Maximizing the probability of Bernoulli Sums Suppose I have $X_1,\cdots X_n$ with $X_i \sim \text{Bernoulli}(p_i)$, here  we only assume $X_1,\cdots X_n$ are independent but not neccessarily identical so there are n degree of freedom.
Now,  I want to show that the probability that
$$\mathbb{P}(\sum_{i=1}^n X_i = m)$$ is maximized at $p_i = \frac{m}{n}$. How do I do that? It seems intuitive proving it is hard.

Your question does not make sense. Consider these changes to you exercise and tell me if it can be what you are looking for:


*

*the $n$ Bernulli's iid rv $X_i$ 

*$$Y= \sum_i X_i \sim Bin(n;p)$$
So Y is again a rv taking values in $y=\{0;1;2;...;n\}$. Let's suppose that the result of the sum is a fixed $0 \leq m \leq n$ 
The goal is to maximize the probability
$$\mathbb{P}[Y=m]=\binom{n}{m}p^m(1-p)^{n-m}$$
as we have to maximize the function with respect to $p$ we can take into consideration only
$$\mathbb{P}[Y=m]=p^m(1-p)^{n-m}$$
Let's take the log (log is a monotone transformation so it doesn't change the argmax)
$$log P=mlog p+(n-m)log(1-p)$$
Let's derive with respect to $p$
$$\frac{\partial}{\partial p}log P=\frac{m}{p}-\frac{n-m}{1-p}$$
setting it =0 and solving with respect to $p$ you find
$p=\frac{m}{n}$
It is obviuosly a maximum, but you can check it with calculating the second derivative
