I have a question.
Assume I carry out an experiment by Bernoulli process.
I repeat the tests until the number of successful outcomes
exceed the number of unsuccessful outcomes by m.
What will be the distribution function f = f(m) of the number of tests?
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Legotin
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@hardmath, sure, it's about Bernoulli process – Legotin May 12 '16 at 20:45
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@SangchulLee, the Pascal distribution it's "number (K) of successes in a sequence of independent and identically distributed Bernoulli trials before a specified (non-random) number (R) of failures occurs". But in my case R is random because depends on K. Or I'm wrong? – Legotin May 12 '16 at 20:52
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Oh, I must be severely sleep deprived to misread your problem. You are right. – Sangchul Lee May 12 '16 at 20:55
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2Now it seems to me that your question boils down to the distribution of some hitting time of a biased random walk, because the process $$S_n = [\text{number of successes in the first }n\text{ tests}] - [\text{number of fails in the first }n\text{ tests}]$$ can ne realized by a biased random walk. – Sangchul Lee May 12 '16 at 21:01
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@SangchulLee interesting idea. But what is the final formula for f(m)? – Legotin May 12 '16 at 21:08
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It will involve Catalan numbers, as in this answer. – Sangchul Lee May 12 '16 at 21:12