the beta distribution

logistic_guy

Full Member
Joined
Apr 17, 2024
Messages
304
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
 
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?
 
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 [imath] \alpha,\beta [/imath] 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
 
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.
 
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 [imath]x[/imath]. Do it, what do you get?
 
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 [imath]x[/imath]. 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 \(\displaystyle x\)
 
[imath]\alpha,~\beta[/imath] 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.
 
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:

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

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

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


In the OP, you have told us that:

[imath]\displaystyle B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)}[/imath]

Instead of that, you should have told us that:

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


Now compare between:

First integral:

[imath]\displaystyle \int_{0}^{1}x^{\alpha}(1 - x)^{\beta - 1} \ dx[/imath]

And

Second integral:

[imath]\displaystyle \int_{0}^{1}x^{\alpha - 1}(1 - x)^{\beta - 1} \ dx[/imath]

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)!

💡
 
Top