How to show $I_p(a,b) = \sum_{j=a}^{a+b-1}{a+b-1 \choose j} p^j(1-p)^{a+b-1-j}$

Question

Show that $$I_p(a,b) = \frac{1}{B(a,b)}\int_0^p u^{a-1}(1-u)^{b-1}~du\\= \sum_{j=a}^{a+b-1}{a+b-1 \choose j} p^j(1-p)^{a+b-1-j}$$ when $a,b$ are positive integers.

I have no idea how to proceed. Please help.

Answer 1

Hint: Use $\frac{1}{B(a,b)}=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}=\frac{(a+b-1)!}{(a-1)!(b-1)!}$ and integration by parts to evaluate the integral:

$$I=\int_{0}^{1}u^{a-1}(1-u)^{b-1} = \left.\frac{1}{a}u^a(1-u)^{b-1}\right|_{0}^{p}+\frac{b-1}{a}\int_{0}^{p}u^{a}(1-u)^{b-2}\,du, $$ $$ I= \frac{1}{a}p^a(1-p)^{b-1}+\frac{b-1}{a}\int_{0}^{p}u^{a}(1-u)^{b-2}\,du, $$ $$ I = \frac{1}{a}p^a(1-p)^{b-1} + \frac{b-1}{a(a+1)}p^{a+1}(1-p)^{b-2} +\frac{(b-1)(b-2)}{a(a+1)}\int_{0}^{p}u^{a+2}(1-u)^{b-3}\,du = \ldots $$

Answer 2

Hint:

1) Convince yourself, from its integral representation, that $$ I_{p}(a, b){}={}I_{p}(a, b-1){}+{}\dfrac{p^a (1-p)^{b-1}}{(b-1)B(a,b-1)}\,. $$

2) $$ I_{p}(a, b){}={}I_{p}(a+1, b){}+{}\dfrac{p^a (1-p)^{b}}{(a)B(a,b)}\,. $$

3) $B(a,b){}={}\dfrac{(a-1)!(b-1)!}{(a+b-1)!}\,$.

4) Show the result recursively!

Answer 3

Hint: Assume $a,b > 0$. First, Show that $$ I_p(a,b)= \int_0^p B(a,b)^{-1} t^{a-1}(1-t)^{b-1} dt = {a+b-1 \choose a}p^a(1-p)^{b-1} + I_p(a+1,b-1) $$ You can do this many ways and one way is by denoting $$ f:=(1-t)^b-1 \qquad\qquad g:=t^{a}/a $$ and apply integration by parts with the following identity $$ B(a,b)^{-1} = \frac{a}{b-1}B(a+1,b-1)^{-1} $$

This should be straightforward to work out and just solve it recursively with the final summation term $$ I_p(a+b-1,1) = p^{a+b-1} $$ and you are done.

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\,{\rm I}_{p}\pars{a,b}={1 \over \,{\rm B}\pars{a,b}} \int_{0}^{p}u^{a - 1}\ \pars{1 - u}^{b - 1}\,\dd u =\sum_{j\ =\ a}^{a + b - 1} {a + b - 1 \choose j}p^{j}\pars{1 - p}^{a + b - 1 -j}\,\,\,:\ {\large ?}}$.

We'll perform a 'direct evaluation' of the integral. That is accomplished with a suitable change of variable: $\ds{u\ \mapsto\ p - u}$:

\begin{align}&\int_{0}^{p}u^{a - 1}\ \pars{1 - u}^{b - 1}\,\dd u =\int_{0}^{p}\pars{p - u}^{a - 1}\ \pars{1 - p + u}^{b - 1}\,\dd u \\[5mm]&=\int_{0}^{p}\sum_{k\ =\ 0}^{a - 1}{a - 1 \choose k} p^{a - 1 - k}\,\,\pars{-1}^{k}\,u^{k}\ \sum_{j\ =\ 0}^{b - 1}{b - 1 \choose j} \pars{1 - p}^{b - 1 - j}\,\,\,\,u^{j}\,\,\,\dd u \\[5mm]&=\sum_{k\ =\ 0}^{a - 1}\ \sum_{j\ =\ 0}^{b - 1} {a - 1 \choose k}{b - 1 \choose j}\pars{-1}^{k}\,p^{a - 1 - k}\,\, \pars{1 - p}^{b - 1 - j}\,\,\,{p^{k + j + 1} \over k + j + 1} \\[5mm]&=\sum_{k\ =\ 0}^{a - 1}\ \sum_{j\ =\ 0}^{b - 1} {a - 1 \choose k}{b - 1 \choose j} {\pars{-1}^{k}\, \over k + j + 1}\,p^{a + j}\,\,\pars{1 - p}^{b - 1 - j} \\[5mm]&=\sum_{j\ =\ a}^{a + b - 1}\ \sum_{k\ =\ 0}^{a - 1} {a - 1 \choose k}{b - 1 \choose j - a} {\pars{-1}^{k} \over k + j - a + 1}\,p^{j}\,\,\pars{1 - p}^{a + b - 1 - j} \\[5mm]&=\sum_{j\ =\ a}^{a + b - 1}\bracks{{b - 1 \choose j - a} \dsc{\sum_{k\ =\ 0}^{a - 1} {a - 1 \choose k}{\pars{-1}^{k} \over k + j - a + 1}}} p^{j}\,\,\pars{1 - p}^{a + b - 1 - j}\tag{1} \end{align}

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\begin{align} &\dsc{\sum_{k\ =\ 0}^{a - 1} {a - 1 \choose k}{\pars{-1}^{k} \over k + j - a + 1}} =\sum_{k\ =\ 0}^{a - 1}{a - 1 \choose k}\pars{-1}^{k} \int_{0}^{1}t^{k + j - a}\,\,\,\dd t \\[5mm]&=\int_{0}^{1}t^{j - a} \sum_{k\ =\ 0}^{a - 1} {a - 1 \choose k}\pars{-t}^{k}\,\,\dd t =\int_{0}^{1}t^{j - a}\,\,\pars{1 - t}^{a - 1}\,\,\dd t =\,{\rm B}\pars{j -a + 1,a} \\[5mm]&={\Gamma\pars{j - a + 1}\Gamma\pars{a} \over \Gamma\pars{j + 1}} ={\pars{j - a}!\,\pars{a - 1}! \over j!}\tag{2} \end{align}

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With $\pars{1}$ and $\pars{2}$: \begin{align}&{b - 1 \choose j - a} \dsc{\sum_{k\ =\ 0}^{a - 1} {a - 1 \choose k}{\pars{-1}^{k} \over k + j - a + 1}} ={\pars{b - 1}! \over \pars{j - a}!\pars{b - 1 - j + a}!}\, {\pars{j - a}!\,\pars{a - 1}! \over j!} \\[5mm]&={\pars{a + b - 1}! \over j!\pars{a + b - 1 - j}!}\, {\pars{a - 1}!\pars{b - 1}! \over \pars{a + b - 1}!} ={a + b - 1 \choose j}\,{\Gamma\pars{a}\Gamma\pars{b} \over \Gamma\pars{a + b}} \\[5mm]&={a + b - 1 \choose j}\,\,{\rm B}\pars{a,b} \end{align}

Replace this result in $\pars{1}$ to get the final result.