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I'm using R and would like to calculate the probability of an event happening. I used the ecdf function on daily revenue (month-to-date).

I have a target and a daily number necessary to hit target and I want to calculate based upon historical performance, the probability of hitting target.

Say my g= ecdf(x). Would g(daily target)-g(median performance) give me the probability that I am looking for?

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Define $f(x)$ as the ecdf of $x,$ that is, the probability of observing a value less than or equal to $x$. What you want is probably $1-f(\text{target})$, the probability of observing a value larger than the target. Note that this makes some strong assumptions, like that each day is an iid random variable. If there's some kind of seasonal or day-of-week effect, your model won't be a great representation of reality.

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  • $\begingroup$ Good point on the seasonality, but atm I don't want to make it too complicated. So in my example, if I do 1- g(daily target)-g(median performance) I will get the number I want? The reason why I am doing this is to take into account where we are now in relation to where we need to go in order to hit target. What is the difference between what I am trying to calculate and what you are recommending? Thanks! $\endgroup$ – Hidden Markov Model Aug 18 '15 at 20:18
  • $\begingroup$ @HiddenMarkovModel It's not clear to me at all why you'd subtract f(median). For example, some values of your target and your median, this will give you a negative number, not a probability! I interpreted your question as "Before tomorrow happens, what's the probability of exceeding the target tomorrow?"; your question in this comment appears to be "It's noon right now and we have D dollars, whats' the probability we have $target by the end of the day?" $\endgroup$ – Reinstate Monica Aug 18 '15 at 21:12
  • $\begingroup$ Good point, actually let me clarify. The number needed to hit target is x dollars per day for the next 10 days. Thus, wouldn't the probability be [1-F(x)]^10 since each day is independent (assuming Independence)? So if 1-F(x) is 0.12, then this would mean 0.12^10? This probability seems reasonable I guess since [1-F(x)]^10 is essentially saying what is the probability of some rare event occurring every day for the next 10 days. One issue I have with my idea is that say we set 1-F(X)=0.99, then 0.99^10 is 0.90, but if say 1-F(x)=0.9 then 0.9^10 = 0.34. This seems weird. What am I doing wrong? $\endgroup$ – Hidden Markov Model Aug 19 '15 at 10:50

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