Sufficient condition for $ \sigma_{X}^{2} \leq \sigma_{Y}^{2}$ Suppose $X$ and $Y$ are random variables whose expected values are $\mu_X$ and $\mu_Y$, and variances are $\sigma_{X}^{2}$ and $\sigma_{Y}^{2}$, respectively.
Also, we suppose $F_x$ and $F_Y$ are the distribution function of $X$ and $Y$.
Say that the inquality
$$ F_{X}(\mu_X+t) - F_{X}(\mu_X-t) \geq F_{Y}(\mu_Y+t) -  F_{Y}(\mu_Y-t)$$
holds for all $t$.
Then, is it true to be
$$ \sigma_{X}^{2} \leq \sigma_{Y}^{2}$$?
The only thing that I was able to do is deriving the inequalities below, but I don't think these work for the problem.
\begin{align*}
\sigma_{X}^{2} 
&\geq t^2 \left\{ F_{X}(\mu_X - t) - F_{X}(\mu_X + t) \right\} \\
\sigma_{Y}^{2} 
&\geq t^2 \left\{F_{Y}(\mu_Y - t) - F_{Y}(\mu_Y + t)\right\}
\end{align*}
I would appreciate if you could help me.
 A: Let us assume, without loss of generality, that $\mu_X = \mu_y = 0$.  We then have:
$$F_X(t)  - F_X(-t)  \geq F_Y(t) - F_Y(-t)$$
We observe that $\sigma^2_X = \mathbb{E}x^2$ (as we have set the expectation equal to zero) and similarly for $\sigma^2_Y$; therefore we want to work with squares instead of the initial variables.   We denote the cumulative distributions of $x^2$ and $y^2$ by $F^{'}_X$ and $F^{'}_Y$,and define $t^{'} = t^2$.  Since $x^2 \leq $ any arbitrary $c^2$ is equivalent to $-c \leq x \leq c$, it's clear that $F^{'}_X(t^{'}) = F_X(t) - F_X(-t)$, which leads directly to:
$$F^{'}_X(t^{'}) \geq F^{'}_Y(t^{'})$$
Now, as $t^{'}$ is non-negative, being a square, we know that $\mathbb{E}t^{'} = \int_0^{\infty}(1-F_{t^{'}}(t^{'}))dt^{'}$.  Substituting and engaging in some minor algebra gives us:
$$\sigma_Y^2 - \sigma_X^2 = \int_0^{\infty}(1-F_Y^{'}(y))dy - \int_0^{\infty}(1-F_X^{'}(x))dx$$
which can be rearranged as:
$$\sigma_Y^2 - \sigma_X^2 = \int_0^{\infty}(F^{'}_X(t) - F^{'}_Y(t)) dt$$
Since $F^{'}_X(t) - F^{'}_Y(t) \geq 0$ for all $t$, the integral on the right is bounded below by $\int_0^{\infty}0dt = 0$; therefore:
$$\sigma_Y^2 - \sigma_X^2 = \int_0^{\infty}(F^{'}_X(t) - F^{'}_Y(t)) dt \geq 0$$
and $\sigma^2_X \leq \sigma^2_Y$.
A: For simplicity and without loss of generality we can set $\mu_X$ and $\mu_Y$ to zero (the variances of the shifted variables is the same), then we can write
$$\sigma_X^2 = E[x^2] = \int_{-\infty}^{\infty}x^2f_X(x)dx = \int_{0}^{\infty}x^2(f_X(x)+f_X(-x))dx = \int_{0}^{\infty}x^2\frac{d}{dx}\left(F_X(x)-F_X(-x) \right)dx$$
denoting $A_X(t) := F_X(t)-F_X(-t)$ and using integration by parts:
$$\sigma_Y^2 - \sigma_X^2 = \int_{0}^{\infty}x^2\frac{d}{dx}\left(A_Y(x)-A_X(x) \right)dx \\ = x^2(A_Y(x)-A_X(x))\biggr\rvert_0^{\infty} + \int_{0}^{\infty}2x(A_X(x)-A_Y(x))dx$$
The integral in the second term is non negative because of the assumption $A_X(t) \ge A_Y(t)$, so if we could show that $\lim_{t \to \infty}t^2(A_Y(t)-A_X(t))=0$ we we will have the result $\sigma_Y^2 \ge \sigma_X^2$.
Notice that if the variables were bounded then both $A_Y$ and $A_X$ evaluated at the upper bound will give 1, making the first term identically zero and therefore establishing the result.
(Chebyshev's inequality doesn't quite get us there in the unbounded case, because it only bounds $A(x)$ by something that goes like $1/x^2$, but that might be the direction to go)
