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I have a uniform random number: $U \in (0,1)$ and a constant scalar, $c$. I want to evaluate the expectation:

$$E([c-U])$$

where $[x]$ is the greatest integer less than or equal to $x$. Simulation shows that this equals $c-1$ (Python code below). How do I prove this?

import numpy as np
np.mean([int(i) for i in 3.2-np.random.uniform(size=10000)])

Replace the $3.2$ with any other number, $c$ and you always get something very close to $c-1$.

Rohit Pandey
  • 6,803

2 Answers2

4

The expected value is $$\mathbb{E}[\textrm{floor}(c-U)]=\int_0^1\textrm{floor}(c-u)du$$ Let us assume $c \in \mathbb{N}$. Then $\textrm{floor}(c-u)$ is almost surely equal to $c-1$, then $\mathbb{E}[\textrm{floor}(c-U)]=c-1$. This is because $\textrm{floor}(c-u)$ can take a value different from $c-1$ only if $u=0$ but such event has probability zero: $$\mathbb{E}[\textrm{floor}(c-U)]=(c-1)P(u\in(0,1])+cP(u\in\{0\})=c-1$$

Let us assume a more general $c \in \mathbb{R}$ such that $n<c<n+1$. We have that $\textrm{floor}(c-u)=n$ if $c-u\geq n$ and $\textrm{floor}(c-u)=n-1$ if $c-u<n$. Therefore $$\mathbb{E}[\textrm{floor}(c-U)]=nP(u\in[0,c-n])+(n-1)P(u \in (c-n,1])=$$ $$=n(c-n)+(n-1)(1-c+n)=nc-n^2+n-nc+n^2-1+c-n=c-1$$

Snoop
  • 15,214
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I'd like to prove a generalization which is very helpful in various contexts (like here: Need help validating a proof that for any point process with MTBF $t$, the events in an interval sized $u$ will be $\frac{u}{t}$). I'll prove (if $\eta$ is a random variable with mean $0$ and that's all we know about it):

$$E([c+\eta-U]) = c-1$$

using the result from @Snoop above.

First, we condition on $\eta$ to get:

$$E([c+\eta-U]|\eta) = c+\eta-1$$

Now, take expectation again

$$E([c+\eta-U]) = E(E([c+\eta-U]|\eta)) = E(c+\eta-1) = c-1$$

Rohit Pandey
  • 6,803