This exercise asks us to prove that if $\mathbb{E}[|X|] = 0$, then $X = 0$ almost surely ($a.s.$).

Solution:

1. Given:

   • $X$ is a random variable (r.v.).

   • The expectation of the absolute value of $X$, $\mathbb{E}[|X|]$, is defined as: $\mathbb{E}[|X|] = \int_{\Omega} |X(\omega)| \, \mathrm{d}\mathbb{P}(\omega),$ where $\Omega$ is the probability space and $\mathbb{P}$ is the probability measure.

   • We are given $\mathbb{E}[|X|] = 0$, which implies: $\int_{\Omega} |X(\omega)| \, \mathrm{d}\mathbb{P}(\omega) = 0.$

2. Key Property of Integrals:

   • The integral of a non-negative function (like $|X(\omega)|$) with respect to a measure $\mathbb{P}$ is zero if and only if the function is equal to zero almost everywhere with respect to $\mathbb{P}$. Formally: $\int_{\Omega} f(\omega) \, \mathrm{d}\mathbb{P}(\omega) = 0 \quad \iff \quad f(\omega) = 0 \,\, \text{for } \mathbb{P}\text{-almost all } \omega.$

3. Application:

   • Since $\mathbb{E}[|X|] = 0$, it follows that $|X(\omega)| = 0$ for $\mathbb{P}$-almost all $\omega \in \Omega$.
   • But $|X(\omega)| = 0$ implies $X(\omega) = 0$ for $\mathbb{P}$-almost all $\omega$.

4. Conclusion:

   • Therefore, $X = 0 \, a.s.$, meaning $X(\omega) = 0$ except on a set of probability zero.

Final Result:

We have shown that if $\mathbb{E}[|X|] = 0$, then $X = 0 \, a.s.$.

Would you like a more detailed explanation or an example? Let me know!

Related Questions:

1. What is the definition of "almost surely" ($a.s.$) in probability theory?
2. How does the expectation $\mathbb{E}[X]$ differ from $\mathbb{E}[|X|]$?
3. Can $\mathbb{E}[X] = 0$ imply $X = 0 \, a.s.$? Why or why not?
4. What is the difference between almost sure convergence and convergence in probability?
5. What happens if $\mathbb{E}[|X|] > 0$? Does $X$ have to be nonzero?

Tip:

Always verify whether integrals involving random variables are non-negative when applying properties like $\int f \, d\mathbb{P} = 0 \implies f = 0 \, a.s.$. It’s a key tool in probability theory!