Sum of probabilities of all samples gives the total volume? I am working on a churn problem (binary classification - whether a customer will churn or not). Now using logistic regression, I get the probability whether the customer will churn or not.
Can I add up these probabilities (rows are distinct per user) to get the total volume of predicted churn? (expected value of the total number of users that will churn)?
How would I prove this statistically?
$$ E(X) = \sum_{x=1}^{n} x_i P(x_i)$$
$$ where  \ X:churn\ volume,  x_i = 1 \ (1 \ for \ every \ user), P(x_i):probability \ from \ the \ classification \ model $$

*

*Does this look right? Or what might be the right working?

*Does the equations change when there is a class imbalance.
Please correct me or point me in the right direction


You have $n$ customers, for each customer $i$ you have the value $c_i$ telling us whether the customer churned ($c_i=1$) or not ($c_i=0$), and you have the belonging probabilities $p_i := p(c_i=1)$ and $1-p_i = p(c_i = 0)$. As always with binary random variables, the expectation $E[c_i]$ is $p_i$:
$$
E[c_i] = 0\cdot (1-p_i) + 1\cdot p_i = p_i.
$$
Now, you want to compute the expectation of the sum $\sum_{i=1}^n c_i$. But the expectation operation is linear, thus:
$$
E[\sum_{i=1}^n c_i] = \sum_{i=1}^n E[c_i] = \sum_{i=1}^n p_i.
$$
Thus, in summary, you can indeed sum up all the probabilities to obtain the expected total amount of churning.
This result is independent of whether there is a class imbalance or not.
