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Given a formula to calculate instantaneous probability of an event.
f(i) = 0.0222 * e ^ (-­i / 11.5)

For instance 0.0222 * e ^ (­-4 / 11.5) is the probability of the event occurring exactly during the fourth months given that it hasn’t happened before. Calculate the cumulative probability in 18 months, the probability of the event happens within 18 months.

i attempted it concluded here - 0.0222 * ( e ^ (­-1 / 11.5) + e ^ (­-2 / 11.5) + e ^ (­-3 / 11.5) + e ^ (­-4 / 11.5) +..... + e ^ (­-17 / 11.5) + e ^ (­-18 / 11.5) )

Is this right and what is the best and most efficient way to solve this.

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    $\begingroup$ Hi Amol! This seems like it's a textbook-style problem, so we ask that you add the self-study tag and read its wiki. $\endgroup$
    – Danica
    Commented Jan 28, 2016 at 16:56
  • $\begingroup$ Your question also seems to be somewhat unclear. Is there missing context? $\endgroup$
    – Glen_b
    Commented Jan 28, 2016 at 16:57
  • $\begingroup$ was asked this during an interview for data scientist role..question is complete. solving with code was also expected but first step is reaching to the correct expression and simplifying it. $\endgroup$ Commented Jan 28, 2016 at 17:01
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    $\begingroup$ You are asking how to sum a geometric series. When you do that, you will discover that this formula cannot possibly describe a set of probabilities because they will not sum to unity. $\endgroup$
    – whuber
    Commented Jan 28, 2016 at 18:15

1 Answer 1

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The cumulative distribution is simply the integral of the pdf

$F(i)=\int_{-\infty}^i f(k) dk$

in this case

$F(i)=\int_{-\infty}^i \frac{1}{11.5}e^{-\frac{k}{11.5}} dk$

$F(i)=-\frac{11.5}{11.5}e^{-\frac{i}{11.5}} - -\frac{11.5}{11.5}e^{-\frac{-infty}{11.5}} $

$F(i)=1-e^{-\frac{i}{11.5}} $

Oops, didn't realize this was discrete - which is why 0.0222 is used instead of 1/11.5. Give me 5 and I'll try again

I don't think that the function is a valid pmf. Doing a quick numerical example in MATLAB, I get a sum of 0.26666. Perhaps there is a 73.3% chance the event never occurs?

Assuming the pmf is meant as given, the cmf is just the sum (versus integral above - where I was thinking pdf).

$F(i)=\sum_{-\infty}^i f(k) dk$

in this case

$F(i)=\sum_{k=0}^i 0.0222*e^{-\frac{k}{11.5}} $

$F(i)=0.0222*\frac{1-e^{-\frac{i+1}{11.5}}}{1-e^{-\frac{1}{11.5}}} $

$F(i)=0.2666-0.2666*e^{-\frac{i+1}{11.5}} $

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