6
$\begingroup$

It came up when finding a constant such that the integral is equal to 1 and thus behaves like a pdf. I used the parts method but have made an error, just curious how others might approach the problem.

$\endgroup$
3
  • 7
    $\begingroup$ I'd probably either use integration by parts directly or take $u=1-x$ so that I could expand a binomial to the third power instead of the sixth power. $\endgroup$
    – Ian
    Dec 31 '14 at 21:00
  • 2
    $\begingroup$ As a side note this is a Beta Distribution with $cf(x),\alpha=4,\beta=7$ where $c=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}$. $\endgroup$
    – Chris
    Dec 31 '14 at 21:05
  • 2
    $\begingroup$ See beta function. $\endgroup$
    – Lucian
    Dec 31 '14 at 21:16
19
$\begingroup$

The beta function is the best idea.

$$\beta(a, b) = \int_{0}^{1} x^{a-1}(1-x)^{b-1} dx$$

For here $I$, let $a = 4, b = 7$

Using:

$$\beta(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$

$$\beta(4, 7) = \frac{\Gamma(4)\Gamma(7)}{\Gamma(11)} = \frac{3!6!}{10!}$$

$$= \frac{3\cdot2\cdot1}{10\cdot9\cdot8\cdot7} = \frac{1}{840}$$

$\endgroup$
2
  • $\begingroup$ Nice, was just working on this answer after thinking about it :) $\endgroup$
    – Chris
    Dec 31 '14 at 21:22
  • $\begingroup$ @Chris, thanks =) $\endgroup$
    – Amad27
    Dec 31 '14 at 21:23
11
$\begingroup$

This way might be faster, but it ultimately depends on your personal preference.

Let $u = 1-x$. Then the integral becomes:

$$\int (u-1)^3u^6 \ \text{d}u$$

And that expansion is much easier to work with than the original!

$\endgroup$
5
  • $\begingroup$ Be careful with the sign of the definite integral if you do this method though. $\endgroup$
    – GFauxPas
    Dec 31 '14 at 21:07
  • $\begingroup$ Indeed @GFauxPas. We picked up a negative since $du = -dx$, which I incorporated into the binomial. $\endgroup$
    – Kaj Hansen
    Dec 31 '14 at 21:10
  • $\begingroup$ But also, the transformation will have the bounds be from $1$ to $0$ rather than from $0$ to $1$. $\endgroup$
    – GFauxPas
    Dec 31 '14 at 21:16
  • 1
    $\begingroup$ Oh sure. I usually do the integral as is, then at the end sub back in whatever $u$ was and evaluate at the original bounds. Just a weird habit of mine. $\endgroup$
    – Kaj Hansen
    Dec 31 '14 at 21:19
  • $\begingroup$ I also tend to sub back and evaluate the original bounds. $\endgroup$
    – Chris
    Dec 31 '14 at 21:31
4
$\begingroup$

By symmetry,

\begin{align} \int_0^1 x^3(1-x)^6\,\mathrm{d}x &= -\int_1^0x^6(1-x)^3\mathrm{d}x\\ &=-\int_0^1 (x-1)^3x^6\,\mathrm{d}x \\ &=-\int_0^1 x^9-3x^8+3x^7-x^6 \,\mathrm{d}x\\ &=-\left(\frac{1}{10}-\frac{1}{3}+\frac{3}{8}-\frac{1}{7}\right)+0\\\\ &=-\frac{84-280+315-120}{840}\\ &=\boxed{\displaystyle\frac{1}{840}} \end{align}

$\endgroup$
2
  • $\begingroup$ Check your signs; you're integrating something positive. Also this is the solution I just posted... $\endgroup$
    – pre-kidney
    Dec 31 '14 at 21:11
  • 1
    $\begingroup$ @pre-kidney Thanks. I didn't see yours when I posted mine (look at the differences in time - I don't think anyone can type that fast). Also it's the solution Kaj posted 15 minutes ago :P $\endgroup$ Dec 31 '14 at 21:14
2
$\begingroup$

Along the lines of Kaj, I think the fastest is to just expand: $$ \int x^3(1-x)^6 = \int (1-u)^3u^6=\int u^6-3u^7+3u^8-u^9=\frac{1}{7}-\frac{3}{8}+\frac{3}{9}-\frac{1}{10}. $$

$\endgroup$
2
$\begingroup$

If we realize the integral is part of the Beta Distribution: $$\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1},0<x<1$$ We find $\alpha=4,\beta=7$. We also know the integral of any pdf equals $1$ and thus this our coefficient must be the inverse of the result.

$\endgroup$
0
$\begingroup$

using that $$x^3(1-x)^6=x^9-6 x^8+15 x^7-20 x^6+15 x^5-6 x^4+x^3$$ we can calculate the integral without the beta function, we get $$\int_{0}^{1} x^9-6 x^8+15 x^7-20 x^6+15 x^5-6 x^4+x^3dx=\frac{1}{840}$$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.