the beta distribution

[logistic_guy]

here is the question

Derive the mean and variance of the beta distribution.


my attemb
i want to show the mean = $\displaystyle \mu = \frac{\alpha}{\alpha + \beta}$
i want to show the variance = $\displaystyle \sigma^2 = \frac{\alpha \beta}{(\alpha + \beta)^2(\alpha + \beta + 1)}$

the beta function $\displaystyle B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)}$
how this information help me driving the answer

 

[khansaheb]

here is the question

Derive the mean and variance of the beta distribution.


my attemb
i want to show the mean = $\displaystyle \mu = \frac{\alpha}{\alpha + \beta}$
i want to show the variance = $\displaystyle \sigma^2 = \frac{\alpha \beta}{(\alpha + \beta)^2(\alpha + \beta + 1)}$

the beta function $\displaystyle B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)}$
how this information help me driving the answer

Very "poor" question!!

what is α ? what the definition of α?

what is ß ? what the definition of ß?

 

[logistic_guy]

Very "poor" question!!

why?

what is α ? what the definition of α?

what is ß ? what the definition of ß?

$\displaystyle \alpha$ and $\displaystyle \beta$ isn't part of the question

 

[fresh_42]

why?


$\displaystyle \alpha$ and $\displaystyle \beta$ isn't part of the question

That's not true. The cumulative distribution function (CDF) and/or the density function is part of the question since it is necessary to compute mean and variance, and the parameters $\alpha,\beta$ are part of these functions. How do you define the beta-distribution?

Wikipedia has quite some information about it: https://en.wikipedia.org/wiki/Beta_distribution

 

[logistic_guy]

That's not true

before i answer your question, show me where in the question it talk about $\displaystyle \alpha$ and $\displaystyle \beta$

 

[fresh_42]

before i answer your question, show me where in the question it talk about $\displaystyle \alpha$ and $\displaystyle \beta$

You did not, that's the point. You used it but did not explain it.

 

[logistic_guy]

You did not, that's the point. You used it but did not explain it.

how did you know it if i don't explaint it?

 

[BeansNRice]

https://math.stackexchange.com/questions/497577/mean-and-variance-of-beta-distributions

 

[logistic_guy]

https://math.stackexchange.com/questions/497577/mean-and-variance-of-beta-distributions

thank

$\displaystyle \mu = \int_{0}^{1}x\frac{x^{\alpha - 1}(x - 1)^{\beta - 1}}{B(\alpha,\beta)}dx = \frac{\alpha}{\alpha + \beta}$

it don't explain how he did this step

 

[BeansNRice]

https://statproofbook.github.io/P/beta-mean.html

 

[logistic_guy]

https://statproofbook.github.io/P/beta-mean.html

i'm still lost

explain how it go from integration to mean $\displaystyle \mu$

 

[BeansNRice]

I don't know what to tell you. The integration is a bit complicated but what's shown there is straightforward.

If you're at the level where you've been tasked to find the mean and variance of the Beta distribution you should be able to follow that integration.

 

[mario99]

thank

$\displaystyle \mu = \int_{0}^{1}x\frac{x^{\alpha - 1}(x - 1)^{\beta - 1}}{B(\alpha,\beta)}dx = \frac{\alpha}{\alpha + \beta}$

it don't explain how he did this step

If you followed what professor fresh_42 has given you in post #4, you would have already solved this integral by now. Instead, you have involved in an unnecessary conversation.

How to solve this integral?
The first step is to pull out any constants and to leave inside the integral only the things that have the variable $x$. Do it, what do you get?

 

[BeansNRice]

Very "poor" question!!

what is α ? what the definition of α?

what is ß ? what the definition of ß?

$\alpha,~\beta$ are clearly the parameters of the beta distribution.

Further explanation shouldn't be necessary.

 

[logistic_guy]

If you followed what professor fresh_42 has given you in post #4, you would have already solved this integral by now. Instead, you have involved in an unnecessary conversation.

How to solve this integral?
The first step is to pull out any constants and to leave inside the integral only the things that have the variable $x$. Do it, what do you get?

thank

$\displaystyle \mu = \frac{1}{B(\alpha,\beta)}\int_{0}^{1}x\frac{x^{\alpha - 1}(x - 1)^{\beta - 1}}{1}dx$

$\displaystyle B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)}$ so it don't have $\displaystyle x$

 

[fresh_42]

$\alpha,~\beta$ are clearly the parameters of the beta distribution.

Further explanation shouldn't be necessary.

I thought the point was that the OP understands why there are such parameters and their role in the distribution. Otherwise, he could have just asked for the integrals.

 

[mario99]

thank

$\displaystyle \mu = \frac{1}{B(\alpha,\beta)}\int_{0}^{1}x\frac{x^{\alpha - 1}(x - 1)^{\beta - 1}}{1}dx$

$\displaystyle B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)}$ so it don't have $\displaystyle x$

Very good. And when you simplify it, you will get:

$\displaystyle \mu = \frac{1}{B(\alpha,\beta)}\int_{0}^{1}x^{\alpha}(x - 1)^{\beta - 1} \ dx$

I am sure that by mistake you wrote $(1 - x)^{\beta - 1}$ as $(x - 1)^{\beta - 1}$, so the correct integral is:

$\displaystyle \mu = \frac{1}{B(\alpha,\beta)}\int_{0}^{1}x^{\alpha}(1 - x)^{\beta - 1} \ dx$


In the OP, you have told us that:

$\displaystyle B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)}$

Instead of that, you should have told us that:

$\displaystyle B(\alpha,\beta) = \int_{0}^{1}x^{\alpha - 1}(1 - x)^{\beta - 1} \ dx = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)}$


Now compare between:

First integral:

$\displaystyle \int_{0}^{1}x^{\alpha}(1 - x)^{\beta - 1} \ dx$

And

Second integral:

$\displaystyle \int_{0}^{1}x^{\alpha - 1}(1 - x)^{\beta - 1} \ dx$

What do you notice? And what should you do to make the first integral to be a beta function? Think a little and I am sure that you will see the trick (the idea)!

 

[logistic_guy]

What do you notice?

i notice the difference $\displaystyle x^{\alpha}$ and $\displaystyle x^{\alpha - 1}$

And what should you do to make the first integral to be a beta function?

i think i need -1 to make it beta function. i think it's wrong to just add -1
is the trick $\displaystyle u$ subtution?

$\displaystyle u = x^{\alpha}$
$\displaystyle du = \alpha x^{\alpha - 1} dx$

$\displaystyle \int_{0}^{1}x^{\alpha}(1 - x)^{\beta - 1}dx = \int_{0}^{1}u(1 - \sqrt[\alpha]{u})^{\beta - 1}\frac{du}{\alpha x^{\alpha - 1}}$

it become more complicated i don't see the trick
i study integration long time ago i forgot all tricks

 

[mario99]

i notice the difference $\displaystyle x^{\alpha}$ and $\displaystyle x^{\alpha - 1}$


i think i need -1 to make it beta function. i think it's wrong to just add -1
is the trick $\displaystyle u$ subtution?

$\displaystyle u = x^{\alpha}$
$\displaystyle du = \alpha x^{\alpha - 1} dx$

$\displaystyle \int_{0}^{1}x^{\alpha}(1 - x)^{\beta - 1}dx = \int_{0}^{1}u(1 - \sqrt[\alpha]{u})^{\beta - 1}\frac{du}{\alpha x^{\alpha - 1}}$

it become more complicated i don't see the trick
i study integration long time ago i forgot all tricks

Nice try, but that was not the trick. The trick is:

$x^{\alpha} = x^{\alpha + 1 - 1}$

Let $k = \alpha + 1$

Then

$x^{\alpha} = x^{k - 1}$

And the integral will be:

$\displaystyle\int_{0}^{1} x^{k - 1}(1 - x)^{\beta - 1}\ dx$

This is just a beta function.

$\displaystyle\int_{0}^{1} x^{k - 1}(1 - x)^{\beta - 1}\ dx = B(k,\beta) = \frac{\Gamma(k)\Gamma(\beta)}{\Gamma(k + \beta)}$

Or

$\displaystyle B(k,\beta) = B(\alpha + 1,\beta) = \frac{\Gamma(\alpha + 1)\Gamma(\beta)}{\Gamma(\alpha + \beta + 1)}$

To simplify the gamma function, visit this thread.

the gamma distribution

here is the question Show that the rth moment about the origin of the gamma distribution is \mu'_r = \frac{\beta^r\Gamma(\alpha + r)}{\Gamma(\alpha)}. my attemb i think \alpha and \beta are the parameter of the gamma distribution first moment \mu'_1 = \frac{\beta^1\Gamma(\alpha +...

www.freemathhelp.com

Go to post #5 written by professor fresh_42.

 

[logistic_guy]

Nice try, but that was not the trick. The trick is:

$x^{\alpha} = x^{\alpha + 1 - 1}$

Let $k = \alpha + 1$

Then

$x^{\alpha} = x^{k - 1}$

And the integral will be:

$\displaystyle\int_{0}^{1} x^{k - 1}(1 - x)^{\beta - 1}\ dx$

This is just a beta function.

$\displaystyle\int_{0}^{1} x^{k - 1}(1 - x)^{\beta - 1}\ dx = B(k,\beta) = \frac{\Gamma(k)\Gamma(\beta)}{\Gamma(k + \beta)}$

Or

$\displaystyle B(k,\beta) = B(\alpha + 1,\beta) = \frac{\Gamma(\alpha + 1)\Gamma(\beta)}{\Gamma(\alpha + \beta + 1)}$

To simplify the gamma function, visit this thread.

the gamma distribution

here is the question Show that the rth moment about the origin of the gamma distribution is \mu'_r = \frac{\beta^r\Gamma(\alpha + r)}{\Gamma(\alpha)}. my attemb i think \alpha and \beta are the parameter of the gamma distribution first moment \mu'_1 = \frac{\beta^1\Gamma(\alpha +...

www.freemathhelp.com

Go to post #5 written by professor fresh_42.

thank mario99 very much

you explain it nicely

fresh_42 say $\displaystyle \Gamma(x + 1) = x\Gamma(x)$

so $\displaystyle \frac{\Gamma(\alpha + 1)\Gamma(\beta)}{\Gamma(\alpha + \beta + 1)} = \frac{\alpha\Gamma(\alpha)\Gamma(\beta)}{(\alpha + \beta)\Gamma(\alpha + \beta)}$

still $\displaystyle \frac{\alpha\Gamma(\alpha)\Gamma(\beta)}{(\alpha + \beta)\Gamma(\alpha + \beta)} \neq \frac{\alpha}{\alpha + \beta}$