Is a sigma algebra created by a function of a random variable (vector) smaller than the sigma algebra of that random variable(vector)?

Question

I want to prove the following inequality, where $V$ is a Gaussian random variable with law $~ N(0,\sigma_V)$: \begin{align} VAR[V|\mathcal{F}^B]>VAR[V|\mathcal{F}^A] \end{align} where $ \mathcal{F}^A = {\sigma(\mu,\theta)} $ and $ \mathcal{F}^B = {\sigma(a*\mu+b*\theta)} $ are $\sigma$ algebra and $a$ and $b$ are just some constants. Here $\mu$ and $\theta$ are also Gaussian random variables with zero mean and some variance. My assumption is that $\mathcal{F}^B \subseteq\mathcal{F}^A $, since $a*\mu+b*\theta$ is a just a function of $(\mu,\theta)$. Is this assumtion correct?

If yes, I would prove the inequality above using the following lemma: For jointly normal random variables $X$ and $Y$, if $\mathcal{F}^B \subseteq \mathcal{F}^A$ it holds: \begin{align} COV[E[X|\mathcal{F}^A],E[Y|\mathcal{F}^A]|\mathcal{F}^B] = COV[X,Y|\mathcal{F}^B] - COV[X,Y|\mathcal{F}^A]\end{align} (this lemma can be proven by using law of total covariance and the fact that, conditional covariance is constant for jointly normal random variables, thus I omit the proof).

By using the lemma given above I can rewrite $VAR[V|\mathcal{F}^A]$ as $VAR[V|\mathcal{F}^B]- VAR[E[V|\mathcal{F}^A]|\mathcal{F}^B]$. Thus,

\begin{align} VAR[V|\mathcal{F}^B] > VAR[V|\mathcal{F}^B]- VAR[E[V|\mathcal{F}^A]|\mathcal{F}^B] \end{align} \begin{align} 0 >- VAR[E[V|\mathcal{F}^A]|\mathcal{F}^B] .\end{align}

So, my question is if the assumtion $\mathcal{F}^B \subseteq\mathcal{F}^A $ is correct? If yes, where could I find a proof for this assumption. Thanks in advance!

Answer 1

The assumption is correct under the condition that the function involved is measurable.

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In general if $X$ is a random variable and $f:\mathbb R\to\mathbb R$ is a Borel measurable function then $f(X)$ is a random variable and this with $\sigma(f(X))\subseteq\sigma(X)$.

Here $\sigma(X)=\{\{X\in B\}\mid B\in\mathcal B\}$ and similarly $\sigma(f(X))=\{\{f(X)\in B\}\mid B\in\mathcal B\}$.

Essential is that $f$ is Borel measurable so that: $$B\in\mathcal B\implies f^{-1}(B)\in\mathcal B$$

That leads to the observation: $$\{f(X)\in B\}=\{X\in f^{-1}(B)\}\in\sigma(X)$$for every $B\in\mathcal B$.