# The difference between $\mathbb{P}$ and $\mathbb{P}_X$ measures

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:

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

• 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) $\mathbb R$. – Math1000 Jan 5 '20 at 19:39
• $\mathbb{P}_X$ 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 $\mathcal{F}$). – angryavian Jan 5 '20 at 19:41
• 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? – Blg Khalil Jan 5 '20 at 19:51
• Yes, I think that's correct. – angryavian Jan 5 '20 at 20:06

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 $$\mathbb P_X.$$

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