Why is it "obvious" that the expected value of a continuous uniform distribution is (a+b)/2?

Question

For a continuous uniform random variable X with support on an interval [a,b], where a<b, one can always calculate the expected value, by integrating, to arrive at the value of (a+b)/2.

However, since, I have been, for better or worse, "corrupted" by formal mathematics, I seem to no longer trust my intuition; even worse, I seem to question even the most "obvious" mathematical conclusions, given that I have been humbled one too many times whenever I have thought things to be "obvious".

However, I keep running into the fact that it is "obviously obvious" that the expected value of a continuous uniform distribution is (a+b)/2 time and time again. Everyone I talk to, literally everyone, no matter their backgrounds, their occupations, can see this to be true. Even middle school children can see this to be true (when uniform distribution is explained in an informal way). I, however, can't. I seem to be completely oblivious to its obviousness.

So, could someone please explain why it is "obvious" that the expected value of a continuous random variable on an interval [a,b], where a<b, is (a+b)/2? I understand the 'why' of it all when one formally defines a uniform distribution and integrates it. But I would be grateful if someone could provide an intuitive explanation for the same, preferably through an example.

Answer 1

Basic idea: for a uniform distribution on an interval the expectation is going to be the center of the interval since the distribution is symmetric about the center.

More rigorously: if a random variable $X$ is uniform on $[-c,c]$ for some $c > 0$ then $X$ has the same distribution as $-X$. This gives $E(X) = E(-X) = -E(X)$, leading to $E(X) = 0$.

Next, if $X$ is uniform on $[-c + d, c + d]$, then we add $d$ to the above random variable, giving $E(X) = d$, the center of the interval $[-c + d, c + d]$.

If $X$ is uniform on $[a,b]$ it falls into the situation of the last paragraph: solving $ - c + d = a$ and $c + d = b$ gives $d = {a + b \over 2}$ and $c = {b - a \over 2}$. Thus here $E(X) = d = {a + b \over 2}$.

Answer 2

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It's a good question. Let $X$ be a random variable with uniform distribution $\pdf$ in the closed interval $\intcc{a}{b}$.

\begin{align*} \pE\brp{X} &\eqd \int_{x\in\R} x \pdf(x) \dx &&\eqd \int_{x=a}^{x=b} x \frac{1}{b-a} \dx &&= \brs{\frac{1}{b-a}} \int_{x=a}^{x=b} x \dx \\&= \brs{\frac{1}{b-a}} \brlr{\frac{1}{2}x^2}_{x=a}^{x=b} &&= \brs{\frac{1}{b-a}} \frac{1}{2} \brp{ b^2 - a^2 } &&= \brs{\frac{1}{b-a}} \frac{1}{2} \brs{ (b+a)(b-a) } &&= \frac{b + a}{2} \end{align*}

The proof states, in part, that $$\pE\brp{X}=\brs{\frac{1}{b-a}} \int_{x=a}^{x=b} x \dx$$ which implies that

1. All the values of $x$ are being summed in the interval $\intcc{a}{b}$ and
2. Each value of $x$ in $\intcc{a}{b}$ is equally weighted (by $\frac{1}{b-a}$).

As such, because each value is equally weighted, it is somewhat intuitive that

1. The interval $\intcc{a}{b}$ is like a plank with sand evenly distributed along its surface, and
2. The balancing point of this sand covered plank should be halfway between (or the average of) the two points $a$ and $b$
3. This average of $a$ and $b$ (or halfway between $a$ and $b$) is $\frac{a+b}{2}$.

Answer 3

For a continuous random variable with density $f$, the expectation is $\int xf(x)\,dx$. This formula is exactly the calculation for the center of mass of a univariate mass distribution given by $f$. (Normally the calculation requires dividing by the total mass $\int f(x)\,dx$, but here this equals unity.)

If you're comfortable with an appeal to physical intuition, the center of mass is the balance point of a mass distribution. And for a uniform distribution the mass distribution balances at the center of the distribution.

Granted, the physical intuition fails when the expectation does not exist; and if $E(X)=\infty$ we get the dubious assertion that the distribution balances at infinity. But the uniform distribution has a continuous density defined over a bounded interval, so physical intuition leads to the "obvious" conclusion without resorting to an explicit integration.