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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 $(\Omega', \mathcal{A}', \mathbb{P})$?

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    $\begingroup$ $\mathbb P:\mathcal A \to [0,1]$ 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 $\mathcal A' \not=\mathcal A$ $\endgroup$
    – Henry
    Commented Nov 17, 2021 at 22:47

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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.

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  • $\begingroup$ thank you so much, that was exactly what i was looking for. $\endgroup$ Commented Nov 17, 2021 at 23:27
  • $\begingroup$ 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 $\omega$ $\endgroup$ Commented Nov 18, 2021 at 6:53
  • $\begingroup$ 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. $\endgroup$ Commented Nov 18, 2021 at 9:08
  • $\begingroup$ alright, got it. many thanks. $\endgroup$ Commented Nov 18, 2021 at 9:31

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