Let $ X \sim \text{Bin}(n,p) $ and $ Y \sim \text{Beta}(r,n-r+1) $. Show , without integration by parts, that $P(X \ge r ) = P (Y \le p)$.
From which point of view I answer this question.
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A.D
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@Xi'an: The question says show the equality without integration by parts. Then how can I show? I don't understand. – A.D Jan 08 '15 at 16:47
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can you write down any of the two probabilities? – Xi'an Jan 08 '15 at 16:48
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@Xi'an Yes. $P[X \ge r] = \sum_{x=r}^n {n \choose x} p ^x (1-p)^{n-x}$ – A.D Jan 08 '15 at 16:51
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Try to do the same with $$\frac{(n+r)!}{(r-1)!(n-r)!}\int_0^p y^{r-1}(1-y)^{n-r} \text{d}y$$ – Xi'an Jan 08 '15 at 17:50
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1@Xi'an I'm curious where you would go with that integral, given that you are enjoined from integrating by parts. (That injunction probably should be broadly interpreted as an extremely strong hint that no integration at all will be necessary.) Did you have some particularly simple or elegant manipulation in mind? – whuber Jan 08 '15 at 18:58
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@whuber: yes, I would simply expand the $(1-y)^{n-r}$ using the binomial formula$$(1-y)^{n-r}=\sum_{k=0}^{n-r} (-1)^k {n-r\choose k} y^k$$ and from there the integral is trivial. – Xi'an Jan 08 '15 at 19:34