Consider $n$ independent trials, each of which results in any of the outcomes $i$,$\ i=1,2,3$, with respective probabilities $p_1,p_2,p_3,\ p_1+p_2+p_3=1$. Let $N_i$ denote the number of trials that result in outcome $i$. How do I show that $Cov(N_1,N_2)=-np_1p_2$? Why is it intuitive that the covariance is negative?
Currently, I have the following.
For $i=1,...,n$ let
$X_i = \begin{cases} 1 & \text{if trial} \ i \text{ results in outcome 1} \\ 0 & \text{if trial } i \text{ does not result in outcome 1}\end{cases}$
Similarly, for $j=1,...,n$, let
$Y_j = \begin{cases} 1 & \text{if trial} \ j \text{ results in outcome 2} \\ 0 & \text{if trial } j \text{ does not result in outcome 2}\end{cases}$
How do I argue the following?
$N_1 = \sum_{i=1}^{n}X_i,\ N_2=\sum_{j=1}^{n}Y_j$
How should I proceed using the properties of covariance from this point?