3

I am currently hospitalised and reading a queueing theory book. I encountered in a proof this, and I fail to understand how this is true: $$E[R_j]=\int_0^\infty{P(R_j>u)du}$$

Other than the fact that $R_j$ is a random variable defined in $[0,\infty)$ I dont think that any further context is needed for my question.

Due to my hospitalisation I don't have good access to my more basic probability books but I really don't recall reading any similar alternative definition for the expected value.

If someone is curious or believes that the context is important, in a stochastic renewal process with holding times $X_j$, for a given value $x>0$, $R_j$ is defined as $R_j=X_j$ when $X_j\le x$, and $R_j=0$ otherwise.

Regardless of context, however, I find the line in question hard to comprehend.

Panagiotis Basiouras Serrano
  • 55
  • 4
  • $R_j$ is indeed continuous and positive, I just don't understand why the formula is true. The only definition for expected values of continuous random variables that I am familiar with is $E(R_j)=\int uP(R_j=u)du$ – Panagiotis Basiouras Serrano Nov 20 '21 at 13:21
  • 1
    It can be proved (by Fubinii) that for non-negative random variable $X$, both $\int_{\Omega} X(\omega)d\mathbb P(\omega)$ (i.e - the standard definition of expectation) and $\int_0^\infty \mathbb P(X >t)dt$ (i.e your definition of expectation) gives the same result. The latter is known as layer cake representation and can be generalized to $\mathbb E[|X|^p] = \int_0^\infty pt^{p-1}\mathbb P (|X| > t)dt$ (or even more, for example Young functions) – Presage Nov 20 '21 at 13:23
  • Thank you for your answer, I didn't know about the layer cake representation, I will have to study about that! – Panagiotis Basiouras Serrano Nov 20 '21 at 19:16
  • 1
    @Surb why does $R_j$ have to be continuous? It has to be non-negative, because the more general expression would be $E[R_j]=\int_0^\infty{P(R_j>u)du}-\int_{-\infty}^0{P(R_j<u)du}$ – Henry Nov 20 '21 at 22:23
  • @Henry: Indeed, continuity is not mandatory. – Surb Nov 21 '21 at 08:02
  • Also related https://math.stackexchange.com/questions/172841/explain-why-ex-int-0-infty-1-f-x-t-dt-for-every-nonnegative-rando – Henry Nov 21 '21 at 10:00

2 Answers2

4

If $X\geqslant 0$ you have that

$$ \begin{align*} \mathrm{E}[X]&=\int_{\Omega }X dP=\int_{\mathbb{R}}t \,dF_X(t)\\ &=\int_{[0,\infty )}t \,dF_X(t)\\ &=\int_{[0,\infty )}\int_{[0,\infty )}[s\leqslant t]\,d s\,dF_X(t)\\ &=\int_{[0,\infty )}\int_{[0,\infty )}[s\leqslant t]\,dF_X(t)\,d s\\ &=\int_{[0,\infty )}\Pr [X\geqslant s]\,d s \end{align*} $$

where $[s\leqslant t]$ is an Iverson bracket. In general you have that

$$ \begin{align*} \mathrm{E}[X]&=\int_{\Omega }X dP=\int_{\mathbb{R}}t \,dF_X(t)\\ &=\int_{[0,\infty )}t \,dF_X(t)+\int_{(-\infty,0)}t \,dF_X(t)\\ &=\int_{[0,\infty )}\Pr [X\geqslant s]\,d s+\int_{(-\infty,0)}\int_{(-\infty,0)}-[s\geqslant t]\,d s\,dF_X(t)\\ &=\int_{[0,\infty )}\Pr [X\geqslant s]\,d s-\int_{(-\infty,0)}\int_{(-\infty,0)}[s\geqslant t]\,dF_X(t)\,d s\\ &=\int_{[0,\infty )}\Pr [X\geqslant s]\,d s-\int_{(-\infty,0)}\Pr [X\leqslant s]\,d s \end{align*} $$

Masacroso
  • 30,417
3

Let me make a hand-waving proof that gets at the intuition - a discrete case to make it clearer. Say you have a simple process that returns X which is 1, 2 or 3 with probabilities $p_1$, $p_2$ and $p_3$. The expected value is $1p_1+2p_2+3p_3$. Another way to write this is like this:
$$1p_1+\hspace{45pt}$$ $$1p_2+1p_2+\hspace{15pt}$$ $$1p_3+1p_3+1p_3$$ If you read this sum by column, you get three terms: $1p_1+1p_2+1p_3$, and $1p_2+1p_3$ and then $1p_3$. The first part is $P(X\ge 1)$, the second part is $P(X\ge 2)$ and the third part is $P(X\ge 3)$. So the expected value is the sum of the exceedance probabilities.

TickaJules
  • 170