# Induced Probability Random Variable

I am trying to understand the concept of induced probability spaces in particular the relation between the induced and the "original" space and probability measure.

Given $$(\Omega, \mathcal{A}, \mathbb{P})$$ and a random variable $$X:\Omega \mapsto \Omega '$$, then $$X$$ induces the probability space $$(\Omega', \mathcal{A}', \mathbb{P}_X)$$. Now since $$X$$ is a random variable, by definition it is $$(\Omega, \mathcal{A}), (\Omega', \mathcal{A}')$$ measurable. Also since $$X$$ is measurable and there exists a probability measure $$\mathbb{P}$$ on $$(\Omega, \mathcal{A}, \mathbb{P})$$, then $$\mathbb{P}(X^{-1}(\omega' \in \Omega'))$$ is well defined.

If $$X$$ is a Bernoulli random variable for example, then by definition $$P_X(\omega ') = p^{\omega '}(1-p)^{1 - \omega '}, \ \omega ' \in \{0,1\}$$ but also since $$\mathbb{P}_X(\omega ') = \mathbb{P}(X^{-1}(\omega ')) = p^{\omega '}(1-p)^{1 - \omega '}$$ but this is the same distribution function. So what does $$X$$ actually induce? Does it change $$\mathbb{P}$$ in any way? In other words why don't we simply write $$(\Omega', \mathcal{A}', \mathbb{P})$$?

• $\mathbb P:\mathcal A \to [0,1]$ is a probability function. So too is $\mathbb P_X:\mathcal A' \to [0,1]$. While they are related in the way you describe, they are different functions, most obviously when $\mathcal A' \not=\mathcal A$ Commented Nov 17, 2021 at 22:47

It might be simpler to explain things with an example. Let $$\Omega = \mathbb{R}$$, $$\Omega'=\{0,1\}$$ and $$\mathbb{P}=N(0,1)$$, that is $$\mathbb{P}(A) = \frac{1}{\sqrt{2\pi}} \int_A e^{-\frac{x^2}{2}} \: dx$$ for $$A \in \mathcal{A}$$, where $$\mathcal{A}$$ is the borel $$\sigma$$-algebra on $$\mathbb{R}$$. To construct a $$\operatorname{Bernoulli}$$ variable we can consider $$X: \Omega \rightarrow \Omega'$$ as $$X(\omega) = \begin{cases} 1 &\omega > 0 \\ 0 &\omega\leq 0 \end{cases}.$$ We can now compute the induced measure on $$\Omega'$$, which will then be $$\mathbb{P}(X^{-1}(\{0\})) = \mathbb{P}(\{\omega \in \Omega \: : \: \omega \leq 0)\}) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^0 e^{-\frac{x^2}{2}}\: dx = \frac{1}{2}$$ and similarly $$\mathbb{P}(X^{-1}(\{1\})) = \frac{1}{2}$$. We may thus conclude that $$X$$ has a $$\operatorname{Bernoulli}(\frac{1}{2})$$ distribution, since that is the distribution that it induces. In fact you may take it as a definition, that the distribution of a random variable $$X$$ is the measure $$\mathbb{P}_X$$ that it induces.
• just one follow up question, thus this imply that any random variable that maps to {0,1} is a bernoulli random variable? because the way i understand your response is that the actual probabilities of 0 or 1 are determined by $\mathbb{P}$ and of course $\omega$ Commented Nov 18, 2021 at 6:53
• Yes any random variable, that only takes the values $\{0,1\}$ is necessarily a bernoulli random variable. Also if we change $\Omega'$ to $\mathbb{R}$ and $$X(\omega) = \begin{cases} 1 & \omega > 0 \\ \pi & \omega = 0 \\ 0 & \omega < 0 \end{cases},$$ then $X$ is still $\operatorname{Bernoulli}(\frac{1}{2})$, since the event that $\omega = 0$ has probability $0$, so it has no influence on the distrubution. Commented Nov 18, 2021 at 9:08