# This expected value has a minimum!

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?

• You know that the optimal pdf is $g:=𝟙_{[0,1]}$. Now note that $g\ge f$ for $x\in [0,1]$ and $g\le f$ for $x>1$. Use that to show $\int x g(x) dx \le \int x f(x) dx$ via $\int x(g(x)-f(x))dx$ since $\int_ℝ f-g =0$ by splitting the integral in the regions $[0,1]$ and $(1,\infty)$ Commented Jul 17 at 13:28

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

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.

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 $$0 \leq a, $$\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.