Given i.i.d. random variables $\xi_i$ where $E\xi_i^2<\infty,E\xi_i=0$, and real bounded sequence $\{a_i\}$, Suppose $\bar{\xi}_i:= \xi_i\mathbf{1}[|\xi_i|\leq i^{3/4}]$, I want to show that $E(\sum\limits_{i=1}^n a_i\bar{\xi}_i)^4\leq C n^{5/2}$ is true where $C$ is some constant.
Here is my try.
$E(\sum\limits_{i=1}^n a_i\bar{\xi}_i)^4 =E[ \sum\limits_{i=1}^n a_i^4 \bar{\xi}_i^4+C_1\sum\limits_{i\neq j}^n a_i \bar{\xi}_ia_j^3\bar{\xi}_j^3+C_2\sum\limits_{i\neq j}^n a_i^2 \bar{\xi}_i^2a_j^2 \bar{\xi}_j^2+C_3\sum\limits_{i,j,k \ \text{different}}^n a_i \bar{\xi}_ia_j \bar{\xi}_ja_k^2 \bar{\xi}_k^2+C_4\sum\limits_{i,j,k,l \ \text{different}}^n a_i \bar{\xi}_ia_j \bar{\xi}_ja_k\bar{\xi}_ka_l \bar{\xi}_l]$.
The first integral is trivial because $\leq K_1 n^{3/2}\sum\limits_{i=1}^n E\bar{\xi}_i^2\leq K_1n^{5/2}$. $K_1$ is the symbol of constant, absorbing any constant popping up in this term.
The third integral is also trivial because $\leq K_3\sum\limits_{i\neq j}1\leq K_3 n^2$.
Now I get stuck in bounding the remainder terms. I do not know how to deal with first order and third order terms (relating to the second order moment) and still finally get an upper bound of at most $n^{5/2}$.
Actually, this inequality is from this paper: http://www.meyn.ece.ufl.edu/archive/spm_files/Courses/ECE555-2011/555media/poljud92.pdf Page 848 second half to Page 849 first half.
Observe that the sequence of random variables $\bar{\xi}_i$ is still independent. Using $\mathbb{E} \xi_i = 0 $ we have $$ \mathbb{E} \bar{\xi_i} = \int_{|\xi_i|\leq i^{3/4}} \xi_i d \mathbb{P} = - \int_{|\xi_i|> i^{3/4}} \xi_i d \mathbb{P} = -\int \xi_i \chi_{ \{ |\xi_i|>i^{3/4} \}} d \mathbb{P}, $$ hence, applying Cauchy-Schwarz, $$ |\mathbb{E} \bar{\xi}_i| \leq (\mathbb{E} \xi_{i}^2 )^{1/2} (\mathbb{P}(|\xi_i| > i^{3/4}))^{1/2}. $$ Next, from Chebichev's inequality, we have $$ \mathbb{P}(|\xi_i| > i^{3/4}) \leq i^{-3/4} \mathbb{E} |\xi_i| \leq i^{-3/4} \mathbb{E}|\xi_i|^2. $$ Combining this from the previous estimate, we obtain $$ (1) \qquad |\mathbb{E} \bar{\xi}_i| \leq i^{-3/8} (\mathbb{E} \xi_i^2)^{1/2} \leq C i^{-3/8}, $$ where we used the fact that $\xi_i$ are i.i.d. and hence all have the same variance.
We thus get $$ \mathbb{E}(\sum_{i=1}^n a_i \bar{\xi}_i)^4 = \mathbb{E}\sum_{1\leq i_1,i_2,i_3,i_4 \leq n}a_{i_1}a_{i_2} a_{i_3}a_{i_4} \bar{\xi}_{i_1} \bar{\xi}_{i_2} \bar{\xi}_{i_3} \bar{\xi}_{i_4} \leq \\ (\text{using independence of } \bar{\xi}_i \text{ and boundedness of } a_i) \\ C \sum_{1\leq i_1,i_2,i_3,i_4 \leq n} i_1^{-3/8}i_2^{-3/8}i_3^{-3/8}i_4^{-3/8} =\\ C (\sum_{i = 1}^n i^{-3/8})^4 \leq C n^{4(1-3/8)} = C n^{5/2}, $$ where we used $(1)$ to bound the expectations in the sum.