We know that if

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


$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?


1 Answer 1


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.


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