Given $n = n_0 + n_1 + \cdots+ n_{m-1}$ and the average value of $n_{j}$ is $E[n_{j}] = n/m$. (Here $E$ is for expectation in context of probability). Here $n$ is not random variable, but $n_{j}$ is random variable.
Does that mean $E[\log(n_{j})] = \log(n)/m$ ? If no, can we find a function $f$ such that $E[\log(n_{j})] = f(n,m)$ ?
PS This is in context of analysis of hashing with chaining with Simple uniform hashing assumption i.e. any given element is equally likely to hash into any of the $m$ slots, independently of where any other element has hashed to. Also $\alpha = n/m$ is called load factor.