I have two questions as I read over Example 3.4.13 (Infinite Variance) of Rick Durrett's Probability: Theory and Examples.
The example assumes $X_1, X_2, \cdots$ to be i.i.d. R.V.'s and have distribution functions $P(|X_1| > x) = x^{-2}$ for $x \geq 1$ and $P(X_1 > x) = P(X_1 < -x)$.
The questions are (also highlighted in blue in the screenshot below):
(i) How do I derive that $E|X_1|^2 = \int_0^{\infty} 2x P(|X_1| > x) dx$?
(ii) In the example, $X_m$ is truncated using a threshold $c_n = n^{1/2} \log \log n$ and set to $Y_{n,m} = X_m 1_{(|X_m| \leq c_n)}$ so that $\Sigma_{m=1}^n P(Y_{n,m} \neq X_m) \leq nP(|X_1| > c_n) \rightarrow 0$.
My question is, why is it $\Sigma_{m=1}^n P(Y_{n,m} \neq X_m) \leq nP(|X_1| > c_n)$ instead of $\Sigma_{m=1}^n P(Y_{n,m} \neq X_m) = nP(|X_1| > c_n)$? Isn't $P(Y_{n,m} \neq X_m) = P(|X_m| > c_n)$, so $\Sigma_{m=1}^n P(Y_{n,m} \neq X_m) = \Sigma_{m=1}^n P(|X_m| > c_n) = \Sigma_{m=1}^n P(|X_1| > c_n) = nP(|X_1| > c_n)$ because $X_1, X_2, \cdots$ are i.i.d.?
Thank you very much in advance.
Here's the screenshot of the example:
Regarding question (i), this identity is a consequence of the LOTUS and integration by parts applied to the nonnegative random variable $|X_1|$ and the function $g:x\mapsto x^2$. If we let $F$ be the cumulative distribution function of $|X_1|$, we have : $$\begin{align}\mathbb E[g(|X_1|)]&=\int_0^\infty g(x)dF(x)\\ &=\int_0^\infty -g(x)d[1-F](x)\\ &=\underbrace{\left[-g(x)(1-F(x))\right]_0^\infty}_{(*)} - \int_0^\infty(1-F(x))d[-g](x)\\ &=0 + \int_0^\infty(1-F(x))dg(x)\\ &=\int_0^\infty(1-F(x))g'(x)dx \\ &=\int_0^\infty2x\mathbb P(|X_1|>x)dx\end{align} $$ Here, $(*)$ is equal to $0$ because, since $g$ is nondecreasing we have $$\begin{align}g(x)(1-F(x)) &= g(x)\mathbb E[\mathbf 1_{|X_1|\ge x}]\\ &= \mathbb E[g(x)\mathbf 1_{|X_1|\ge x}]\\ &\le \mathbb E[g(|X_1|)\mathbf 1_{|X_1|\ge x}] \end{align}$$ You can then apply DCT and take the limit as $x\to\infty$ to get the desired conclusion.
Regarding your question (ii), your calculations are absolutely correct, but remember that "equal" implies "less than or equal", so it is always correct to write $``a\le b"$ when $a=b$. Here the author probably chose to write it this way because we usually prove that nonnegative quantities converge to zero by upper bounding them. In any case, the conclusion remains the same.