I have this equation
$\lambda_i=\frac{w_i h_i}{g_i \sum_{j=1}h_j p_j(1+\frac{h_ip_i}{\sum_{j=1}{h_jp_j +\sigma^2}})}$.
I want to solve this equation in terms of $p_i$ but I am not sure how to do it because of the summation signs. Can anyone give me a clue ?
First, factor out all the terms that do not depend on $j$.
Note: I am not sure if the $\sigma^2$ is in the sum or not; I will assume not.
$\lambda_i =\dfrac{w_i h_i}{g_i \sum_{j=1}h_j p_j(1+\frac{h_ip_i}{\sum_{j=1}h_jp_j +\sigma^2})} =\dfrac{w_i h_i}{g_i(1+\frac{h_ip_i}{\sigma^2+\sum_{j=1}h_jp_j})} \dfrac1{\sum_{j=1}h_j p_j} $.
If we now let $S = \sum_{j=1}h_j p_j $, this becomes $\lambda_i =\dfrac{w_i h_i}{g_i(1+\frac{h_ip_i}{\sigma^2+S})} \dfrac1{S} =\dfrac{w_i h_i(\sigma^2+S)}{Sg_i(\sigma^2+S+h_ip_i)} $.
From this, $\sigma^2+S+h_ip_i =\dfrac{w_i h_i(\sigma^2+S)}{g_i\lambda_iS} $ so
$\begin{array}\\ h_ip_i &=\dfrac{w_i h_i(\sigma^2+S)}{g_i\lambda_iS}-(\sigma^2+S)\\ &=\dfrac{w_i h_i\sigma^2}{g_i\lambda_iS}+\dfrac{w_i h_i}{g_i\lambda_i}-(\sigma^2+S)\\ &=\dfrac{w_i h_i}{g_i\lambda_i}(\dfrac{\sigma^2}{S}+1)-(\sigma^2+S)\\ \end{array} $
There are a number of things you can do from here.
For example, if we sum over $i$, we get
$\begin{array}\\ S &=\sum_ih_ip_i\\ &=\sum_i\dfrac{w_i h_i}{g_i\lambda_i}(\dfrac{\sigma^2}{S}+1)-\sum_i(\sigma^2+S)\\ &=(\dfrac{\sigma^2}{S}+1)\sum_i\dfrac{w_i h_i}{g_i\lambda_i}-n\sigma^2-S^2\\ &=(\dfrac{\sigma^2}{S}+1)T-n\sigma^2-S^2\\ \end{array} $
where $T = \sum_i\dfrac{w_i h_i}{g_i\lambda_i} $.
This gives a cubic equation for $S$.
Once $S$ is gotten, the $p_i$ can be gotten from $h_ip_i =\dfrac{w_i h_i}{g_i\lambda_i}(\dfrac{\sigma^2}{S}+1)-(\sigma^2+S) $.
Anyway, hope this helps.