Let $\{\mathcal{F}_n \}_{n \in \mathbb{N}}$ denotee a filtration with $\mathcal{F}_n \subset \mathcal{F}$ for all $n\in \mathbb{N}$, on the probability space $(\Omega,\mathcal{F},P)$. If $X$ is a square integrable random variable, what can one say, for $m<n$, about the ratio $$\frac{\mathrm{E}[\mathrm{Var}[X\mid \mathcal{F}_n]]} {\mathrm{E}[\mathrm{Var}[X\mid \mathcal{F}_m]]}\ ?$$
My try:
Consider $Y_n=\mathrm{Var}(X \mid \mathcal{F}_{n})$ for every $n \in\mathbb N$, then
$$\mathrm{E}[Y_{n+1}\mid\mathcal{F}_{n}]=\mathrm{E}[\mathrm{Var}(X \mid \mathcal{F}_{n+1})\mid\mathcal{F}_{n}]=\mathrm{E}[\mathrm{E}[(X-\mathrm{E}[X\mid\mathcal{F}_{n+1})^2\mid\mathcal{F}_{n+1}]\mid\mathcal{F}_n]$$ $$=\mathrm{E}[(X-\mathrm{E}[X\mid\mathcal{F}_{n+1}])^2\mid\mathcal{F}_n]= \mathrm{E}[X^2-2X\mathrm{E}[X\mid \mathcal{F}_{n+1}]+\mathrm{E}[X\mid\mathcal{F}_{n+1}]^2\mid\mathcal{F}_n]= $$ $$\mathrm{E}[X^2\mid\mathcal{F}_n]-\mathrm{E}[2X\mathrm{E}[X\mid \mathcal{F}_{n+1}]\mid\mathcal{F}_n]+\mathrm{E}[\mathrm{E}[X\mid\mathcal{F}_{n+1}]^2\mid\mathcal{F}_n]= X^2-2X^2+\mathrm{E}[\mathrm{E}[X\mid\mathcal{F}_{n+1}]^2\mid\mathcal{F}_n] $$ I don't know how to go on
Recall that, for a general $\mathcal{A}$ sigma algebra, $$Var[X\vert \mathcal{A}] = \mathbb{E}[(X-\mathbb{E}[X\vert \mathcal{A}])^2\vert \mathcal{A}] = \mathbb{E}[X^2\vert \mathcal{A}] - \mathbb{E}[X\vert\mathcal{A}]^2$$ So you have $$\mathbb{E}[Var[X\vert \mathcal{F}_n]] = \mathbb{E}[X^2]-\mathbb{E}[\mathbb{E}[X\vert\mathcal{F}_n]^2]$$ Now let $Z_n=\mathbb{E}[X\vert \mathcal{F}_n]^2$. Since $\{\mathbb{E}[X\vert \mathcal{F}_n]\}_n$ is a martingale, $Z_n$ is a submartingale, so $\mathbb{E}[Z_n]\geq \mathbb{E}[Z_m]$ for $n>m$. Therefore $$\mathbb{E}[X^2]-\mathbb{E}[Z_n] \leq \mathbb{E}[X^2]-\mathbb{E}[Z_m]$$ which gives you $$ \dfrac{\mathbb{E}[Var[X\vert \mathcal{F}_n]]}{\mathbb{E}[Var[X\vert \mathcal{F}_m]]}\leq 1$$ I don't know if this answers your problem completely. I think better results can be achieved, for example probably the sequence $Y_n=Var[X\vert \mathcal{F}_n]$ is a supermartingale, and maybe under suitable hypothesis it can also be shown that it converges a.s. to $Var(X)$, which implies that the ratio tends to $1$ as $m, n\to \infty$.