# Measure vs Random variable (P vs Px) [Measure Theory]

I am given a probability space $$(\mathbb{R}, \mathcal{B}(\mathbb{R}), \mathbb{P})$$, where $$\mathbb{P}=0.1 \delta_{-2}+0.7 \delta_{1}+0.2 \delta_{10}$$.

I am also given a X$$(\omega)=2 \omega^{3} I_{(-\infty, 3]}(\omega)$$ and the question is to determine $$\mathbb{P}_{X}([0,2])$$.

Now, my understanding is that $$\mathbb{P}$$ is probability measure that maps $$\mathcal{B}(\mathbb{R})$$ to [0,1].

What I don't understand is how X$$(\omega)$$ connects $$\mathbb{P}_{X}$$ and $$\mathbb{P}$$. I read What is the difference between a probability distribution on events and random variables? but the answer was not very helpful. I read Jacod & Protter and but still cannot figure out how the two are related and honestly feel quite dumb.

P.S. I am also looking for a reference book that is more focused on problem solving than theory to grasp mechanics. So far I tried Çınlar, Jacod & Protter, Shiryaev and Schilling. Schilling is the most gentle but all of them are still more theory focused.

• Evaluate $X(-2)$, $X(1)$, and $X(10)$. That is to say, at the values of $\omega$ that carry all the mass in $\mathbb P$. The numerical values you get, associated with the corresponding probabilities for $\omega$, contain your answer. Nov 10, 2019 at 13:32

Random variable $$X$$ induces a new probability measure on $$(\mathbb R,\mathcal B(\mathbb R))$$.
The measure is denoted as $$\mathbb P_X$$ and is prescribed by:$$B\mapsto\mathbb P(X\in B)$$ So based on that we find that: $$\mathbb P_X([0,2])=\mathbb P(X\in[0,2])=\mathbb P(\{\omega\in\mathbb R\mid X(\omega)\in[0,2]\})=\mathbb P(\{\omega\in\mathbb R\mid2\omega^2\mathbf1_{(-\infty,3]}(\omega)\in[0,2]\})$$ Now observe that: $$2\omega^2\mathbf1_{(-\infty,3]}(\omega)\in[0,2]\iff\omega\in[-1,1]\cup(3,\infty)$$
So it remains to find:$$\mathbb P([-1,1]\cup(3,\infty))$$Can you take it from here?
• Thank you @drhab, Yes, it is almost clear now. I get the [-1.1] part, but where does the union with (3, $\infty$) come from? Nov 10, 2019 at 14:22
• If $\omega \in (3,\infty)$ then $X(\omega ) = 0 \in [0,2]$ Nov 10, 2019 at 14:46
• What is the role of index $\mathbf1$ ($\mathbf{I}$) here? Nov 11, 2019 at 11:56
• What if a random variable is defined as X$(\omega)=2I_{(-\infty, 3]}(\omega)$? Nov 11, 2019 at 12:49
• That is correct. However you contradict yourself if you are saying that the function does not depend on the $\omega$. The function can take two values ($0$ and $2$) and which of these values will be taken depends clearly on the value taken by $\omega$. Nov 11, 2019 at 19:38