Why the local Bayes fdr is greater than the Bayes FDR? My question is related to empirical Bayes and large-scale inference. It is explained that the local Bayes false discovery rate (fdr) is greater than the Bayes false discovery rate (FDR). 
It is argued that the FDR is the tangent that connects the origin and the point and fdr the slope of the secant at that point.  
But, I am stack in solving the next "simple" problem. Given $F_1(z) = F_0(z)^\alpha$ for $\alpha < 1$ show that:
$log\left ( \frac{fdr(z)}{1-fdr(z)} \right ) = log\left ( \frac{FDR(z)}{1-FDR(z)} \right ) + log\left ( \frac{1}{\alpha} \right ) $
From the figure seem pretty clear that indeed fdr must be greater than the FDR.

But I'd find the answer to the problem very reassuring.
---------- EDIT ---------
I want to add two more equations, given in large-scale inference, that make my question more complete:
the distribution of the mixture:
$F(Z) = \pi _0F_0(Z)+(1-\pi _0)F_1(Z)$
The pdf of the mixtures:
$f(z) = \pi _0f_0(z)+(1-\pi _0)f_1(z)$
 A: I think there are really two questions here
Question 1: Formula for FDR and local fdr under Lehmann alternatives
Note that from the problem setup we have to assume differentiability (existence of densities $f_1$ resp $f_0$ with respect to the Lebesgue measure) for both distributions $F_1$, resp $F_0$.
Since $F_1(z) = F_0(z)^{\alpha}$ we get by the chain rule that $f_1(z) = \alpha f_0(z) F_0(z)^{\alpha-1}$. $F_0$ and $F_1$ are mixed at proportions $\pi_0, \pi_1$ to produce $F$.
Further let us recall the definition of Bayesian (or marginal) FDR and local fdr,
$$\text{FDR}(z) = \frac{\pi_0 F_0(z)}{F(z)},\;\;\text{fdr}(z)=\frac{\pi_0 f_0(z)}{f(z)}$$
By direct calculation and letting $\pi_1 = 1-\pi_0$, we see that
$$\text{FDR}(z)/(1-\text{FDR}(z))= (\pi_0 F_0(z))/(\pi_1 F_1(z)) = \pi_0/\pi_1 F_0(z)^{1-\alpha}$$
Similarly:
$$\frac{\text{fdr}(z)}{1-\text{fdr}(z)} = \frac{\pi_0 f_0(z)}{\pi_1 f_1(z)}=\frac{\pi_0 f_0(z)}{\pi_1 \alpha f_0(z) F_0(z)^{\alpha-1}} = \frac{\pi_0 F_0(z)^{1-\alpha}}{\alpha \pi_1}$$
Now take logarithms of the two expressions and subtract them to get:
$$\log\left(\frac{\text{fdr}(z)}{1-\text{fdr}(z)}\right)-\log\left(\frac{\text{FDR}(z)}{1-\text{FDR}(z)}\right) = -\log(\alpha)$$
This is the result we were looking for.
Question 2: Why is the local fdr typically larger than FDR?
For Lehmann Alternatives, this follows immediately from the preceding display, recalling that $\alpha < 1$ and that $u \mapsto \log(u/(1-u))$ is strictly increasing.
Another case where the local fdr is seen to be larger than the FDR is the following: Let us assume our test statistic is a p-value, so take $F_0$ to be $U[0,1]$, i.e. $F_0(p) = p, p \in [0,1]$. Assume the alternative $F_1$ is concave, i.e. its density $f_1$ is non-increasing (if you look at p-value histograms from real data, they will often seem like a mixture of a uniform distribution plus a peak to the left which decreases as $p$ becomes larger). Notice that then also $F(z)$ will be concave. Then by concavity, we may check that:
$$ F(p) \geq F(0) + p f(p) \geq p f(p) \Rightarrow \frac{1}{f(p)} \geq \frac{p}{F(p)}$$
Multiplying both sides by $\pi_0$ and noting $f_0(p) = 1$, we get:
$$ \frac{\pi_0 f_0(p)}{f(p)} \geq \frac{\pi_0 F_0(p)}{F(p)} $$
But this of course, is the same as $\text{fdr}(p) \geq \text{FDR}(p)$.
