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Given beta distribution as: $$ \mathcal{B}(x;a,b) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{a-1} (1-x)^{b-1} $$

I am trying to show: $$ \int_0^1 x^{a-1} (1-x)^{b-1}\,dx = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} $$

which can be considered as the normalizing constant.

Here is what I have done so far (using the hints at Bishop's). Write $\Gamma(a)\Gamma(b)$ and change variable $t=y+x$ fixing $x$:

\begin{align*} \newcommand{\ints}{\int_0^\infty\!\!\int_0^\infty} \Gamma(a)\Gamma(b) &= \int_0^\infty \exp(-x)x^{a-1} \, dx \int_0^\infty \exp(-y)y^{b-1} \, dy \\ &= \ints \exp(-x)x^{a-1} \exp(-y)y^{b-1} \, dx\,dy \\ &= \ints e^{-t} x^{a-1}(t-x)^{b-1} \, dt\,dx \end{align*}

Now if we make the variable change $x=\mu t$ \begin{align*} \ints e^{-t} x^{a-1}(t-x)^{b-1} \, dt\,dx &= \ints e^{-t} (\mu t)^{a-1} (t-\mu t)^{b-1} t \, dt\, d\mu \\ &= \ints e^{-t} t^{a+b-1} \mu^{a-1} (1-\mu)^{b-1} \, d\mu \, dt \end{align*}

We can split the integrals: \begin{align*} &= \underbrace{\int_0^\infty e^{-t} t^{a+b-1} \,dt}_{\Gamma(a+b)} \underbrace{\int_0^\infty \mu^{a-1} (1-\mu)^{b-1} \, d\mu}_{\text{limits are not from 0 to 1}} \end{align*}

What is wrong with the limits? Is it my mistake? Or can it be substituted to from $0$ to $1$?

PS. sorry for the big space between double integrals, I don't know the right way to do them.

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Use \int_a^b\!\!\int_c^d. –  Did Dec 5 '12 at 15:43

1 Answer 1

up vote 1 down vote accepted

The problem is in the last step here:

$$ \begin{align*} \newcommand{\ints}{\int_0^\infty\int_0^\infty} \Gamma(a)\Gamma(b) &= \int_0^\infty \exp(-x)x^{a-1} \, dx \int_0^\infty \exp(-y)y^{b-1} \, dy \\ &= \ints \exp(-x)x^{a-1} \exp(-y)y^{b-1} \, dx\,dy \\ &= \ints e^{-t} x^{a-1}(t-x)^{b-1} \, dt\,dx \end{align*} $$

As you made change of variables $t=x+y$, naturally $t \geqslant x$, thus the last integral should be $$ \int_0^\infty \exp(-t)\left( \int_0^t x^{a-1} (t-x)^{b-1} \mathrm{d}x\right) \mathrm{d}t \stackrel{x=\mu t}{=} \int_0^\infty \exp(-t) \left( t^{a+b-1} \int_0^t \mu^{a-1} (1-\mu)^{b-1} \mathrm{d}\mu\right) \mathrm{d}t = \\ \int_0^\infty t^{a+b-1}\mathrm{e}^{-t} \mathrm{d}t \cdot \int_0^1 \mu^{a-1} (1-\mu)^{b-1} \mathrm{d} \mu $$

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Thank you for the answer. Can you explain the limit change from (0,t) to (0,1) a little bit in detail? It has been years since I have taken the calculus course :) A link is fine too. –  nimcap Dec 5 '12 at 20:33
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With $x = \mu t$, $\mathrm{d} x = t \mathrm{d} \mu$, integration interval $ 0 < x < t$ translates into $0 < \mu t < t$ impliying $0<\mu<1$ for $t>0$. Thus, for $t>0$: $$\int_0^t x^{a-1} (t-x)^{b-1} \mathrm{d}x = \int_0^1 (\mu t)^{a-1} (t - \mu t)^{b-1} t \mathrm{d} \mu = t^{a+b-1} \int_0^1 \mu^{a-1} (1-\mu)^{b-1} \mathrm{d} \mu $$ –  Sasha Dec 5 '12 at 21:39
    
thank you very much! –  nimcap Dec 5 '12 at 21:48
    
@Sasha: hi. Are you around? –  Chris's sis Dec 8 '12 at 20:55
    
@Chris'ssister Hi. I am. –  Sasha Dec 8 '12 at 21:08

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