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In order to make some predictions for my work I have modeled a process using a binomial distribution, but in my case every single experiment must be a success and I am just changing the probability that each individual trial is a success. Effectively what I am looking at is:

$$ P({\rm success}) = p^m $$

Where $m$ is the number of trials I need to succeed at. If $m$ and $p$ are known with certainty is there any way to estimate the variance of $P({\rm success})$, or am I fundamentally misunderstanding something?

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  • $\begingroup$ If $m$ and $p$ are known with certainty, $p^m$ is a constant and thus its variance is $0$. I suppose you are looking for something else? $\endgroup$ Commented Jul 2, 2014 at 18:04
  • $\begingroup$ @JuhoKokkala. Success is Bernoulli with success probability $p^m$. So the variance of having a success is $p^m (1-p^m)$. $\endgroup$
    – Michael M
    Commented Jul 2, 2014 at 20:22
  • $\begingroup$ @MichaelMayer Correct, but the question as written asks about 'the variance of $P(\textrm{Success})$', not the variance of the Bernoulli random variale Success. Furthermore, the title is about uncertainty in a probability estimate, so to me it is not clear what this question is about. $\endgroup$ Commented Jul 2, 2014 at 20:29
  • $\begingroup$ @JuhoKokkala. You are right, I was guessing. $\endgroup$
    – Michael M
    Commented Jul 2, 2014 at 20:46
  • $\begingroup$ Do you mean the variance of the outcome (where overall success=1 or 0), or as your text says the variance of the probability? $\endgroup$
    – Glen_b
    Commented Jul 2, 2014 at 23:15

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If $m$ and $p$ are known with certainty, and the Bernoulli model you used is also a given (basically, adding independence), then the probability of succeeding every trial is $p^m$ and that probability of overall success is certain (variance is 0).

Do you instead mean something else? If you want the variance of the outcome, that's different (it's like tossing a fair coin - given you somehow know it to be fair, the probability of a head might be known with certainty, but the outcome is still variable).

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  • $\begingroup$ Yes, I was mistaken in asking the question. The variance I was looking for was indeed zero. I was overcomplicating things! $\endgroup$
    – mjnichol
    Commented Jul 3, 2014 at 21:02

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