# What is an induced probability function?

My textbook defined the probability function of a random variable as:

the function $$P_X$$ is an induced probability function on $$X(\Omega)$$, defined in terms of the original function P.

In other words, the function $$P_X$$ induces a surjection by restricting its co-domain to the image of its domain $$X(\Omega)$$?

Am I correct in stating this?

edit: Textbook is Statistical Inference by Casella and Berger page 29

• For completeness sake: Could you add a reference to the textbook? – COOLSerdash Feb 27 at 14:15
• I've never seen the wording "probability function". Given the ambient probability space $(\Omega,\mathfrak{F},\mathbb{P})$, the probability induced by $X$ is the law of $X$ (under $\mathbb{P}$), or the distribution of $X$. This is also called the image probability of $\mathbb{P}$ by $X$ (@whuber says push-forward). – Stéphane Laurent Feb 29 at 10:29
• @StéphaneLaurent The textbook I'm using doesn't touch upon probability spaces. It is Casella and Berger' Statistical Inference. It defines the probability measure P, which must follow the Kolmogorov Axioms. – DataInTheStone Feb 29 at 16:01

Not quite. The setting is a probability space $$(\Omega,\mathfrak{F},\mathbb{P})$$ and a measurable function $$X$$ whose domain is $$\Omega$$ and whose codomain usually is $$\mathbb{R}$$ with its Borel sigma-algebra $$\mathfrak{B}$$ (but generally could be any measurable space).

$$X$$ induces a probability distribution $$\mathbb{P}_X$$ as the push-forward of $$\mathbb{P}$$ via $$X$$, sometimes written $$X_{*}\mathbb{P},$$ defined as

$$\mathbb{P}_X(E) = (X_{*})\mathbb{P}(E) = \mathbb{P}(X^{-1}(E)) = \mathbb{P}\left(\{\omega\in\Omega\mid X(\omega)\in E\}\right)$$

for any event $$E\in\mathfrak{B}.$$

Let's do a simple example. Let $$\Omega$$ be the set of the three possible ways a flipped coin may land: heads, tails, or on its edge. Let its sigma-algebra $$\mathfrak{F}$$ consist of all subsets of $$\Omega.$$ Let the probability distribution $$\mathbb{P}$$ assign the value $$p$$ to $$\{\text{Heads}\},$$ $$1-p$$ to $$\{\text{Tails}\},$$ and $$0$$ to $$\{\text{Side}\}.$$ This determines $$\mathbb P$$ on every subset of $$\Omega$$ according to the laws of probability.

The function $$X:\Omega\to\mathbb{R}$$ that equals $$1$$ for $$\omega=\text{Heads}$$ and otherwise equals $$0$$ is the indicator of $$\text{Heads}.$$ $$X$$ is obviously measurable (because every subset of $$\Omega$$ is measurable). To figure out what $$\mathbb{P}_X$$ is, let $$E\subset\mathfrak{B}$$ be a Borel-measurable set. $$X_{*}\mathbb{P}(E)$$ is the sum of up to three values: $$p$$ if $$X(\text{Heads})\in E,$$ plus $$1-p$$ if $$X(\text{Tails})\in E,$$ plus $$0$$ if $$X(\text{Side})\in E$$.

One convenient way to express $$\mathbb{P}_X$$ uses the "one-point" measures $$\delta_a$$ defined on the Borel sets of $$\mathbb{R}.$$ These assign the value $$1$$ to an event $$E$$ when $$a\in E$$ and otherwise assign the value $$0.$$ It's easy to check that they are indeed measures.

The random variable $$X$$ thereby pushes $$\mathbb P$$ into the induced measure (or "induced probability function") $$\mathbb{P}_X = (1-p)\delta_0 + p\delta_1.$$

Another description of the induced measure considers only events of the form $$E(x)=(-\infty, x]$$ for $$x\in \mathbb{R},$$ because these determine the entire Borel sigma algebra of $$\mathbb R.$$ The formula $$F_X: x\to \mathbb{P}_X(E(x)) = \mathbb{P}(X\le x) = \mathbb{P}\left(\{\omega\in\Omega\mid X(\omega)\le x\}\right)$$

defines a function on $$\mathbb R,$$ the cumulative distribution function of $$X.$$ It equals $$0$$ for $$x\lt 0,$$ jumps up to a constant value of $$1-p$$ for $$0\le x \lt 1,$$ and then jumps (by an amount $$p$$) up to $$1$$ for $$x\ge 1.$$

In this example of a Bernoulli$$(p)$$ random variable, please notice that

• $$X$$ is neither an injection nor a surjection from $$\Omega$$ to $$\mathbb R.$$ Its image is merely the set $$\{0,1\}.$$

• $$F_X$$ is neither an injection nor a surjection from $$\Omega$$ to the set of possible probabilities $$[0,1].$$ Its image is the set $$\{0,1-p,1\}.$$