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.

  • $\begingroup$ 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. $\endgroup$ Nov 10, 2019 at 13:32

1 Answer 1


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?

  • $\begingroup$ 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? $\endgroup$ Nov 10, 2019 at 14:22
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    $\begingroup$ If $\omega \in (3,\infty)$ then $X(\omega ) = 0 \in [0,2]$ $\endgroup$
    – Falrach
    Nov 10, 2019 at 14:46
  • $\begingroup$ What is the role of index $\mathbf1$ ($\mathbf{I}$) here? $\endgroup$ Nov 11, 2019 at 11:56
  • $\begingroup$ What if a random variable is defined as X$(\omega)=2I_{(-\infty, 3]}(\omega)$? $\endgroup$ Nov 11, 2019 at 12:49
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    $\begingroup$ 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 $\omega$. $\endgroup$
    – drhab
    Nov 11, 2019 at 19:38

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