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Why is the probability function of a discrete random variable called a probability "mass" function? What does the word "mass" mean here?

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I think this is a nice question. "Mass" refers to the weight that is given to each event in your sample space. Since the random variable is discrete, the sample space is finite (or at least countable) and hence each event gets attributed a "mass".

The analogous probability "density" for continuous random variables can be interpreted in the same way. If $f\left(x\right)$ is your probability density, then

$$ \int_a^b f\left(x\right) dx $$

can be interpreted as a "mass".

Edit:

Here's an example. If you are rolling a six sided die, then your sample space is $\Omega = \left\{1,2,3,4,5,6\right\}$. The "mass" associated with rolling $\omega \in \Omega$ is $F\left(\omega\right)$, where $F$ is your probability mass function. If the die is fair, a sensible probability mass function to use is $F\left(x\right)=\frac{1}{6}$.

Now let's say that you are instead rolling a magical die that takes values from $\Omega^\prime = \left(-\infty,\infty\right)$ (all real numbers; e.g. $\pi=3.14\ldots$). Let $f\left(\omega\right)$ be the probability density function. The probability that you roll a number between $-1$ and $1$ is:

$$ \int_{-1}^1 f\left(x\right)dx $$

which can be interpreted as a mass.

Note that the probability of rolling exactly $a$ in this setting (where $a$ is any number) is

$$ \int_{a}^a f\left(x\right)dx = 0. $$

This is because our sample space $\Omega^\prime$ in the continuous example has "a lot more stuff in it" than the sample space $\Omega$ in the discrete example.

If you are interested, you might enjoy reading the blurb on Wikipedia about measure theoretic probability.

parsiad
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