Kurtosis of linear combination of independent variables Dear fellow statisticians, mathematicians,
I struggle to find proof of one seemingly simple statement. I would be very grateful for any help, I am quite desperate finding the proof. I thank you very much in advance for your help. Ask whatever you want in return, I'd be happy to help you with anything in my capabilities. :)
The statement
Suppose random vector $\tilde{X}=(\tilde{X}_1,\dots,\tilde{X}_m)$ has following attributes. 


*

*$\mathbb{E}[ \tilde{X}^k_i]$ exists for $k=1,\dots, 4, i=1,\dots,m$.

*$\mathbb{E}[\tilde{X}_i]=0, \mathbb{E}[\tilde{X}_i^2]=1$ for $i=1,\dots,m$.

*The components of the random vector $\tilde{X}$ are stochasticly independent.


Suppose $L$ is lower triangular matrix $m\times m$, $R= L L^T$ (from Cholesky decomposition), where $R$ is symetric positive definite correlation matrix of $\tilde{X}$, diag($R$)=$(1,\dots,1)^T$. We define $\tilde{Y}$ as $\tilde{Y}:= L \tilde{X}$. Then $\tilde{Y}$ has following property
$$
 \mathbb{E}[\tilde{Y}^4_i]-3= \sum_{j=1}^i L_{ij}^4 ([\mathbb{E}[\tilde{X}_j^4]-3)
$$
A hint
As $L$ comes from Choleski decomposition of a matrix $R$, which has 1 on it's diagonal, it follows:
$$
\sum_{j=1}^{i} L^2_{ij}=1, i =1,\dots,m
$$
The source
The source of this statement is from a paper of Kaut, Hoyland, Wallace (Pages 183-184). Paper is discussing a heuristic for moment matching scenario generation.
 A: Found the proof. Hope it'll help.
$$
 \mathbb{E} [\tilde{Y}^4_i]-3= \mathbb{E} [(\sum_{j=1}^i L_{ij} \tilde{X}_j)^4]-3
$$
$$
=  \sum_{j=1}^i\sum_{k=1}^i\sum_{l=1}^i\sum_{r=1}^i(L_{ij}L_{ik}L_{il}L_{ir} \mathbb{E}[\tilde{X}_{j}\tilde{X}_{k}\tilde{X}_{l} \tilde{X}_{r} ])-3
$$
$$
\mathbb{E}[\tilde{X}_{j}\tilde{X}_{k}\tilde{X}_{l} \tilde{X}_{r} ]=0
$$
if one of the indexes is different from all the other. There are two possibilities that 
$$
\mathbb{E}[\tilde{X}_{j}\tilde{X}_{k}\tilde{X}_{l} \tilde{X}_{r} ] \neq 0
$$
Either all of the indexes are the same 
$$
\mathbb{E}[\tilde{X}_{j}\tilde{X}_{k}\tilde{X}_{l} \tilde{X}_{r} ]= \mathbb{E}[\tilde{X}_j^4]
$$
Or $j=k,l=r,j\neq l$.
$$
\mathbb{E}[\tilde{X}_{j}\tilde{X}_{k}\tilde{X}_{l} \tilde{X}_{r} ]=\mathbb{E}[\tilde{X}_j^2\mathbb{E}[\tilde{X}_l^2]=1
$$
We have 6 different possibilities of that.
$$
\mathbb{E} [\tilde{Y}^4_i]-3=\sum_{j=1}^i L_{ij}^4 \mathbb{E}[\tilde{X}_j^4] + 6\sum_{j=1}^i \sum_{l=j+1}^i L_{ij}^2 L_{il}^2-3
$$
$$
\sum_{j=1}^i \sum_{l=j+1}^i L_{ij}^2 L_{il}^2=\sum_{j=1}^i  L_{ij}^2 \sum_{l=j+1}^i L_{il}^2
$$
$$
\sum_{j=1}^i \sum_{l=j+1}^i L_{ij}^2 L_{il}^2=\sum_{l=1}^i \sum_{j=1}^{l-1} L_{ij}^2 L_{il}^2
$$
$$
=\sum_{j=1}^iL_{ij}^2 \sum_{l=1}^{j-1}  L_{il}^2
$$
Adding the expressions together we get
$$
2\sum_{j=1}^i \sum_{l=j+1}^i L_{ij}^2 L_{il}^2= \sum_{j=1}^i L_{ij} \Big ( \sum_{l=1}^{j-1} L_{il}^2 + \sum_{l=j+1}^i L_{il}^2 \Big )
$$
$$
= \sum_{j=1}^i L_{ij}^2 \Big ( \sum_{l=1}^i L_{il}^2 - L_{ij}^2 \Big ) 
$$
$$
= \sum_{j=1}^i L_{ij}^2 \sum_{l=1}^{i} L_{il}^2 -\sum_{j=1}^i L_{ij}^4= 1- \sum_{j=1}^i L_{ij}^4
$$
Together we get
$$
\mathbb{E} [\tilde{Y}^4_i]-3 =\sum_{j=1}^i L_{ij}^4 \mathbb{E}[\tilde{X}_j^4] + 3(1- \sum_{j=1}^i L_{ij}^4)-3
$$
$$
= \sum_{j=1}^i L_{ij}^4 (\mathbb{E}[\tilde{X}_j^4]-3) 
$$                        
