For X,Y random variables, with pdfs that are symmetric around 0, does $V(X)\geq V(Y) \Rightarrow E(|X|)\geq E(|Y|)$? I need to show the following thing. 
Consider two continuous random variables $X,Y$ which take values in $[-1,1]$ and are have pdf's that are symmetric around zero. How can I show that $V(X)\geq V(Y) \Rightarrow E(|X|)\geq E(|Y|)$ ? I have tried a number of examples where this holds, but I straggle finding a general answer. As an example one can try $f_1(x)=\frac{1}{2}$ (uniform) and $f_2(x)=\frac{3}{2}x^2$ that satisfy the above conditions and the result holds.
 A: Here's an explicit counterexample.
Let $U=|X|$ and $V=|Y|$. 
Then $\text{Var}(X)\geq \text{Var}(Y) \Rightarrow E(|X|)\geq E(|Y|)$ is equivalent to 
$E(U^2)\geq E(V^2) \Rightarrow E(U)\geq E(V)$ for $U,V$ continuous in [0,1]. 
Consider a $\text{beta}(\alpha,\beta)$; it has mean $\mu=\frac{\alpha}{\alpha+\beta}$
and second raw moment $\mu_2'=\mu \frac{\alpha+1}{\alpha+\beta+1}$
Let $U\sim \text{beta}(1,1)$ and let $V\sim \text{beta}(6,5)$.
Then $E(U) = \frac{1}{2}$ and $E(U^2)=\frac{1}{3}$ 
and $E(V) = \frac{6}{11}$ and $E(V^2)=\frac{6}{11} \cdot \frac{7}{12}=\frac{7}{22}$ 
Now $E(U^2)>E(V^2)$ but $E(U)<E(V)$.
I'll leave it to you to construct the example explicitly in terms of $X$ and $Y$.
A: I will set out a solution strategy and some hints, in keeping with the site's homework policy. It's quite an effective one in general - in this case, checking the back of my envelope, I used four simple lines of algebra and three rough sketch plots (two PMFs and an overlaid contour plot) - but other approaches are available. In particular it can make life easier to use symmetry considerations: let $U=|X|$ and $V=|Y|$ so our variables are now on on $[0,1]$, and we instead consider whether $\mathbb{E}(U^2) \geq \mathbb{E}(V^2)$ implies $\mathbb{E}(U) \geq \mathbb{E}(V)$. This gets rid of the modulus and means we only need consider positive variables, both of which are nice. However since this would have made only a negligible difference to my answer (the most visible would be a change in notation from the original question) I present my answer without applying this trick.
Simplify, simplify, simplify... reduce the problem to one involving as few "moving parts" as possible. A discrete PMF is easier to deal with than a continuous PDF so let's stick to that; let $X$ be non-zero with probability $p$ and place that probability at two point masses (at $\pm t$, by symmetry) with $\Pr(X=0)=1-p$. We can rescale, so may as well have the point masses at $\pm 1$. Now our system has only one "moving part", the probability $p$. 
$\Pr(X=x) =  
   \begin{cases}
    p/2  & \text{if } x = -1 \\
    1 - p &\text{if } x = 0 \\
    p/2  & \text{if } x = -1 \\
    0 & \text{otherwise}
   \end{cases}$
Then since $\mathbb{E}(X)=0$, we have $V(X)=\mathbb{E}(X^2)=p(1)^2+(1-p)(0)^2=p$. Let two variables $X_1$ and $X_2$ have the above distribution but non-zero probabilities of $p_1$ and $p_2$ respectively; we only have $V(X_1) > V(X_2)$ if $p_1 > p_2$. We see $\mathbb{E}(|X|)=p$ also, so in this case $V(X_1) > V(X_2)$ would guarantee $\mathbb{E}(|X_1|) > \mathbb{E}(|X_2|)$. No counterexample here.
We need more room to manoeuvre. We can gain more flexibility if we move the point-mass we previously had at zero; by the symmetry restriction we will have to split it up into two masses at $\pm t$ with $t \in [0,1]$, each with probability $\frac{1-p}{2}$. 
$\Pr(X=x) =  
   \begin{cases}
    p/2  & \text{if } x = -1 \\
    (1 - p)/2 &\text{if } x = -t \\
    (1 - p)/2 &\text{if } x = t \\
    p/2  & \text{if } x = -1 \\
    0 & \text{otherwise}
   \end{cases}$
Now there are two moving parts: $t$ moves the two inner probability masses around, while $p$ allocates the weights between the outer masses at $\pm 1$ (fully when $p=1$) and the inner masses at $\pm t$ (fully when $p=0$). Then $V(X)=\mathbb{E}(X^2)=p(1)^2+(1-p)(t)^2=p+(1-p)t^2$ while $\mathbb{E}(|X|)=p(1)+(1-p)(t)= p + (1-p)t$.
To visualise $\mathbb{E}(|X|)$  in Wolfram Alphra,  use the command plot[p+(1-p)t, {p,0,1}, {t,0,1}] (left), and for $\mathbb{E}(X^2)$ run plot[p+(1-p)t^2, {p,0,1}, {t,0,1}] (right).

Let $X_1$ have this distribution with parameters $p_1$ and $t_1$, and $X_2$ with parameters $p_2$ and $t_2$. Can we find two pairs $(p_1, t_1)$ and $(p_2, t_2)$ such that $(p_1, t_1)$ is "uphill" of $(p_2, t_2)$ in the plot of $\mathbb{E}(|X|)$ but is "downhill" of $(p_2, t_2)$ in the plot of $\mathbb{E}(X^2)$? The surfaces have such different shapes that this looks possible, even more obviously so when we look at ContourPlot[p+(1-p)t, {p,0,1}, {t,0,1}] and ContourPlot[p+(1-p)t^2, {p,0,1}, {t,0,1}].
 
It is clear the contours will intersect each other; let's overlay a particular example with ContourPlot[p+(1-p)t^2==0.5, p+(1-p)t==0.6, {p,0,1}, {t,0,1}].

Observe that $V(X) > 0.5$ if $(p,t)$ is on the "high" (north-east) side of the blue contour, and $V(X) < 0.5$ on the "low" (south-west) side. Similarly the purple contour tells you how $\mathbb{E}(|X|)$ compares to $0.6$. If $(p_1, t_1)$ is on the high side of the purple contour while $(p_2, t_2)$ is on the low side, then $\mathbb{E}(|X_1|) > \mathbb{E}(|X_2|)$. Now, can you also get them on opposite sides of the blue contour, but in such a way that $V(X_1) < V(X_2)$?
You only need to know roughly what the contour plots are going to look like to realise that a counterexample will exist, which is why my back-of-envelope sketch sufficed to satisfy me, though the accurate plots make numerical choices for $p$ and $t$ straightforward for verification. It gives a PMF not a PDF, but if you make a suitably "spiky" PDF to approximate the PMF (very thin rectangular spikes will do) then you can find a continuous counterexample also.
