Non-trivial bound for $E[\exp(Z^2)]$ when $Z \sim {\rm Bin}(n, n^{-\beta})$ with $\beta \in (0,1)$ How to find a non-trivial upper bound on $E[\exp(Z^2)]$ when $Z \sim {\rm Bin}(n, n^{-\beta})$ with $\beta \in (0,1)$? A trivial bound is obtained for substituting $Z$ with $n$.
A background on this question. In the paper by Baraud, 2002 -- Non-asymptotic minimax rates of testing in signal detection, if one is to substitute the model in Eq. (1), by a random effects model, then the above quantity appears in the computation of a lower bound. 
 A: (I will delete my other non-answer, the edit of which had this nugget in it)
No. Asymptotically, the 'trivial' upper bound is the least upper bound. To see this, Let $P_n = P(Z = n)$. Trivially, $E[\exp{(Z^2)}] \ge P_n \exp{(n^2)} = L$, where $L$ is the lower bound of interest. Since $Z$ is binomial, we have $P_n = {n\choose n} (n^{-\beta})^n (1-n^{-\beta})^0 = n^{-n\beta}$. Then $\log{L} = -\beta n \log{n} + n^2$. It is easily shown that this is $\Omega{(n^2)}$, and thus, $L$ is $\Omega{(\exp{(n^2)})}$. Thus the trivial upper bound is, asymptotically, the least upper bound, i.e. $E[\exp{(Z^2)}] \in \Theta{(\exp{(n^2)})}$.
A: Numerical experiments (for 2 <= n <= 4000 and all values of beta) indicate the estimate n^2 - (n Ln(n))*beta exceeds the logarithm of the expectation by an amount on the order of beta*Exp(-n).  The error appears to increase monotonically in beta for each n.  This should provide some useful clues about how to proceed (for those with the time and interest).  In particular, an upper bound for the expectation exists of the form Exp(n^2 - (n Ln(n))*beta + C*Exp(-n)*beta) with C << 1. 
Update
After staring at the summation expression for the expectation, it became evident where the nLn(n)*beta term comes from: break each binomial coefficient Comb(n,k) into its sum Comb(n-1,k) + Comb(n-1,k-1) and write Exp(k^2) = Exp((k-1)^2)*Exp(2k-1).  This decomposes the expectation e(n,beta) into the sum of two parts, one of which looks like the expectation e(n-1,beta) and the other of which is messy (because each term is multiplied by Exp(2k-1)) but can be bounded above by replacing all those exponential terms by their obvious upper bound Exp(2n-1).  (This is not too bad, because the last term with the highest exponent strongly dominates the entire sum.)  This gives a recursive inequality for the expectation,

e(n,beta) <= (n^-beta * Exp(2n-1) + 1 - n^-beta) * e(n-1,beta)

Doing this n times creates a polynomial whose highest term is in fact Exp(n^2)*n^(-n*beta), with the remaining terms decreasing fairly rapidly.  At this point any reasonable bound on the remainder will produce an improved bound for the expectation essentially of the form suggested by the numerical experiments.  At this point you have to decide how hard you want to work to obtain a tighter upper bound; the numerical experiments suggest this additional work is not going to pay off unless you're interested in the smallest values of n.
A: This answer is inspired by shabbychef's answer using the median. By definition:
$E[exp(Z^2)] = \sum_{z=1}^{z=n} exp(z^2) P(z;n,n^{-\beta})$ 
where,
$P(z;n,n^{-\beta})$ is the binomial probability.
Denote the mode of this binomial distribution by: $m(n,n^{-\beta})$. Thus, by definition we have:
$P(z;n,n^{-\beta}) \le P(m(n,n^{-\beta});n,n^{-\beta}) \ \ \forall z$
Let,
$\bar{P} = P(m(n,n^{-\beta});n,n^{-\beta})$
Thus,
$E[exp(Z^2] \le \sum_{z=1}^{z=n} exp(z^2) \bar{P}$
This upper bound is a function of $n$ and $\beta$ as desired. 
Hopefully, this in the right track unlike my previous attempt.
This approach is technically not ok as $Z$ is a discrete variable but can be justified if we take the normal approximation to the binomial.
I am not sure to what extent this is a better bound then the trivial bound but here is one approach. Take the taylor series expansion of $exp(z^2)$ and ignoring terms higher than the second term, you get:
$\int e^{z^2} f(z) dz < \int (1 + z^2) f(z) dz$
Now,
$\int (1 + z^2) f(z) dz = 1 + \int z^2 f(z) dz$
But, we know that:
$\int z^2 f(z) dz = Var(z) + E(z)^2$
Substituting for the variance and mean of the binomial distribution and simplifying, we get:
$\int e^{z^2} f(z) dz < 1 +  n^{1-\beta} (1-n^{-\beta} + n^{1-\beta})$
PS: Please check the math as I corrected one error.

