Is probability density itself a random variable?

Question

I'm trying to understand random variables a bit better. As I spent time thinking about, it occurred to me that the probability density of a random variable could itself be considered a random variable.

My reasoning for this is as follows:

A random variable is a function mapping each element of the sample space to the reals.

A probability density is a function mapping each element from the sample space of the random variable to the reals.

This implies that the probability density of a random variable is itself a random variable.

Is this correct?

Edit: Ok, I figured out what the thought process behind the question was. Let us that we have a random variable $X$ with an associated PDF $f_X(x)$. Clearly $f_X(x)$ is not random. However, we could define a new random variable $Y = f_X(X)$. Then $Y$ is a random variable, since it is a function of the random variable $X$. Essentially, the situation is analogous to that between the CDF $F_X(x)$ and the probability integral transform $Y = F_X(X)$

Answer 1

No.

A probability density is a function mapping each element from the sample space of the random variable to the reals.

This sounds more like the definition of the random variable itself, not its density. The random variable is the mapping of events to the numbers. For instance, you could map two heads to 1 and any other combo of tails and heads in two coin tosses to 0. This would define a random variable.

The density is function that associated these random variable values (like 0 and 1's in my example) to the probabilities. Yes, it's not a random variable. If you say that probability of 1 is 1/4, and of 0 is 3/4, then there's nothing random about this mapping.

The randomness in the density may come in the situation when you don't know the density and try to estimate it from the sample. In this case you don't know the true parameters of the density function, so you estimate them from the sample, and the parameters become random. In this sense you could say that the sample density function (through its parameters) is a random entity.

For instance, let's say you observe {0,0,1,0,1,0,0,0,1}. You could infer that $P[0]=2/3, P[1]=1/3$.

Then you get another sample: {0,1,0,1,0,0,0}, and come to a different density: $P[0]=5/7, P[1]=2/7$

Answer 2

Kind of (not really).

A random variable formally maps $\Omega$, your sample space, to the reals $\mathbb{R}$. The usual probability density is defined from the reals to the reals, so from this point of view, it's not a random variable. For example, you can define $\Omega$ to be the sigma-algebra of coin flips $H,T$ and then $X(H)=1$ and $X(T)=0$ would define a Bernoulli random variable. Its density (pmf in this case on $\mathbb{Z}$) $f(x)=\frac{1}{2}\delta(x)+\frac{1}{2}\delta(x-1)$ wouldn't be a function from $\Omega\rightarrow \mathbb{R}$, but rather $\mathbb{Z}\rightarrow\mathbb{R}$.

On the other hand, there are high-brow theorems that state that instead of using $\Omega$, you can think of a random variable defined by its push-forward measure, specifically through $P(X\leq x)$. For example when you want to define a standard normal random variable, you can do away with the need for $\Omega$ by instead defining $P(X\leq x)=\int_{-\infty}^x\frac{1}{\sqrt{2\pi}}\exp(-t^2/2)dt$ (and correspondingly it's density).

This procedure actually defines a probability measure on $\mathbb{R}$ through $P_X((-\infty,x)):=P(X\leq x)$. Thus we can assign $\Omega=\mathbb{R}$ and then the normal random variable becomes $X:\Omega\rightarrow \mathbb{R}$, where we equip $\Omega=\mathbb{R}$ with $P_X$. Now when you have the density $f_X(x)$, you can turn it into a random variable on $\Omega=\mathbb{R}$, with measure $P_X$.