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Let $(\Omega,\mathcal{F},P)$ be a probability space and $X$ be a random variable. The $\sigma$-algebra generated by $X$ is defined as

$$\sigma(X):=\{X^{-1}(B)\; | \; B\in B_{\mathbb{R}}\}$$

where $B_{\mathbb{R}}$ is the Borel $\sigma$-algebra of subsets of $\mathbb{R}$. The problem statement is as follows :

Let $(\Omega,\mathcal{F},P)$ be a probability space and $X,Y$ are random variables such that $\sigma(X)=\sigma(Y)$. Show that $\sigma(X+Y)\subseteq \sigma(X)$.

I can show that $X+Y$ is a random variable. So, how do I go about proving this?

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1 Answer 1

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By definition, $\sigma(Z)$ denotes the smallest $\sigma$-algebra $\Sigma$ on $\Omega$ such that $Z: (\Omega, \Sigma) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ is measurable.

This means in particular that $X:(\Omega,\sigma(X)) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ is measurable. Since $\sigma(Y) = \sigma(X)$, we also have that

$$Y: (\Omega,\sigma(X)) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$

is measurable. Consequently, the sum

$$X+Y: (\Omega,\sigma(X)) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$

is measurable as a sum of two measurable random variables. Since $\sigma(X+Y)$ is the smallest $\sigma$-algebra $\Sigma$ such that

$$X+Y: (\Omega,\Sigma) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$

is measurable, this shows $\sigma(X+Y) \subseteq \sigma(X)$.

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