How to show that $\pi^*_i \pi^*_j - \pi^*_{ij} = \pi_i \pi_j - \pi_{ij}$ in a probability sample? Let $\pi^*_i \quad   i=1,2,\dotsc, N$ be the probability of not including the $i$th unit in the sample and let $\pi^*_{ij} \quad i, j = 1, 2,\dotsc, N$ be the probability of including neither the $i$th nor the $j$th units in the sample. Then, show that
$$\pi^*_i  \pi^*_j  - \pi^*_{ij}  =  \pi_i  \pi_j  - \pi_{ij} $$
where is $\pi_i  $   the probability of including the $i$th unit in the sample and $\pi_{ij} $  is the probability of including both the $i$th and the j th units.
Where  $\ P_i$ is the probability of selection of $\ i^{th} $ unit   from the population without replacement .
My solution
$$\pi_i  =  P_i[1 + \sum_ {i=0}^N  \frac {P_j}{1-P_j} ] $$
$$\pi_{ij}  = \ P_i\ P_j [  \frac {1}{1-P_j}  +\frac {1}{1-P_i} ] $$
Now if I work on the lhs of question by using
$$\pi^*_i   = 1- P_i[1 + \sum_ {i=0}^N  \frac {P_j}{1-P_j}  -  \frac{P_1}{1-P_1} ] $$
$$\pi^* _j  = 1-  P_j[1 + \sum_ {j=0}^N  \frac {P_i}{1-P_i}  - \frac{P_1}{1-P_1}] $$
$$\pi^*_{ij}  = 1- \ P_i\ P_j [  \frac {1}{1-P_j}  +\frac {1}{1-P_i} ]$$
But the problem is , if I am putting these values in left hand side then it is  becoming a mess which looks nothing like rhs.
Putting everything together
I am getting
$$- P_j[1 + \sum_ {j=0}^N  \frac {P_i}{1-P_i}  - \frac{P_1}{1-P_1}] - P_i[1 + \sum_ {i=0}^N  \frac {P_j}{1-P_j}  -  \frac{P_1}{1-P_1} ] + \ P_iP_j[1 + \sum_ {j=0}^N  \frac {P_i}{1-P_i}  - \frac{P_1}{1-P_1}]\ ^2  + \ P_i\ P_j [  \frac {1}{1-P_j}  +\frac {1}{1-P_i} ] $$
Need help in further procedure.
 A: Draw a picture.
This is a Venn diagram of the two events $\mathscr{i}$ (unit $i$ is included in a sample) and $\mathscr{j}$ (unit $j$ is included in a sample).  On it I have posted the inclusion probabilities, which I worked out by observing $\pi_i = \pi_{ij} + (\pi_i-\pi_{ij})$ and $\pi_j = \pi_{ij} + (\pi_j-\pi_{ij}).$

One axiom of probability tells us the chance of not including $i$ in the sample is $1$ minus the chance of including $i,$ since those chances must sum to $1.$  The same reasoning applies to $j,$ whence
$$\pi^{*}_i = 1-\pi_i\ \text{and}\ \pi^{*}_j = 1-\pi_j.$$
The chance that neither $i$ nor $j$ are included corresponds to all outcomes not enclosed in one of the circles.  By the same reasoning, that chance ($\pi^{*}_{ij}$) must be equal to $1$ minus the chance of being in one of the circles.  As the diagram shows, the latter is the sum of the three chances posted in the diagram (as asserted by another probability axiom: the chance of a disjoint union of events is the sum of the chances of the individual events).  Therefore
$$\begin{aligned}
\pi^{*}_{ij} &= 1 - \left[\left(\pi_1 - \pi_{ij}\right) + \pi_{ij} + \left(\pi_j - \pi_{ij}\right)\right] \\
&= (1-\pi_i)(1-\pi_j) - \pi_i\pi_j + \pi_{ij}\\
&= \pi^{*}_i\pi^{*}_j + \pi_{ij} - \pi_i\pi_j.
\end{aligned}$$
Subtracting $\pi^{*}_i\pi^{*}_j$ from both sides yields the desired result.
