Calculating a rate with a single summation I have a $n$ events, each with some value $x$ and duration $t$. I can calculate the global rate as follows:
$$\text{rate} = \frac{\sum_{i=1}^nx_i}{\sum_{i=1}^nt_i}$$
Is it possible (or provably impossible) to calculate a global rate with a single summation over the events:
$$\text{rate} = \sum_{i=1}^nf(x_i, t_i)\text{ ?}$$
 A: $$\frac{\partial r}{\partial x_i}=\frac{1}{\sum_{i=1}^n t_i}$$
On the second definition take the derivative respect to $x_i$:
$$\frac{\partial r}{\partial x_i}=\frac{\partial f(x_i,t_i)}{\partial x_i}$$
Equating both for $i=1$:
$$\frac{1}{\sum_{i=1}^n t_i}=\frac{\partial f(x_1,t_1)}{\partial x_i}$$
Integrating respect to $x_1$:
$$\frac{x_1}{\sum_{i=1}^n t_i}+g(x_2,\cdots,x_n)=f(x_1,t_1)$$
Obviously (the derivatives are zero):
$$g(x_2,\cdots,x_n)=c$$
Therefore:
$$\frac{x_1}{\sum_{i=1}^n t_i}+c=f(x_1,t_1)$$
Clearly this give a contradiction (because the value of $t_1$ is conditionated by the other $t_i$)
A: The answer is that it is impossible. 
Let:$${\sum_{i=1}^n{x_i} \over \sum_{i=1}^n{t_i}}=\sum_{i=1}^n{f(x_i,t_i)}$$
We can write it as:$${X_n \over T_n}=F_n$$
Or:$${X_n}=F_n T_n$$
$${\partial {X_n} \over \partial {x_i}}={\partial {F_n} \over \partial {x_i}}T_n+F_n{\partial {T_n} \over \partial {x_i}}$$
$$1={\partial {f_i} \over \partial {x_i}}T_n+0$$
$$\left ({\partial {f_i} \over \partial {x_i}} \right )^{-1}={T_n}$$
Let:
$$u(x_i,t_i)=\left ({\partial {f_i} \over \partial {x_i}} \right )^{-1}$$
$${\partial \over \partial {t_j}}u(x_i,t_i)={\partial {T_n} \over \partial {t_j}}$$
This yields:
$$0=1$$
