This expected value has a minimum!

Question

Problem. Let $X$ be a positive, real random variable whose probability density function is bounded by $1$. Prove that $E[X]\geq \frac 12$.

Hi everyone. This problem is essentially saying that the more spread out a variable is, the higher its expectancy will be. I solved it in a very measure-theoretic way, as follows.

Let $f$ be the p.d.f. of $X$. I start supposing $f$ is a step function of rational sides, i.e. one that takes rational values and whose discontinuities occur at rational points. This allows me to "cut" the area under the graph of $f$ into small squares that (thanks to rationality), moved backwards one by one, reconstruct exactly the unit square. In this process of moving backwards, the expectancy drops a little. In the end, the expectancy is equal to $\int_0^\infty x\chi_{[0,1]}(x)\,dx=1/2$, so $\int_0^\infty xf\geq1/2$. Now we naturally extend this result to an arbitrary step function. Then to an arbitrary $f$ of bounded support and finally to an arbitrary $f$.

This all works well, but I'm sure there is a faster way of showing this given the simplicity with which the problem is stated. Any clue?

Answer 1

Because $X$ is positive, $$ E[X] = \int_0^\infty a(x) \,dx \ge \int_0^1 a(x) \,dx $$

where $a(x) = 1-F(x) = 1 - \int_0^x f(u) \,du$, with $a(0)=1$.

Now, $f(x) \le 1 \implies a'(x) \ge -1 \implies a(x) \ge 1-x$ in the interval $[0,1]$ .

Hence, $$E[X]\ge \int_0^1 (1-x) \, dx = \frac12$$

Answer 2

Here is a variant of @ConnFus's solution.

Let $f(x)$ denote the density function of $X$, and let $g(x) = \mathbf{1}_{[0, 1]}(x)$. Then we note:

1. $\int_{0}^{\infty} [f(x) - g(x)] \, \mathrm{d}x = 0$,

2. $(x - 1)(f(x) - g(x)) \geq 0$ for all $x \geq 0$ by the assumption on $f$.

Hence,

\begin{align*} 0 & \leq \int_{0}^{\infty} (x - 1)(f(x) - g(x)) \, \mathrm{d}x \\ &= \int_{0}^{\infty} x (f(x) - g(x)) \, \mathrm{d}x \\ &= \mathbb{E}[X] - \frac{1}{2}. \end{align*}

This also tells that the minimum is attained if and only if $f = g$ almost everywhere.

Answer 3

Another way to do this arises from stochastic dominance. Given a positive random variable $X$ whose p.d.f is bounded above by $1$, create a uniform random variable $U[0,1]$ on the same probability space such that $X \geq U$ almost surely. Then, we automatically have $\mathbb E[X] \geq \mathbb E[U] \geq \frac 12$.

Here, we use the famous Inverse transform sampling from a first course in statistics. Namely, let $(\Omega, \mathcal F)$ be a sample space and $X : \Omega \to \mathbb R$ be a positive random variable such that its p.d.f $f : [0,\infty) \to \mathbb R_+$ is bounded above by $1$. Let $$F_X(y) = \mathbb P(X \leq y) = \int_0^{y} f(t)dt $$ for $y\geq 0$ be the cumulative distribution of $X$. $F_X$ admits a generalized inverse $$F_X^{-1}(p) = \inf\{x : F_X(x) \geq p\},$$ which satisfies $F_X(F_X^{-1}(y)) = y$ since $F_X$ is continuous. Define $U : \Omega \to [0,1]$ by $$U(\omega) = \int_{0}^{X(\omega)} f(t) dt =F_X(X(\omega))$$ We claim that $U$ is a uniform random variable. Indeed, for any $0 \leq a<b \leq 1$, $$ \mathbb P(a < U \leq b) = \mathbb P(a < F_X(X(\omega)) \leq b) = \mathbb P(F_{X}^{-1}(a) < X(\omega) \leq F_{X}^{-1}(b))\\ = F_X(F_X^{-1}(b)) - F_X(F_X^{-1}(a)) = b-a. $$

Thus, we have established that $X$ and $U$ can be defined on the sample space. However, since $f(t) \leq 1$, we also have $$ U(\omega) = \int_{0}^{X(\omega)} f(t)dt \leq \int_0^{X(\omega)} 1 dt \leq X(\omega). $$ Thus, $X \geq U$, showing that $\mathbb E[X] \geq \mathbb E[U] \geq \frac 12$. We have actually shown something stronger in the above calculation.

If $f(t) \leq 1$ for all $t>0$ then for every $y>0$, $$ \mathbb P(U[0,1] \geq y) \leq \mathbb P(X \geq y). $$

This is called a first order stochastic dominance of $X$ over $U$, which is a widely used concept in probability. It has multiple applications in control theory, finance, statistics and probability theory in general. You can check out this article for more on it.