Solved – For persons aged 25 and over in the US would the average or the median be higher for income

central-tendencyinferencemeanmedian

Question:

For persons aged 25 and over in the US would the average
or the median be higher for income?

I'm not sure how to answer this.

If I assume that the wage increases with age, and that every age range is
equally represented, then perhaps I'll (extremely roughly) have a graph like
this :

There the y axis is the wage,and the x axis is representing age.

Then the median wage is where half people earn more, and half of the people earn
less. And the mean wage is the average wage that people will earn.

In this case the mean will be greater than the median.

But I'm not sure if this is an appropriate way to consider this problem, or how
to explain it clearly.

Best Answer

Yes, the average is higher than the median, and the reason is that the distribution of incomes is not symmetric.

In fact, symmetric distributions have the same value for the mean and the median. But the income in the US is not symmetric: in fact you have a few people making a lot of money and a lot of people making little money. So that the median is pretty low, compared to the mean: the mean is affected by the rich guys but the median is not.

About your argument, it is probably true that the income is correlated with age but that's not necessarily what makes the distribution asymmetric.

Related Solutions

Solved – When would we use tantiles and the medial, rather than quantiles and the median

This is really a comment, but too long for a comment. It is trying to clarify the definition of "tantile" (in the $p=0.5$ case which is analogous to the median). Let $X$ be a (for simplicity) absolutely continuous random variable with density function $f(x)$. We assume that the expectation $\mu= \mathbb E X$ does exist, that is the integral $\mu=\int_{-\infty}^\infty x f(x)\; dx $ converges. Define, analogously with the cumulative distribution function, a "cumulative expectation function" (I have never seen such a concept, does it have an official name?) by $$ G(t) = \int_{-\infty}^t x f(x) \; dx $$ Then the "tantile" is the solution $t^*$ of the equation $G(t^*) = \mu/2$.

Is this interpretation correct? Is this what was intended?

To return to the original question, in the context of an income distribution, the tantile is the value of income such that half of total income is for people with above that income, and half of total income is for people with below that income.

EDIT

These quantities ( function $G(t)$ above) are related to various risk measures used in some financial literature, such as "expected shortfall".

Have a look at the paper A J Ostaszewski & M B Gietzmann: "Value Creation with Dye's Disclosure Option: Optimal Risk-Shielding with an Upper Tailed Disclosure Strategy" (may 2006), especially around page 15, where they define something they call "Hemi-mean" which is related to $G(t)$ above, also "expected shortfall relative to $t$ and also known as $first lower partial moment". It would be interesting to look into these connections ...

Another term used for this idea is "partial expectation". See for instance https://math.stackexchange.com/questions/1080530/the-partial-expectation-mathbbex-xk-for-an-alpha-stable-distributed-r and use google!

Also, the book Kotz & Kleiber:"Statistical Size Distributions in Economics and Actuarial Science" give relevant information, on page 22 they define (Here $X>0$) $$ F_k(x) = \frac1{E X^k} \int_0^x t^k f(t)\; dt $$ which is "the $k$th-moment distribution", note that $G(t)=\mu F_1(t)$ so is basically the first-moment distribution. They refer to Champernowne (1974) who calls $F_1$ the "income curve", and denotes the underlying cdf $F$ by $F_0$. In terms of the first moment distribution the Lorenz curve can be given as $$ \{(u, L(u))\} = \{(u,v)\colon u=F(x),v=F_1(x); x\ge 0\} $$

Solved – What does it mean if the median or average of sums is greater than sum of those of addends

Medians are not linear, so there are a variety of circumstances under which something like that (i.e. $\text{median}(X_1)+\text{median}(X_2)<\text{median}(X_1+X_2)$) might happen.

It's very easy to construct discrete examples where that sort of thing occurs, but it's also common in continuous situations.

For example it can happen with skewed continuous distributions - with a heavy right tail, the medians might both be small but the median of the sum is "pulled up" because there's a good chance that one of the two is large, and a value above the median is typically going to be far above it, making the median of the sum larger than the sum of the medians.

Here's an explicit example: Take $X_1,X_2 \, \stackrel{\text{i.i.d.}}{ \sim} \operatorname{Exp}(1)$. Then $X_1$ and $X_2$ have median $\log(2) \approx 0.693$ so the sum of the medians is less than $1.4$, but $X_1+X_2\sim \operatorname{Gamma}(2,1)$ which has median $\approx 1.678$ (actually $ -W_{-1}(-\frac{1}{2 e}) - 1$ according to Wolfram Alpha)

Best Answer

Related Solutions

Solved – When would we use tantiles and the medial, rather than quantiles and the median

Solved – What does it mean if the median or average of sums is greater than sum of those of addends

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