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I have a forecasting worksheet which describes a set (worksheet) of independent events, all of which have a likelihood of happening given as a probability (e.g. 0.7). Every event also has a yield value associated with it. There are about 30 items in the table.

The table looks like this:

Name    P     Yield-value
Event1  0.3   120
Event2  0.7   70
Event3  0.1   310
Event4  0.9   40

The table is read so that Event1 has 30% chance of happening, which would yield a value of 120. So the event either happens or not, but if it does, it yields a value 120.

My question is how can I know the total yield value of the table, which has a likelihood of a given probability? For example, at 50% probability, what is the total yield of the event set?

I did solve the problem using a Monte Carlo simulation, in where I did Bernoulli trials over the set using a lot (millions) of different scenarios, however I'm interested to know if there is also an analytic approach.

Please share your thoughts :)

Andres.

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  • $\begingroup$ The expected yield is just the sum of p(event i) * yield (event i). Sounds like you want to know something like: in what range is the yield with some "confidence level", yes ? $\endgroup$ Mar 29, 2014 at 8:52
  • $\begingroup$ Yes, exactly, because p(event i) * yield (event i) is not discrete.. If I had Event1 with P=0.3 and Yield=120, then then I would just do P*yield, then it would mean that Event1's yield is linearly dependent of P (which means it occurs always if P > 0, though with less yield, but that is not correct at all). $\endgroup$
    – user315648
    Mar 29, 2014 at 9:37

1 Answer 1

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For each $i = 1, 2, \ldots, n$, let $X_i \sim {\rm Bernoulli}(p_i)$ be a Bernoulli random variable with $\Pr[X_i = 1] = p_i$. Then the yield associated with observing $X_i$ is $Y_i = c_i X_i$, where $c_1, c_2, \ldots, c_n$ are fixed constants. That is to say, we assume the yield is zero if the $i^{\rm th}$ event does not occur.

Then the total yield is simply $S = \sum_{i=1}^n Y_i$ and its expected value is $${\rm E}[S] = \sum_{i=1}^n c_i p_i.$$ The variance, if the individual events are independent, is $${\rm Var}[S] = \sum_{i=1}^n c_i^2 {\rm Var}[X_i] = \sum_{i=1}^n c_i^2 p_i(1-p_i).$$

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  • $\begingroup$ If I do this with Bernoulli random variable (so that yield is 0 when event does not occur), then is there any other way to calculate this other that doing a Monte Carlo simulation? Isn't there a solution that would do this with combinations or something like that? $\endgroup$
    – user315648
    Mar 29, 2014 at 9:48

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