Expected Value in terms of CDF

It is well known that the expectation of a non-negative random variable $X$ can be written as

$\displaystyle \mathbb{E}[X] \overset{\underset{\mathrm{d}}{}}{=} \sum_{k=0}^\infty \mathbb{P}(X>k)  \overset{\underset{\mathrm{c}}{}}{=} \int\limits_0^\infty \mathbb{P}(X>x)\,dx$

for the discrete and continuous cases.

It's quite easy to prove this, at least in the discrete case. It is interesting that, in the same vein, this can be extended to arbitrary functions. That is,

$\begin{align} \displaystyle \mathbb{E}[g(X)]&=\sum_{k=0}^\infty g(k)\mathbb{P}(X=k)\\ &=\sum_{k=0}^\infty \Delta g(k)\mathbb{P}(X>k)\\ \end{align}$

where $\Delta g(k)=g(k+1)-g(k)$ is the forward difference operator. WLOG, We also have made an assumptions that $g(0)=0$.

Comparing this with continuous case, we can make an 'educated guess' that

$\displaystyle \mathbb{E}[g(X)]=\int\limits_0^\infty \mathbb{P}(X>x)\,dg(x)$

We made a post about estimating a sum with probability where we showed the expected error in the approximation is given by

$\displaystyle \mathbb{E}(\delta)=\sum_{k=1}^\infty f(k)\frac{\binom{n-k}{m}}{\binom{n}{m}}$

Note that the term involving the ratio of binomial coefficients can be interpreted as the probability of the minimum-of-the-$n$-tuple being greater than $k$. Therefore,

$\displaystyle \mathbb{E}(\delta)=\sum_{k=1}^\infty f(k)\mathbb{P}(Y>k)$

where $Y$ denotes the smallest order statistic.

Comparing this with our expression for expectation, we see that the expected value of the (probabilistic) Right Riemann sum is

$\displaystyle \mathbb{E}[\text{Right Riemann sum}]  \overset{\underset{\mathrm{d}}{}}{=} \mathbb{E}\left[\sum_{j=Y}^n f(j)\right]  \overset{\underset{\mathrm{c}}{}}{=} \mathbb{E}\left[ \int\limits_Y^1 f(x)\,dx \right]$

Without going into further calculations, I'm guessing that

(i) $\displaystyle \mathbb{E}[\text{Left Riemann sum}]  \overset{\underset{\mathrm{d}}{}}{=} \mathbb{E}\left[\sum_{j=0}^Z f(j)\right]  \overset{\underset{\mathrm{c}}{}}{=} \mathbb{E}\left[ \int\limits_0^Z f(x)\,dx \right]$

(ii) $\displaystyle \mathbb{E}[\text{error in Trapezoidal sum}]  \overset{\underset{\mathrm{d}}{}}{=} \frac{1}{2}\mathbb{E}\left[\sum_{j=Y}^Z f(j)\right]  \overset{\underset{\mathrm{c}}{}}{=} \frac{1}{2}\mathbb{E}\left[ \int\limits_Y^Z f(x)\,dx \right]$

where $Z$ denotes the largest order statistic.

Hope you enjoyed the discussion. See ya in the next post.

Until then

Yours Aye

Me