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Consider the entire population $\mathcal{P}$ of couples (including singles): let $\mathcal{M}$ denote the population of men, let $\mathcal{W}$ denote the population of women; an element of $\mathcal{P}$ can be $(m,w)$ if $m\in \mathcal{M}$ and $w\in \mathcal{W}$ are married, $(m,*)$ if $m\in \mathcal{M}$ is single, $(*,w)$ if $w\in \mathcal{W}$ is single.

Below I will distinguish the couples in $\mathcal{P}$ between singles and matched couples.

Suppose that the econometrician observes a certain characteristic $X$ of each matched couple. $X$ is treated as a continuous random variable with probability density function $h(\cdot)$.

I am confused between the definition of mass function (not to be mixed with probability mass function) and probability density function. Specifically, I found that $$ \frac{\text{Mass of matched couples with $X=x$}}{\text{Mass of couples in the population (including singles)}}=h(x) $$ Could you help me to clarify this relation? What is the definition of mass function and its relation with a pdf?

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What is the definition of mass function and its relation with a pdf?

"Mass" is a measure of amount and, in this case, that is most likely just the count since there is no indication that any couple has any more "weight" than any other.   Well, not in your quote, anyway, although perhaps it is mentioned elsewhere that matched couples are counted as twice much as a single, or some such.

Specifically, I found that $$ \frac{\text{Mass of matched couples with $X=x$}}{\text{Mass of couples in the population (including singles)}}=h(x) $$

Well, no.   If $X$ actually is a continuous random variable, then its probability density function will not exactly equal the ratio of sample masses.   However, if there is a mass of couples whose $X$ quality is approximately $x$ (within some degree of precision) then you have an estimation for the probability density function.

$$\widehat h(x)=\frac{\text{Mass of matched couples with $X\approx x$}}{\text{Mass of couples in the population (including singles)}}$$

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  • $\begingroup$ Thanks. But, if "mass" means "number", $\frac{\text{Mass of matched couples with $X\approx x$}}{\text{Mass of couples in the population}}\in [0,1]$, while instead $h(x)\in \mathbb{R}$ in principle. How can that be a good estimator? $\endgroup$
    – Star
    Jan 26, 2018 at 15:20

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