$X_1,\dots,X_T$ are i.i.d. random variables.
What is the expected value of the following function:
$$ \left (\sum_{t=1}^T X_t^a\right)^{(1/a)} $$
as a function of $a$ and $T$?
a. What is the best way to calculate this value explicitly, when the distribution of the $X_t$'s is known? E.g. suppose each $X_t$ is a discrete variable which equals 3 with probability 0.5 and 7 with probability 0.5, how can I compute the expectation of the above expression as a function of $a$ and $T$?
b. What is a symbolic, approximate expression (e.g. in big-Oh notation) for the expectation, when $a$ is constant and $T\to \infty$?
EDIT (a hint from my advisor): By Hölder's inequality for sequences, if the $X_t$ are non-negative and $a\geq 1$, then for every non-negative sequence $Y_t$:
$$ \left (\sum_{t=1}^T X_t^a\right)^{(1/a)} \cdot \left (\sum_{t=1}^T Y_t^q\right)^{(1/q)} \geq \sum_{t=1}^T X_t Y_t$$
where $\frac{1}{q}+\frac{1}{a}=1$ (so $q=\frac{a}{a-1}$).
If we take the sequence $Y_t=1$ for all $t$, then this becomes:
$$ \left (\sum_{t=1}^T X_t^a\right)^{(1/a)} \geq \frac{\sum_{t=1}^T X_t}{T^{1/q}}$$
So a lower bound on the expected value, when the variables are i.i.d, is:
$$\frac{\sum_{t=1}^T E[X_t]}{T^{1/q}} = \frac{T E[X_1]}{T^{1/q}} = T^{1-1/q} E[X_1] = T^{1/a} E[X_1]$$
This is only a lower bound.