Calculating $E[(\sum X_i)^4]$

Trying to figure where I'm going wrong with the following. My goal is to calculate var$$(\bar X_n^2)$$ using $$E[(\bar X_n)^4]=\frac{1}{n^4}E[(\sum X_i)^4]$$ given that $$X_1,...X_n$$ are iid with $$EX_1=\mu, E[(X_1-\mu)^k]=\alpha_k$$. $$E\left[\left(\sum X_i\right)^4\right]=E\left[\sum_i X_i\sum_jX_j\sum_kX_k\sum_lX_l\right]=\sum_{i,j,k,l}E(X_iX_jX_kX_l)$$ Now, I divide this into 5 categories depending on the indices $$i,j,k,l$$.

1. All distinct. $${n\choose 4}E^4(X_1)=\frac{n(n-1)(n-2)(n-3)}{24}\mu^4$$
2. One pair. $${3\choose 1}{n\choose 3}E(X_1^2)E^2(X_1)=\frac{n(n-1)(n-2)}{2}\mu^2\left[E(X_1-\mu)^2+\mu^2+2\mu E(X_1-\mu)\right]\\=\frac{n(n-1)(n-2)}{2}\mu^2\left[\alpha_2+\mu^2\right]\\=\frac{n(n-1)(n-2)}{2}\mu^2\alpha_2+\frac{n(n-1)(n-2)}{2}\mu^4$$
3. Two pair. $${n \choose 2}E^2(X_1^2)=\frac{n(n-1)}{2}(\alpha_2+\mu^2)^2\\=\frac{n(n-1)}{2}\alpha_2^2+n(n-1)\mu^2\alpha_2+\frac{n(n-1)}{2}\mu^4$$
4. Three of a kind. $${3\choose 1}{n\choose 2}E(X_1)E(X_1^3)=\frac{3n(n-1)}{2}\mu[E(X_1-\mu)^3+\mu^3+3\mu E(X_1-\mu)^2]\\=\frac{3n(n-1)}{2}\mu\alpha_3+\frac{9n(n-1)}{2}\mu^2\alpha_2+\frac{3n(n-1)}{2}\mu^4$$
5. All of a kind. $$nE(X_1^4)=n[E(X_i-\mu)^4+\mu^4+4\mu E(X_i-\mu)^3+6\mu^2 E(X_i-\mu)^2]\\=n\alpha_4+4n\mu\alpha_3+6n\mu^2\alpha_2+n\mu^4$$

I know a final simpler form but things aren't adding up that well. Please point out if there are mistakes.

EDIT. More context on this. I followed the method mentioned here to get $$\text{var}(\bar X_n^2)=\frac{4\mu^2\alpha_2}{n}+\frac{4\mu\alpha_3}{n^2}+\frac{\alpha_4}{n^3}+(\frac{2}{n^2}-\frac{3}{n^3})\alpha_2^2$$ which differs from the desired expression $$\text{var}(\bar X_n^2)=\frac{4\mu^2\alpha_2}{n}+\frac{4\mu\alpha_3}{n^2}+\frac{\alpha_4}{n^3}$$ (Ref. The Jackknife & Bootstap, Jun Shao). This brute force expression also differs upon simplification.

• Hello, zaira. Firstly, please add the self-study tag. Secondly, this is the simplest (brute-force) method you could apply. Where do you think a mistake might be incurred? Meant to say you should perhaps expand on "but things aren't adding up that well." Commented Feb 10 at 8:29
• @User1865345 Had a desired expression mentioned in a reference book. However the simplified expression from here doesn't match Commented Feb 10 at 8:48
• The approach is a good one but it's unclear where your coefficients come from. I suspect that an attempt to justify them combinatorially might resolve your question.
– whuber
Commented Feb 10 at 22:53
• @zaira, can you mention your final expression of expectation of the fourth power of the sum? Point is the easiest (though tedious) is the combinatorial approach. If there is any issue, it must be with the calculations somewhere and not in the method itself. Commented Feb 11 at 9:43
• Related to stats.stackexchange.com/questions/531457/…? Commented Feb 11 at 12:25

In pursuit of calculating the fourth power, it might be apt to have a look at the general form of expectation of $$p$$–th power of sum of iid random variables.

$$\rm [I]$$ deduces $$\mathbf E\left[S_n^p\right]$$ based on a combinatorial reasoning.

Namely,

Result $$1:$$ For $$p\in\mathbb N,$$ with iid $$X_1, X_2, \ldots, X_n,$$ the $$p$$–th moment of sum $$S_n:=\sum_{i=1}^n X_i$$ is given by $$\mathbf E\left[S_n^p\right] = \sum_{q_m\in\mathcal Q^p}a_mq_m,\tag 1\label 1$$ where \begin{align}\mathcal Q^p&:=\left\{\underbrace{\mu_{p_1}\mu_{p_2}\cdots\mu_{p_m}}_{:=q_m}\mid p_1, p_2,\ldots, p_m\in \mathbb N_p~\wedge~\sum_{i=1}^m p_i= p\right\},\\a_m&:=\frac{1}{l_1!\cdots l_h!}\frac{n!}{(n-m)!}\frac{p!}{\prod_{i=1}^m p_i!} ,\end{align}

where $$h$$ is the number of distinct $$p_i$$s; $$l_j$$ is the number of times the $$j$$–th constant appears.

$$\rm[II]$$ also builds upon multinomial theorem a similar formula, that is equivalent to $$\eqref 1.$$

Result $$2:$$ $$\mathbf E\left[S_n^p\right] =\sum_{1\leq r\leq p} A_{r,p}{n\choose r},\tag 2\label 2$$ where $$A_{r,p}:= \sum_{\boldsymbol k\in I_{p, r}}{p\choose {k_1\ldots k_r}}{r\choose {n_1\ldots n_a}}\prod_{i=1}^a \mu_{\kappa_i}^{n_i},$$

where $$I_{p,r}:= \{\boldsymbol k\in\mathbb N^r\mid k_1+\cdots k_r = p~\wedge~k_1\geq\cdots\geq k_r \}$$ is the set of unordered partitioned of $$p.$$

Deducing the two forms aren't hard; I'm outlining the sketch of the proofs of the corresponding two results above; it is then upto OP to utilize those to formally perform the required calculations:

$$1:$$ The strategy is to first calculate the cardinality of $$\mathcal C_{i_{p_1}, j_{p_2}, \ldots k_{p_m}},$$ the set of all unique permutations of the sequence where $$X_i$$ appears $$p_1$$ times and so on. This would be $$\frac{p!}{\prod_{i=1}^m p_i!} .$$ Then we would calculate the cardinality of unions of all such sets with varying $$i, j, \ldots, k,$$ no indices being equal to each other. This would yield the factor $$\frac{n!}{(n-m)!}.$$ Finally we need to take care of the cases where some of $$p_i$$s can be equal to each other. This propels the factor $$\frac{1}{l_1!\cdots l_h!}.$$

$$\blacksquare$$

$$2:$$ Crux is expressing the multinomial theorem (with $$J_{p, r}$$ being the unordered counterpart of $$I_{p, r}$$ ) as $$\left(\sum_{i=1}^n x_i\right)^p = \sum_{1\leq r\leq p}\sum_{\boldsymbol a \in\mathbb N^r_n}\sum_{\boldsymbol k\in J_{p, r}} {p\choose {k_1\ldots k_r}}\prod_{s=1}^r x_{a_s}^{k_s};$$ intimidating it might look, it is nothing but writing the LHS as $$\sum_{\boldsymbol j \in \mathbb N_{n}^p}\prod_{s=1}^p x_{j_s}$$ and breaking the sum based on the distinct elements of $$\boldsymbol j$$ viz. $$\boldsymbol a\in\mathbb N_{n}^r,~r\in \mathbb N_p$$ with corresponding multiplicities $$\boldsymbol k \in J_{p, r}.$$ Now owing to $$X_i$$s being iid, $$\mathbf E\left[X_{a_s}^{k_s}\right] = \mu_{k_s}, ~1\leq s\leq r.$$ Thus the final sum isn't impacted by any arbitrary permutation of $$k_s$$s: take the corresponding $$a$$ unique parts $$\kappa_i$$s which appear with multiplicities $$n_i$$s respectively. This means there are $${r\choose {n_1\ldots n_a}}$$ ordered partitions corresponding to a single $$\boldsymbol k$$ implying $$\sum_{\boldsymbol k\in J_{p, r}} {p\choose {k_1\ldots k_r}}\prod_{s=1}^r \mu_{k_s}=\sum_{\boldsymbol k\in I_{p, r}}{p\choose {k_1\ldots k_r}}{r\choose {n_1\ldots n_a}}\prod_{s=1}^a \mu_{\kappa_s}^{n_s}.$$

$$\blacksquare$$

$$\eqref 1$$ seems more tractable.

References:

$$\rm [I]$$ Moments of Sums of Independent and Identically Distributed Random Variables, Daniel M. Packwood, $$2012,$$ url: http://arxiv.org/abs/1105.6283.

$$\rm [II]$$ Moment equalities for sums of random variables via integer partitions and Faàdi Bruno’s formula, Dietmar Ferger, Turkish Journal of Mathematics: Vol. $$38:$$ No. $$3,$$ Article $$15,~2014,$$ url: https://doi.org/10.3906/mat-1301-6.

I would start by setting $$Y_i=X_i-\mu$$, for $$i=1\ldots,n$$. You then have: $$\mathrm{Var}(\bar{X}^2)=\mathrm{Var}((\bar{Y}+\mu)^2)= \mathrm{Var}(\bar{Y}^2+2\mu\bar{Y})\\ = \mathrm{Var}(\bar{Y}^2)+2\mu\mathrm{Cov}(\bar{Y}^2,\bar{Y})\\=\mathbb{E}(\bar{Y}^4)-\mathbb{E}(\bar{Y}^2)^2+2\mu\mathbb{E}(\bar{Y}^3)$$

These terms should be a lot easier to work out when you expand them because the $$Y_i$$ have mean zero. For example, any term of the form $$\mathbb{E}(Y_i Y_j Y_k Y_l)$$ is equal to zero if at least one of $$\{i,j,k,l\}$$ is different to the others.