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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? $$ E(X) = \sum_{x=1}^{n} x_i P(x_i)$$ $$ where \ X:churn\ volume, x_i = 1 \ (1 \ for \ every \ user), P(x_i):probability \ from \ the \ classification \ model $$

  1. Does this look right? Or what might be the right working?
  2. Does the equations change when there is a class imbalance. Please correct me or point me in the right direction
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1 Answer 1

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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.

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  • $\begingroup$ this makes a lot more sense. thanks for the detailed explanation. $\endgroup$
    – maamli
    Jun 10 at 23:00

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