# Why is multinomial variance different from covariance between the same two random variables?

We know that if

$\big(X_1,X_2...X_k) \sim multinomial(n;p_1,p_2...p_k)$

then

$X_i \sim bin(n;p_i)$

Then, $var(X_i) = np_i(1-p_i)$.

But we have $cov(X_i,X_j) = -np_ip_j$.

So doesnt that imply $var(X_i) = cov(X_i,X_i) = -np_i^2$?

(Which is basically impossible because of the definition of variance.) Can someone explain to me what exactly am I not being able to see?

The covariance formula $$\text{Cov}(X_i, X_j) = -np_ip_j$$ applies IF AND ONLY IF $$i \neq j$$. It does NOT apply if $$i = j$$. If you work out the details, notice that $$i$$ and $$j$$ have to be distinct in order to prove the covariance formula.