If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{var}(X)$? Now I see it can't hold. Thank you for the counter examples... You guys rule!
Thank you very much for your comments!
I added, however, some observations that were missing. Most importantly is the fact that we can assume that there exist a positive covariance between X and Y.
At first, it seemed to me that it would be easy to demonstrate... but I still did not manage to solve this problem. Can you guys give me a hand?
Suppose we make use of
$\mathbf{i)}$ a time series $X = [x_1,...,x_N]$ containing only positive entries (i.e. $0 \leq x_i$ for all $i$), 
$\mathbf{ii)}$ a vector of weights of the same of length given by $Y = [y_1,...,y_N]$ where $0 \leq y_i \leq 1$ for all $i$
to build
$\mathbf{iii)}$ a time series $Z = [z_1,...,z_N]$, where the $i$th term is given by $z_i = x_i y_i$, i.e., $Z = [x_1 y_1,...,x_N y_N]$. Clearly, as $Y \in  [0,1]$, we have that $0 \leq Z \leq X$ for all $i$.
$\mathbf{Question)}$ Can we demonstrate that $\text{var}(Z) \leq \text{var}(X)$?
For example, if 
$X = [2, 6, 99, 12, 3, 1]$ and $Y = [0.34, 0.01, 0.2, 1, 0.3, 0.17]$, we have
$ Z = [x_1 y_1,...,x_N y_N] = [0.68, 0.06, 19.8, 12, 0.9, 0.17]$
$ \widehat{\sigma}^{2}_{X} = 1494.70$
$ \widehat{\sigma}^{2}_{Z} = 69.81$
$\mathbf{Important} \text{ } \mathbf{observations}$:
1) $X$ and $Y$ are stationary, ergodic random processes
2) $X$ is not a constant time series, in a sense that $\text{var}(X) \geq 0$
3) It can be assumed that  $\text{var}(X) \geq \text{var}(Y) \geq 0$
4) There exist a positive covariance $X$ between and $Y$


*

*Possible implication of 4)?


As $0 \leq Z \leq X$, we could define a given time series $W \geq 0$ such as $Z + W = X$. Thus, $\text{var}(X) = \text{var}(Z + W) = \text{var}(Z) + \text{var}(W) + 2\text{cov}(Z,W)$. Note that if $\text{cov}(Z,W) \geq 0$ then $\text{var}(X) \geq \text{var}(Z)$ because $\text{var}(W)$ is also greater than zero.
Does the fact that $\text{cov}(X,Y) \geq 0$ infer that $\text{cov}(Z,W) \geq 0$? There is any condition that guarantees $\text{cov}(Z,W) > 0$
Why I was so convinced about $\text{var}(Z) \leq \text{var}(X)$?
In the application I am interested in, I have observed that the relation $\text{var}(Z) \leq \text{var}(X)$ is attended at every time I run my algorithm. If I cannot demonstrate that $\text{var}(Z) \leq \text{var}(X)$ holds given the observations 1) to 4), I would like to know what is forcing that relation, like, for example, $\text{cov}(Z,W) \geq 0$ as mentioned above.
Thanks again for the replies!
Cheers
 A: I do not think Var$(Z)\le $Var$(X)$. Imagine that $X$ is a time series that meanders about values near 100, almost always between 98 and 102. Now imagine that $Z$ meanders between 0 and 100, but is always less than $X$. The variance of $Z$ is clearly going to be larger in such a case than the variance of $X$. This is an example where $X$ and $Z$ are stationary around some constants, but it could easily be extended to a trend stationary example... I am not sure if it would extend to integrated time series... need to think on that.
A: Clearly not.
An easy counterexample (here done in R), that I think satisfies all your constraints:
 set.seed(239843)
 x=rnorm(100,100,1)
 y=rep(c(0.01,0.99),times=50)
 z=x*y
 var(x)
[1] 0.8413043
var(y)
[1] 0.2425253
 var(z)
[1] 2425.296

What's going on:


*

*x is a series with mean 100 and sd 1.

*y alternates between 0.01 and 0.99. 

*z=xy therefore alternates between (about) 1 and 99, but is always $<x$

Alternative [more general] question) Assuming finite variances, is it true that for any random variable a and b such that 0≤a≤b, we have var(a)≤var(b)?

Even more clearly not; without the need for a "y" like variable, it's pretty obvious:
Consider one set of values that alternates between 1 and 99, and a second one that alternates between 100 and 101.

Adding in the new condition that X and Y have positive covariance:
 set.seed(239843)
 oldx=rnorm(100,100,1)
 y=rep(c(0.01,0.99),times=50)
 x = oldx + y  # oldX and Y are independent, so X and Y now have +ve covariance
 z=x*y
 cov(x,y)
[1] 0.2739745  # sample covariance happens to be positive in this case also
 var(x);var(y);var(z)
[1] 1.065326
[1] 0.2425253
[1] 2481.243

If you work out the answers for this case algebraically (compute the population variances and relevant population covariance), you'll see this isn't just a numerical accident from a fortunate choice of seed.
A: For the general case, the answer is no.  For the specific cases, it is also no.
A simple counter example is take $ y\sim U (0,1) $ and take $ x\sim Gamma (a, a) $ such that we have $ E (x)=1 $ and $ var (x)=a^{-1}$ .  Take $ x $ and $ y $ as independent, and we have:
$$ var (z) =E [var (z|y)]+var [E (z|y)]=E [y^2a^{-1}]+var [y]=var (y) + E (y^2)a^{-1}=\frac {1}{12} +\frac {1}{3} var (x)=var (x)\frac {a+4}{12} $$
Now we just choose any value for $ a $ such that $ a> 8$ and we will have $ var (z)> var (x) $
A: Let us be clear that the "variance" under discussion appears to be a random variable derived from a finite portion of a time series.  Specifically, the raw $k^\text{th}$ moment of $\mathrm{X} =(X_1, X_2, \ldots, X_N)$ is
$$\mu_k(\mathrm{X}) = (X_1^k+X_2^k+\cdots+X_N^k)/N,$$
which is a random variable, and the variance is
$$\text{var}(\mathrm{X}) = \mu_2(\mathrm{X}) - \mu_1^2(\mathrm{X}),$$
which also is a random variable.
Similarly we may define moments $\mu_{jk}$ of the bivariate series $(X_i,Y_i)$ and from those compute a covariance.  All these definitions make sense even when either series is constant (although then the moments and variance may reduce to numbers rather than random variables).
To show that counterexamples exist even when $X$ and $Y$ have positive covariance, let the $Y_i$ be bounded by $0$ and $1$, let $\mathrm{Y}$ have nonzero variance, pick $0 \lt \varepsilon \lt 1$, and define 
$$X_i = 1 + \varepsilon Y_i \ge 0.$$
By construction there is perfect (unit) correlation between each $X_i$ and $Y_i$ as well as between $\mu_k(\mathrm{X})$ and $\mu_k(\mathrm{Y})$ for any $k\gt 0$; certainly the covariances are positive.
Yet, since $Z_i=X_iY_i = Y_i + \varepsilon Y_i^2$,
$$\text{Var}(\mathrm{Z}) = \text{Var}(\mathrm{Y}) + 2\varepsilon\mu_1(\mathrm{Y}^3) + \varepsilon^2 \mu_1(\mathrm{Y}^4) \gt \text{Var}(\mathrm{Y}) \gt \varepsilon^2 \text{Var}(\mathrm{Y}) = \text{Var}(\mathrm{X}),$$
disproving the conjecture in the question.
The same analysis (coupled with the fact that $\mu_1(\mathrm{Y}^4)\lt \mu_1(\mathrm{Y}^2)$)  demonstrates that for sufficiently large $\varepsilon\gt 1$, the inequality must be reversed.  Thus there is no necessary inequality relating $\text{Var}(\mathrm{X})$ and $\text{Var}(\mathrm{Z})$.
A: Assume that the processes $\{X\}$ and $\{Y\}$ are ergodic/stationary with finite moments, and independent. Then $\{XY\}$ is also ergodic and
$$\operatorname{Var(XY)} = E(X^2Y^2) - [E(XY)]^2 = E(X^2)E(Y^2) - [E(X)]^2[E(Y)]^2$$
the break-up of expected values due to independence.  
You are asking
$$E(X^2)E(Y^2) - [E(X)]^2[E(Y)]^2 \leq  E(X^2) - [E(X)]^2\;\;??$$
$$\Rightarrow [E(X)]^2\cdot [1-[E(Y)]^2] \leq E(X^2)\cdot [1-E(Y^2)]\;\; ?? \qquad[1]$$
Since $0\leq Y \leq 1$ we have
$$0\leq E(Y) \leq 1 \Rightarrow 0\leq [E(Y)]^2 \leq1,\;\; 0\leq E(Y^2) \leq1$$
and also
$$E(Y^2) > [E(Y)]^2 \Rightarrow [1-[E(Y)]^2] > [1-E(Y^2)] \qquad[2]$$
Examining the desired inequality $[1]$ and the true inequality $[2]$ one sees that $[1]$ may or may not hold, since $[E(X)]^2 < E(X^2)$.
I would say this is an instructive example of how things change when we move from a deterministic to a stochastic assumption - because if the $y_i$'s are designated as a deterministic sequence, then of course the variance of $X_iy_i$ is no greater than the variance of $X_i$.
