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Let $(\Omega ,\mathcal{F},\mathbb{P})$ be a probability space, and let $X$ be a random variable such that,

$$X: (\Omega ,\mathcal{F},\mathbb{P}) \rightarrow \mathbb (\mathbb{R}, \mathcal{B}(\mathbb{R}), \mathbb{P}_X)$$

where $\mathbb{P}_X$ is the induced measure by $X$ on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$.

We know by definition that,

$$\mathbb{P}_X(B) \triangleq \mathbb{P}(\{X \in B\}), \space \forall B \in \mathcal{B}(\mathbb{R})$$

If $\mathbb{P}_X$ and $\mathbb{P}$ are equivalent as shown above, then why do we need two different probability measures?

Will $\mathbb{P}_X$ and $\mathbb{P}$ be equivalently the same if and only if the random variable $X$ is an identity function such that:

$$X: \omega \rightarrow \omega, \space \forall \omega \in \Omega$$

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    $\begingroup$ The problem with setting $X(\omega)=\omega$ is that in general $X$ is an $\mathbb R$-valued function, and is defined on a sample space $\Omega$ which in general is not (a subset of) $\mathbb R$. $\endgroup$ – Math1000 Jan 5 '20 at 19:39
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    $\begingroup$ $\mathbb{P}_X$ assigns probabilities to subsets of $\mathbb{R}$ that are measurable (i.e. subsets in $\mathcal{B}(\mathbb{R})$) while $\mathbb{P}$ assigns probabilities to subsets of $\Omega$ that are measurable (i.e. subsets in $\mathcal{F}$). $\endgroup$ – angryavian Jan 5 '20 at 19:41
  • $\begingroup$ If I understand both of your answers which clarified a lot, if $\Omega = \mathbb{R}$, $\mathcal{F} = \mathcal{B}(\mathbb{R})$ and $X(\omega)=\omega$ then both $\mathbb{P}_X$ and $\mathbb{P}$ would be equivalent, am I right? $\endgroup$ – Blg Khalil Jan 5 '20 at 19:51
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    $\begingroup$ Yes, I think that's correct. $\endgroup$ – angryavian Jan 5 '20 at 20:06
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Toss two fair coins. The set of possible outcomes is $\Omega = \big\{ tt, tH, Ht, HH\big\}.$

For any set $F\in\mathcal F = 2^\Omega,$ you have $$\mathbb P(F) = \dfrac{\text{the number of outcomes in the set } F} 4.$$

Let $X$ be the number of "heads", so $X\in\{0,1,2\}.$

Then $\mathbb P_X(B) = \begin{cases} 1 & \text{if } \{0,1,2\}\subseteq B, \\[8pt] 3/4 & \text{if } 2\in B \text{ and either } 0\in B \text{ and }1\in B \\ & \text{but not both,} \\[8pt] 1/2 & \text{if } 2\in B \text{ and } 0\notin B \text{ and } 1\notin B, \\[8pt] 1/4 & \text{if } 0 \in B \text{ and } 1\notin B \text{ and } 2\notin B \\ & \text{or } 1 \in B \text{ and } 0\notin B \text{ and } 2\notin B \\[8pt] 0 & \text{if } 0\notin B \text{ and } 1\notin B \text{ and }2\notin B. \end{cases}$

That is the difference between $\mathbb P$ and $\mathbb P_X.$

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  • $\begingroup$ Do we mean by "$\text{the number of outcomes in the set} \mathcal{F}$" any subset in the power set $\mathcal{P}(\Omega) = 2^{\Omega}$? $\endgroup$ – Blg Khalil Jan 5 '20 at 20:19
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    $\begingroup$ @BlgKhalil : No. I do not mean the set $\mathcal F = 2^\Omega;$ rather I mean any set $F\in\mathcal F= 2^\Omega. \qquad$ $\endgroup$ – Michael Hardy Jan 5 '20 at 20:24
  • $\begingroup$ Ah thank you, now I understood. $\endgroup$ – Blg Khalil Jan 5 '20 at 20:25

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