Does $-\int_{[0,\infty)}y^p d(1-F)(y)=-\int_{[0,\infty)}y^p dT(y)$ hold true for $T(y)=P(Y\ge y)$?

Question

There is a calculation step in Lemma 5.4 of Probability Theory by Varadhan I am wondering about.

He defines the distribution function as $F(x):=P(X\le x)$ according to Chapter 1.6 and he also introduces the Riemann-Stieltjes integral later on. In Lemma 5.4 he defines $T(y):=P(Y\ge y)$. For continuous distributions, it holds $0(1-F)(y)=P(Y>y)=P(Y\ge y)=T(y)$. However, this is not true in general and I assume, that $(1-F)(y)\ne T(y)$.

My attempt: Since $Y$ is a nonnegative random variable, it holds $P(Y\le y)$ for any $y<0$. Then it holds

$$ \begin{aligned} E[Y^p]=\int_{\Omega}Y^p dP=\int_\mathbb{R}y^p dF(y)=\int_{[0,\infty)}y^p dF(y)=-\int_{[0,\infty)}y^p d(1-F)(y). \end{aligned} $$

On the other hand, the very first equality in the proof of Lemma 5.4 claims that

$$E[Y^p]=-\int_{[0,\infty)}y^p dT(y).$$

How does this coincide? (Again: It is clear to me, that this is true for continuous distributions.)

Any hint or help is appreciated. Thank you in advance!

EDIT:

Alternatively, just notice, that by assumption

$$0\le(1-F)(y)\le P(Y\ge y) \le \frac{1}{y}\int_{Y\ge y} X dP$$

Then, one finds by using integration by parts for Stieltjes integrals

$$ \begin{aligned} E[Y^p]=-\int_{[0,\infty)}py^{p-1} d(1-F)(y)&=\int_{[0,\infty)}py^{p-1} (1-F)(y)dy\\ &\le \int_{[0,\infty)}py^{p-1} \frac{1}{y}\int_{Y\ge y} X dP dy. \end{aligned} $$

But as Mittens stated, you can also see, that $py^{p-1}P(Y>y)$ and $py^{p-1}P(Y\ge y)$ only differ on a countable set, such that the integral with respect to the Lebesgue measure / the variable $y$ is the same. This in return gives the same Stieltjes integral by using integration by parts for Stieltjes integral again, since $$y^pT(y)\bigg|_{y=0}^{y=\infty}=y^p(1-F)(y)\bigg|_{y=0}^{y=\infty}=0.$$

Answer 1

Here is a quick prof using Fubini's theorem.

First notice that $m(t)=P[Y> t]$ is monotone nonincreasing and right continuous; hence $m(t)=m(t-)=\lim_{s\nearrow t}m(t)=P[T\geq t]$ almost surely with respect to the Lebesgue measure on $[0,\infty)$.

As $P(Y\geq t)\leq\frac1t\int_{T\geq t}X\,dP$

\begin{align} E[Y^p]&=\int^\infty_0 p t^{p-1}P[Y>t]\,dt=\int^\infty_0 p t^{p-1}P[Y\geq t]\,dt\\ &\leq p\int^\infty_0t^{p-2}\int_{Y\geq t}X\,dP=p\int_{\Omega} X\int^Y_0t^{p-2}\,dp\\ &=\frac{p}{p-1}\int_{\Omega} X Y^{p-1}\,dP \end{align} and the rest as in your notes.

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One can also use integration by parts as outlined by the OP. But a few facts about the behavior of the $P[y>t]$ at $0$ and $\infty$ require some attention.

First notice that if $Y\in L^+_p$, then \begin{align} \lim_{t\rightarrow0}t^pP[Y>t]=\lim_{t\rightarrow\infty}t^pP[T>t]=0\tag{1}\label{one} \end{align} The second limit follows from Chebyshev-Markov's inequality $$P[Y>t]\leq\frac{1}{t^p}\int_{\{Y>t\}}Y^p\,dP$$ hence, by monotone convergence $$\lim_{t\rightarrow\infty}t^pP[Y>t]=\lim_{t\rightarrow\infty}\int_{\{Y>t\}}Y^p\,dP=0$$ As for the first limit, the fact that $E[Y^p]=\int^\infty_0pt^{p-1}P[Y>t]\,dt<\infty$ implies that $\int_Apt^{p-1}P[T>t]\,dt$ is small when $|A|$ is small enough. That is, given $\varepsilon>0$, there os $\delta>0$ such that for any measurable set $A\subset[0,\infty)$, $|A|<\delta$ implies that $\int_Apt^{p-1}P[X>t]\,dt<\varepsilon$. In particular, for $0<t<\delta$ $$t^pP[Y>t]\leq \int^t_0ps^{p-1}P[Y>s]\,ds<\varepsilon$$ since $P[Y>t]\leq P[Y>s]$ for all $0\leq s\leq t$. Let $F(t)=P[Y\leq t]$ Let $\nu$ be the negative measure such that $\nu(a,b]=P[T>b]-P[Y>a]=-(1-F(b))-(1-F(a))=-P[a<Y\leq b]$

Lebesgue integration by parts yields

\begin{align} -\int^b_ay^p\,\nu(dy)&=-\int^b_ay^p d(1-F)(y)=-(1-F(y))y^p|^b_a+\int^b_a(1-F(y-))py^{p-1}dy\\ &= (1-F(a))a^p-(1-F(b))b^p+\int^b_aP[Y\geq y] py^{p-1}\,dy \end{align} By \eqref{one}, letting $a\rightarrow0+$ and $b\rightarrow\infty$ gives \begin{align} -\int^\infty_0y^p d(1-F)(y)=p \int^\infty_0P[Y\geq y]y^{p-1}\,dy \end{align} the rest as above (and your notes).