Conditional geometric distributions If $p<1$ and $X$ is a random variable distributed according to the geometric distribution $P(X = k) = p (1-p)^{k-1}$ for all $k\in \mathbb{N}$, then it is easy to show that $E(X) = \frac 1p$, $\mathop{Var}(X)=\frac{1-p}{p^2}$ and $E(X^2) = \frac{2-p}{p^2}$. 
Now consider a "conditional" geometric distribution, defined as follows (if there is standard terminology for this, let me know and I'll call it that):


*

*Fix a set $J\subset \mathbb{N}$ and a number $\mu>0$ (this will eventually be large).

*Let $P(X=k) = C \gamma^k$ if $k\in J$ and $P(X=k)=0$ otherwise, where $C>0$ and $\gamma<1$ are chosen so that probabilities sum to $1$ and $E(X) = \mu$.


I'm trying to understand how $E(X^2)$ (or equivalently, $\mathop{Var}(X)$) depends on $J$ and $\mu$.  In the case where $J=\mathbb{N}$ the standard results show that $p=\frac 1\mu$ and so $E(X^2) = \mu^2(2-\frac 1\mu)$.  I'm interested in the case where $\mu$ becomes very large and would like to obtain a similar estimate $E(X^2) \approx A\mu^2$, for some constant $A>1$, in a more general setting.
The example I'm working with at the moment is $J= \{2^n \mid n\in \mathbb{N}\}$, but ideally I'd like some conditions on the set $J$ that would guarantee an estimate of the above form.
Is there a standard name for these distributions, or a reference where I can read more about them?  Are estimates of this form known?
Edit: As Brendan McKay pointed out below, this boils down to understanding the behaviour of the function $g(\gamma) = \sum_{j\in J} \gamma^j$, and in fact the issue that motivated the question I posed can be stated more directly in terms of this function.
The condition $E(X) = \mu$ is equivalent to the equation $\mu = \gamma g'(\gamma) / g(\gamma)$, which determines $\gamma$ implicitly as a function of $\mu$.  We would like to understand how $g(\gamma)$ grows as $\mu\to\infty$, and hence $\gamma\to 1$.  (In particular, this means we're really interested in the case where $J$ is infinite.)
In the case $J=\mathbb{N}$, one has $g(\gamma) = \frac\gamma{1-\gamma} = 1 - \frac 1{1-\gamma}$, and so $\mu = \gamma (\frac{\gamma}{(1-\gamma)^2}) (\frac{1-\gamma}\gamma) = \frac{\gamma}{1-\gamma}$, so that in fact $g(\gamma(\mu)) = \mu$ and the two quantities go to infinity together.
In the more general case a reasonably simple argument shows that $\lim_{\mu\to\infty} g(\gamma(\mu)) = \infty$ provided $J$ is infinite, but it's not at all clear to me how the rate at which $g$ grows (in terms of $\mu$) depends on $J$ for more general sets.
That's the original motivation -- after some messing around we decided that we could figure out the growth rate if we knew something about $E(X^2)$ as suggested above, and since it was phrased in terms of what seemed to be a reasonably natural probability distribution, we decided to ask it in that form.  But now you have the whole story...
 A: Define $g(\gamma) = \sum_{j\in J} \gamma^j$.  The condition $E(X^2)\sim A\mu^2$ as $\mu\to\infty$ seems to be equivalent to
$$ \frac{g(\gamma) g''(\gamma)}{(g'(\gamma))^2} \to A $$
as $\gamma\to 1$ from below.  Alternatively define $h(x)=\sum_{j\in J} ~e^{-jx}$
and then you want
$$ \frac{h(x)h''(x)}{(h'(x))^2} \to A$$
as $x\to 0$ from above.
Of course these are translations of the problem rather than solutions, but
I mention them as someone will probably see what to do next.
A: I'm not sure what you really want but here is a couple of simple minded inequalities that can serve as a baseline.
Below $g=\sum_{k\in J}\gamma^k$, $M=\sum_{k\in J}k\gamma^k$, so $\mu=\frac Mg$. We'll need the counting function $F(n)=\#\{k\in G: k\le n\}$ of the set $J$. I will assume that $F$ is extended as a continuous increasing function to the set $[1,+\infty)$ and that $g\ge 1$.
1) For every $N$, we have the trivial estimate $g\le F(N)+\frac MN$. Taking $N=2\mu$, we get $g\le F(2\mu)+\frac g2$, i.e.,
$$
g\le 2F(2\mu)
$$ 
2) Let $\nu$ satisfy $F(\nu)=3g$. Since $g\ge F(\nu)\gamma^\nu$, we conclude that $\gamma^\nu\le \frac 13$ so $1-\gamma>\frac 1\nu$. Now, for every $N$, we have 
$$
M\le Ng+(N+\frac 1{1-\gamma})\frac 1{1-\gamma}\gamma^N\le Ng+(N+\nu)\nu e^{-N/\nu}\.
$$
Since we clearly have $\nu\ge F(\nu)=3g$, we can choose $N=\nu\log\frac\nu g\ge \nu$. For this choice, the second term on the right is at most $2Ng$, so, dividing by $g$ we get $\mu\le 3N$, i.e.,
$$
\mu\le 3F^{-1}(3g)\log\frac{F^{-1}(3g)}{g}
$$
Examples of what these inequalities yield:
1) Dense set ($F(n)\approx n$). Then $g\approx\mu$
2) Power lacunarity ($F(n)\approx n^p$, $0<p<1$). Then $g$ is between $\mu^p(\log\mu)^{-p}$ and $\mu^p$ up to a constant factor.
3) Geometric lacunarity ($F(n)\approx\log n$). Then $g\approx \log\mu$.
As you see, one can lose a logarithm sometimes but the advantage is that I do not make any regularity assumptions here. Of course, if $F$ is regular enough, you can, probably, do a bit better.
A: Can you estimate $C$ and $\mu$ etc...using the first term say $j=\min J$?
It seems to me for instance that $1/C=\gamma^j+\gamma^{j_2}+\cdots\leq \sum_{k=j}^\infty \gamma^k=\gamma^j/(1-\gamma)$.
