Are these two equations equivalent Can I say that  this equation

is equivalent to
$$\frac{1}{S}\frac{1}{U} \sum_{p=1}^{S} \sum_{u=1}^{U}  PL .  SF (|\alpha_{0,u,p}|^2+\sum_{n=1}^N\sum_{m=1}^M|\alpha_{n,m,u,p}|^2 )$$
Thank you.
 A: The two expressions are equal.

We have
\begin{align*}
\frac{1}{S}\sum_{p=1}^{S} &\left(PL \cdot  SF \cdot \frac{1}{U}\sum_{u=1}^{U}   \left(|\alpha_{0,u,p}|^2+\sum_{n=1}^N\sum_{m=1}^M|\alpha_{n,m,u,p}|^2 \right)\right)\\
&=\frac{1}{S}\sum_{p=1}^{S} PL \cdot  SF \cdot \frac{1}{U}\sum_{u=1}^{U}   \left(|\alpha_{0,u,p}|^2+\sum_{n=1}^N\sum_{m=1}^M|\alpha_{n,m,u,p}|^2 \right)\tag{1}\\
&=\frac{1}{S}  \frac{1}{U}\sum_{p=1}^{S} PL \cdot  SF \sum_{u=1}^{U}   \left(|\alpha_{0,u,p}|^2+\sum_{n=1}^N\sum_{m=1}^M|\alpha_{n,m,u,p}|^2 \right)\tag{2}\\
&=\frac{1}{S}\frac{1}{U} \sum_{p=1}^{S} \sum_{u=1}^{U}  PL \cdot  SF \left(|\alpha_{0,u,p}|^2+\sum_{n=1}^N\sum_{m=1}^M|\alpha_{n,m,u,p}|^2 \right)\tag{3}
\end{align*}

Comment:

*

*In (1) we skip the outer parentheses. They do not have any effect since  multiplication of terms has a higher precedence level than addition of terms.


*In (2) we factor out $\frac{1}{U}$ using the distributive law: $a(b+c)=ab+ac$.


*In (3) we multiply the terms of the third-left inner sum with $PL \cdot  SF$ again using the distributive law.
