Let $X_{i}$~$Bin(n, p)$.
I know that the moment-generating function of a Binomial distribution is defined as :
$$ \mathbb{E}(e^{t X_{i}}) =(1-p+pe^{t})^{n} $$
But what about $\mathbb{E}(e^{t \max_{i}X_{i}})$ ? Do we need any independence between the variables to compute it ? Or is there any upper bound for it ?
$\mathbb{E}(e^{t\max_{i}X_{i}})=\sum_{k=0}^{n}e^{tk}\mathbb{P}(\max_{i}X_{i}=k).$ Under independence of the $X_{i},$ $1\leq i\leq m$, $\mathbb{P}(\max_{i}X_{i}=k)=\mathbb{P}(X_{1}\leq k)^{m}-\mathbb{P}(X_{1}\leq k-1)^{m},$ for $k=1,\ldots,n,$ and $\mathbb{P}(\max_{i}X_{i}=0)=\mathbb{P}(X_{1}=0)^{m}.$ So \begin{align*}\mathbb{E}(e^{t\max_{i}X_{i}})&=\mathbb{P}(X_{1}=0)^{m}+\sum_{k=1}^{n}e^{tk}(\mathbb{P}(X_{1}\leq k)^{m}-\mathbb{P}(X_{1}\leq k-1)^{m})\\ &=(1-e^{t})\mathbb{P}(X_{1}=0)^{m}+\sum_{k=1}^{n-1}e^{tk}(1-e^{t})\mathbb{P}(X_{1}\leq k)^{m}+e^{tn}\\ &=(1-e^{t})\sum_{k=0}^{n-1}e^{tk}\mathbb{P}(X_{1}\leq k)^{m}+e^{tn}.\end{align*}
This last expression could conceivably be used for computation.