Compute the $2k^{\textrm{th}}$ power of a sum of independent random variables Let $\{T_i\}_{i=1}^d$ a set of i.i.d random variables that are symmetric about the origin. Also,let $T = \frac{1}{d} \sum_{i=1}^d T_i$. Then, for every $k=0,1,\dots$, we have $$\mathbb E[T^{2k}] = \frac{1}{d^{2k}}\sum_{i_1=1}^d\cdots \sum_{i_{2k}=1}^d \mathbb E[T_{i_1}\cdots T_{i_{i_{2k}}}]\tag{1}$$
According to @user170231 comment, $(1)$ looks like a multinational expansion. In addition, for a symmetric distribution whose odd-order moments exist, we have that the odd-order moments are equal to zero (see here). Maybe another useful post is this which I found complicated to use it.
Here I summarize what I need to understand after @Robert Israel's answer. How
$$\mathbb E[T^{2k}] = \sum_{j_1+j_2+\cdots+j_d=2k; \\ j_1, j_2, \cdots, j_d \geq 0}  {2k \choose j_1,j_2,\ldots, j_{d}} \mathbb E[T_{1}^{j_1}] \ldots \mathbb E[T_d^{j_{d}}]$$
is reduced to the even powers of $T_{i}$, which probably reveals the term $1/d^{2k}$ and how we get $\mathbb E[T_{1}^{j_1}\cdots T_{d}^{l_{d}}]= \mathbb E[T_{i_1}\cdots T_{i_{2k}}]$.
Any help is highly appreciated.
The equation is from here proof of lemma 8.
 A: Using the multinomial expansion, linearity of expectation, and independence,
$$ \mathbb E[T^{2k}] = \frac{1}{d^{2k}} \sum_{j} {2k \choose j_1,j_2,\ldots, j_{d}} \mathbb E[T_{1}^{j_1}] \ldots \mathbb E[T_d^{j_{d}}] $$
where the sum is over all ordered $d$-tuples $(j_1,\ldots,j_d)$ of nonnegative integers such that $j_1 + \ldots + j_d = 2k$.  Since your random variables are identically distributed, you can replace $\mathbb E[T_i^{j_i}]$ with $\mathbb E[T_1^{j_i}]$; since they are symmetric, only $j$'s that are all even need be considered. Finally, you can collect together all terms for different permutations of one $(j_1, j_2, \ldots, j_d)$: take, say, the one where $j_1 \ge j_2 \ge \ldots \ge j_d$ as representative, and with a coefficient of the number of permutations, which is the multinomial coefficient
${d \choose d_0,d_2,\ldots, d_{2k}}$ where $d_i$ is the number of times $i$ appears in $j_1,\ldots,j_{d}$. Thus the formula is
$$  \mathbb E[T^{2k}] = \frac{1}{d^{2k}} \sum_{j} {2k \choose j_1,j_2,\ldots, j_{d}} {d \choose d_0,d_2,\ldots,d_{2k}} \mathbb E[T_{1}^{j_1}] \ldots \mathbb E[T_1^{j_{d}}]$$
where the sum is over $d$-tuples of even integers $(j_1,\ldots,j_d)$
with $j_1 \ge j_2 \ge \ldots j_d \ge 0$, and $d_i$ the number of times $i$ appears in $(j_1,\ldots,j_d)$.
For example, if $k=3$ and $d=4$ the possible $(j_1,\ldots,j_4)$ are
$(6,0,0,0)$, $(4,2,0,0)$ and $(2,2,2,0)$, resulting in
$$\eqalign{\mathbb E[T^6] &= \frac{1}{4^6} \left( {6 \choose 6,0,0,0} {4 \choose 3,0,0,1}  \;\mathbb E[T_1^6] + {6 \choose 4,2,0,0} {4 \choose 2,1,1,0} \;\mathbb E[T_1^4] \mathbb E[T_1^2] + {6 \choose 2,2,2,0} {4 \choose 1,3,0,0} \;  \mathbb E[T_1^2]^3\right)\cr
&= \frac{1}{4^6} \left( 4 \;\mathbb E[T_1^6] + 180 \;\mathbb E[T_1^4] \mathbb E[T_1^2] + 360\; \mathbb E[T_1^2]^3\right)}$$
A: Start with a simple case, say $d=3$ and $k=1$. Then
$$T=\frac13 \sum_{i=1}^3 T_i = \frac{T_1+T_2+T_3}3$$
$$\implies \Bbb E[T^2] = \Bbb E\left[\frac{{T_1}^2+{T_2}^2+{T_3}^2 + 2T_1T_2 + 2T_1T_3 + 2T_2T_3}9\right]$$
For larger values of $k$, we'll again have a sum of the $(2k)^{\rm th}$ powers of the $T_i$, followed by a sum of products of variables with total degree $2k$.
Compare this to the given identity, which for this case says
$$\Bbb E [T^2] = \frac{1}{3^2} \sum_{i_1=1}^3 \sum_{i_2=1}^3 \Bbb E\left[T_{i_1} T_{i_2}\right] \tag{2}$$
The indices $i_1$ and $i_2$ act independently of one another, so the complete expansion of the double sum includes $3\times3=9$ terms,
$$T^2 = \sum_{i_1=1}^3 \sum_{i_2=1}^3 T_{i_1} T_{i_2} = \frac{\overbrace{T_1T_1 + T_1T_2 + T_1T_3}^{i_1=1} + \overbrace{T_2T_1 + T_2T_2 + T_2T_3}^{i_1=2} + \overbrace{T_3T_1 + T_3T_2 + T_3T_3}^{i_1=3}}9$$
The multisum is a more compact way of writing $\Bbb E\left[T^{2k}\right]$ that covers each case of $\Bbb E\left[T_{i_1}T_{i_2}\cdots T_{i_{2k}}\right]$ that shows up in the numerator.
