# Show that, for any real numbers a and b such that m ≤ a ≤ b or m ≥ a ≥ b, E|Y − a| ≤ E|Y − b| ,where Y be a random variable with finite expectation

Let $$Y$$ be a random variable with finite expectation, and $$m$$ be a median of $$Y;$$ i.e., $$P(Y \le m) \ge 1/2$$ and $$P(Y \ge m) \ge 1/2.$$ Show that, for any real numbers $$a$$ and $$b$$ such that $$m\le a \le b$$ or $$m \ge a \ge b,$$ $$E|Y − a| \le E|Y − b|.$$

It is known than the mean absolute deviation is minimum about the median. If $$m$$ is the median then for $$m\le a \le b$$ or $$m \ge a \ge b,$$ $$a$$ is closer to $$m$$ than $$b.$$ So, the expected absolute deviation for $$a$$ will be less than $$b.$$ The inequality is intuitively clear to me. Please prove it mathematically.

• The first paragraph looks like a homework/self-study question. What are you quoting there? Mar 13, 2022 at 23:31
• @Glen_b Yes, actually it is from a previous year question of an entrance examination. Mar 14, 2022 at 1:47
• Is $E|Y-a|$ shorthand for $E[|Y-a|]$? Mar 14, 2022 at 1:51
• @Acccumulation Yes. Mar 14, 2022 at 1:54
• Please see stats.stackexchange.com/tags/self-study/info, add the tag and adjust the post in accordance. Mar 14, 2022 at 5:54

### Intuition

As explained at Expectation of a function of a random variable from CDF, an integration by parts shows that when a random variable $$X$$ has a (cumulative) distribution function $$F,$$ the expectation of $$|X-a|$$ is the sum of the shaded areas shown:

The left hand region is the area under $$F$$ to the left of $$a$$ while the right hand region is the area above $$F$$ to the right of $$a:$$ that is, it's the area under $$1-F$$ to the right of $$a.$$ This is an extremely useful picture to have in mind when thinking about expectations. It can make seemingly complicated relationships intuitively obvious.

The height of $$F$$ is at least $$1/2$$ at the median $$m,$$ as shown by the dotted lines.

When $$a \ge m$$ is increased to $$b,$$ these areas change: the left hand area grows while the right hand area shrinks. The resulting region is shown here in blue.

The old area was $$I+II+IV$$ while the new one is $$II + III + IV.$$ Their difference therefore is $$III - I.$$ (This is the integral of $$F - (1 - F) = 1 - 2F$$ between $$a$$ and $$b.$$) Because $$III$$ includes the entire rectangle between $$a$$ and $$b$$ below the height $$1/2$$ and $$I$$ lies within the rectangle between $$a$$ and $$b$$ above the height $$1/2,$$ the difference $$III - I$$ cannot be negative (and actually is positive in the figure). Thus, as $$a\ge m$$ increases, the expectation of $$|X-a|$$ cannot shrink -- it can only grow.

A similar geometric argument can be used to analyze any quantile, not just the median, by means of a suitable weighting of the upper and lower areas. See https://stats.stackexchange.com/a/252043/919 for some analysis of this.

### Formal Solution

Fix an arbitrary (but finite) $$b \gt m,$$ suppose $$a$$ is any number for which $$m \le a \le b,$$ and consider the difference

$$h(a) = E\left[|Y-b| - |Y-a|\right] = E\left[|Y-b|\right] - E\left[|Y-a|\right].$$

We aim to show $$h(a) \ge 0.$$ We will do this by computing the expectation and finding a simple lower bound for it. The only technique we need is a straightforward integration by parts.

Writing $$F$$ for the distribution of $$Y$$ and integrating in the sense of Lebesgue-Stieltjes, compute this expectation by breaking the integral into three regions bounded by $$a$$ and $$b$$ so that the absolute values can be expressed more simply within each region (look closely at how the signs in the integrands vary in the third line):

\begin{aligned} h(a) &= \int \left(|x-b| - |x-a|\right)\,\mathrm{d}F(x) \\ &= \left(\int_{-\infty}^a + \int_a^b + \int_b^\infty\right) \left(|x-b| - |x-a|\right)\,\mathrm{d}F(x) \\ &= \int_{-\infty}^a(b-x)-(a-x)\,\mathrm{d}F(x) + \int_a^b(b-x)-(x-a) \,\mathrm{d}F(x) \\&\quad+ \int_b^\infty(x-b)-(x-a)\,\mathrm{d}F(x) \\ &= (b-a)\int_{-\infty}^a\,\mathrm{d}F(x) + \int_a^b(a+b-2x) \,\mathrm{d}F(x) + (a-b)\int_b^\infty\,\mathrm{d}F(x). \\ \end{aligned}

Evaluate the middle integral by parts:

\begin{aligned} \int_a^b(a+b-2x) \,\mathrm{d}F(x) &= (a+b-2x)F(x)\bigg|^b_a + 2\int_a^b F(x)\,\mathrm{d}x\\ &= (a-b)F(b) - (b-a)F(a)+ 2\int_a^b F(x)\,\mathrm{d}x \\ &= (a-b)(F(b) + F(a))+ 2\int_a^b F(x)\,\mathrm{d}x. \end{aligned}

Plug this into the previous expression for $$h(a)$$ and note that because $$a \ge m,$$ $$F(x) \ge 1/2$$ throughout this integral:

\begin{aligned} h(a) & = (b-a)F(a) + \left[(a-b)(F(b)+F(a)) + 2\int_a^b F(x)\,\mathrm{d}x\right] + (a-b)(1 - F(b)) \\ &= a-b + 2\int_a^b F(x)\,\mathrm{d}x\\ &= 2\int_a^b \left(F(x) - \frac{1}{2}\right)\,\mathrm{d}x \\ &\ge 2\int_a^b \left(\frac{1}{2}-\frac{1}{2}\right)\,\mathrm{d}x \\ &=0, \end{aligned}

QED.

The demonstration for the other case $$b \le a \le m$$ follows by applying this result to the random variable $$-Y$$ and the values $$-m\le-a\le-b$$ (because $$-m$$ is a median of $$-Y.$$)

• +1 The second chart is more helpful as an aid to understanding than the formal solution. Mar 14, 2022 at 15:58
• @Henry Agreed. But it takes more effort to create :-(.
– whuber
Mar 14, 2022 at 16:20