Finite mean does not give a finite-variance Monte Carlo error

For $x\ge 1$, $x\mapsto x^{3\mathrm{/}2}$ is strictly increasing by [F2], so its positive reciprocal $x^{-3\mathrm{/}2}$ is decreasing; it is continuous by [F1]. Thus the displayed $F$ is nondecreasing and right-continuous, including at $x=1$ where its two formulas agree. For $x\ge 1$, [F2] gives $x^{1\mathrm{/}2}\ge 1$, hence $0\le x^{-3\mathrm{/}2}=x^{-1}{x}^{-1\mathrm{/}2}\le x^{-1}$. Since $x^{-1}\to 0$, $F(x)\to 1$ as $x\to +\mathrm{\infty }$; also $F(x)\to 0$ as $x\to -\mathrm{\infty }$. By [A1] and [F3], $F$ defines a Borel probability law $\mu$. Since $F(1)=0$, $\mu ((-\mathrm{\infty },1])=0$ and hence $\mu ([1,\mathrm{\infty }))=1$. Restricting $\mu$ to $[1,\mathrm{\infty })$ gives a probability space on which the identity random variable $Y(x)=x$ is measurable by [F4], has CDF $F$, and satisfies $Y\ge 1$.

If $0\le t<1$, $F(t)=0$ and hence $\mathbb{P}(Y>t)=1$. For $t\ge 1$, the CDF gives $\mathbb{P}(Y>t)=1-F(t)=t^{-3\mathrm{/}2}$. By the nonnegative expectation definition [F5] and layer-cake formula [F6], $\mathbb{E}[Y]=1+{\int }_1^{\mathrm{\infty }}{t}^{-3\mathrm{/}2}dt$. The function $t^{-3\mathrm{/}2}$ is nonnegative and continuous on every compact $[1,R]$ by [F1]. For $R>1$, [F8] gives ${\int }_1^R{t}^{-3\mathrm{/}2}dt=2(1-R^{-1\mathrm{/}2})\to 2$; [F9] identifies this finite improper integral with its Lebesgue value. Therefore $\mathbb{E}[Y]=3$.

Since $Y\ge 1$ on the restricted space, $\mathbb{P}(\mathrm{\mid }Y\mathrm{\mid }>t)=\mathbb{P}(Y>t)$. By the nonnegative expectation definition [F5] and [F7] with $p=2$, $\mathbb{E}[Y^2]=1+2{\int }_1^{\mathrm{\infty }}{t}^{-1\mathrm{/}2}dt$. Extend $h(t)=t^{-1\mathrm{/}2}$ from $[1,\mathrm{\infty })$ by $h(t)=1$ for $t<1$; it is continuous at $1$ and on both pieces, so [F11] makes it Borel. Since $[1,N]$ is Borel, [F12] makes each $h{\mathbf{1}}_{[1,N]}$ measurable. For each integer $N\ge 2$, [F1] gives continuity of $t^{-1\mathrm{/}2}$ on $[1,N]$, so [F17] makes it bounded and Riemann integrable there; [F8] and [F10] then give ${\int }_{[1,N]}{t}^{-1\mathrm{/}2}d\lambda =2(\sqrt{N}-1)$. These values diverge because the natural numbers are unbounded. The truncations increase pointwise to $t^{-1\mathrm{/}2}{\mathbf{1}}_{[1,\mathrm{\infty })}$, so [F13] gives ${\int }_{[1,\mathrm{\infty })}{t}^{-1\mathrm{/}2}d\lambda =+\mathrm{\infty }$. Hence $\mathbb{E}[Y^2]=+\mathrm{\infty }$.

The given sequence has common law $\mu$. By [F11], the functions $y\mapsto \mathrm{\mid }y\mathrm{\mid }$ and $y\mapsto y^2$ are Borel; applying [F16] to these nonnegative functions gives $\mathbb{E}\mathrm{\mid }{Y}_1\mathrm{\mid }=\mathbb{E}\mathrm{\mid }Y\mathrm{\mid }=3$ and $\mathbb{E}[Y_1^2]=\mathbb{E}[Y^2]=+\mathrm{\infty }$. Thus $Y_1$ is integrable, and [F16] applied to the identity gives $\mathbb{E}[Y_1]=\mathbb{E}[Y]=3$.

By step 4.1, $\mathbb{E}\mathrm{\mid }{Y}_1\mathrm{\mid }=3<\mathrm{\infty }$, so [F14] proves strong and weak consistency of ${\bar{Y}}_n$. Step 4.1 shows $\mathbb{E}[Y_1^2]=+\mathrm{\infty }$, so the finite-second-moment condition in [F15] fails and no finite population standard error $\sigma \mathrm{/}\sqrt{n}$ is available. For each realized sample of size $n\ge 2$, the empirical formula remains a statistic by [F15].