)}{x} \, dx \\&\geq \int_{x_i}^{x_j} \frac{g'(x)}{r+1} \, dx & \text{by step 2} \\&= \frac{g_j - g_i}{r+1} \end{align*}$$ and result follows immediately.

4. $\left\{ g_n \right\}_n$ is a Cauchy sequence. We need to show that for any $\epsilon > 0$ there exists $N$ such that if $m,n > N$ then $|g_n - g_m| < \epsilon$. So let $\epsilon > 0$. By step 3, there exists $t > 0$ such that $$ \text{if}\ t \leq x_i \leq x_j \ \text{then}\ g_j - g_i < \frac{\epsilon}{2} \tag{1}\label{eqn:boundjumpup} $$ Let $\ell = \inf\left\{ g_n : x_n \geq t \right\}$. Now $g(x)$ is bounded below (by zero) so $\ell$ is finite. Hence by the definition of infimum, there exists $N$ such that $x_N \geq t$ and $\displaystyle g_N < \ell + \frac{\epsilon}{2}$. Indeed $\displaystyle g_N - \frac{\epsilon}{2} < \ell$. Moreover, to emphasize the construction of $\ell$, if $k \geq N$ then $\ell \leq g_k$. Consequently $$ \text{if}\ k \geq N \ \text{then}\ g_N - \frac{\epsilon}{2} < g_k \tag{2}\label{eqn:boundbelowgN} $$ Meanwhile, applying (\ref{eqn:boundjumpup}) with $i = N$ and $j = k$ yields $$ \text{if}\ k \geq N \ \text{then}\ g_k - g_N < \frac{\epsilon}{2} $$ or equivalently $$ \text{if}\ k \geq N \ \text{then}\ g_k < g_N + \frac{\epsilon}{2} \tag{3}\label{eqn:boundabovegN} $$ Now suppose $m,n > N$. Then (\ref{eqn:boundbelowgN}) and (\ref{eqn:boundabovegN}) yield $$\begin{align*} g_N - \frac{\epsilon}{2} &< g_m < g_N + \frac{\epsilon}{2} \quad \text{and} \\ g_N - \frac{\epsilon}{2} &< g_n < g_N + \frac{\epsilon}{2} \end{align*}$$ therefore $|g_n - g_m| < \epsilon$ as required.

Having established that $\left\{ g_n \right\}_n$ is a Cauchy sequence of real numbers, it follows immediately that there exists $c$ finite such that $g_n \to c$ as $n \to \infty$. Result follows from step 1. $\blacksquare$

Proposition. Let function $f(x)$ be real-valued and nonnegative on $(a,\infty)$, and suppose that for $r > -1$ the integral $\int_a^\infty x^r f(x) \, dx$ is convergent. Write $$\displaystyle f(x) = \frac{g(x)}{x^{r+1}}$$ Suppose further that $f(x)$ is differentiable and nonincreasing on its domain. Then $g(x) \to 0$ as $x \to \infty$.

Proof of Proposition. Apply Lemma 2 and then Lemma 1. $\blacksquare$