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Math Help - Deriving expected value and variance of beta distribution

  1. #1
    Junior Member platinumpimp68plus1's Avatar
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    Deriving expected value and variance of beta distribution

    I know what they are, but I don't understand how to derive them from the probability mass function. Any help/hints would be much appreciated!
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    Quote Originally Posted by platinumpimp68plus1 View Post
    I know what they are, but I don't understand how to derive them from the probability mass function. Any help/hints would be much appreciated!
    Start by reading this thread: http://www.mathhelpforum.com/math-he...ion-62210.html
    Last edited by CaptainBlack; February 15th 2011 at 12:30 AM.
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  3. #3
    Junior Member platinumpimp68plus1's Avatar
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    All right thanks, I read through that and it shed some light. At least now I understand where the normalizing constant comes from! But I tried solving again using similar methods and I'm still stuck. The problem for me is that the solution in that thread is working from the gamma->beta direction but I have to solve (or at least I think?) from the opposite direction and I can't seem to wrap my head around it. I'm in the mind frame of E[X] = integral xf(x) and I don't know where I'm supposed to put the x or really how to proceed from there. If you could give me a push in the right direction (ie. where do I start?) I'd be very thankful.
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    Quote Originally Posted by platinumpimp68plus1 View Post
    All right thanks, I read through that and it shed some light. At least now I understand where the normalizing constant comes from! But I tried solving again using similar methods and I'm still stuck. The problem for me is that the solution in that thread is working from the gamma->beta direction but I have to solve (or at least I think?) from the opposite direction and I can't seem to wrap my head around it. I'm in the mind frame of E[X] = integral xf(x) and I don't know where I'm supposed to put the x or really how to proceed from there. If you could give me a push in the right direction (ie. where do I start?) I'd be very thankful.
    I will do \mu = E(X):

    E(X) = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \int_0^1 x \cdot x^{\alpha - 1} (1 - x)^{\beta - 1} \, dx


    = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \int_0^1 x^{\alpha} (1 - x)^{\beta - 1} \, dx


    = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \cdot \frac{\Gamma(\alpha + 1) \, \Gamma(\beta)}{\Gamma(\alpha + \beta + 1)} (using the proven result at the link I gave you)


    = \frac{\Gamma(\alpha + \beta)}{ \Gamma(\alpha + \beta + 1)} \cdot \frac{\Gamma(\alpha + 1)}{\Gamma(\alpha)}


    = \frac{\alpha}{\alpha + \beta} (using the well known property of the Gamma function).


    The calculation of E(X^2) and hence Var(X) = E(X^2) - [E(X)]^2 is left for you.
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    Dear Mr Fantastic.

    Would you please be so kind and give me a hint how you solved your last step in the proof.

    The one with the comment: "using the well known property of the Gamma function"

    Thanks a lot,
    Ziguri

    BTW: i already read through all the links you mentioned in this post.
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    Quote Originally Posted by ziguri View Post
    Dear Mr Fantastic.

    Would you please be so kind and give me a hint how you solved your last step in the proof.

    The one with the comment: "using the well known property of the Gamma function"

    Thanks a lot,
    Ziguri

    BTW: i already read through all the links you mentioned in this post.
    The well known property of the Gamma function I refer to is \Gamma(z + 1) = z \Gamma(z).
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    Quote Originally Posted by mr fantastic View Post
    I will do \mu = E(X):

    E(X) = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \int_0^1 x \cdot x^{\alpha - 1} (1 - x)^{\beta - 1} \, dx


    = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \int_0^1 x^{\alpha} (1 - x)^{\beta - 1} \, dx


    = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \cdot \frac{\Gamma(\alpha + 1) \, \Gamma(\beta)}{\Gamma(\alpha + \beta + 1)} (using the proven result at the link I gave you)


    = \frac{\Gamma(\alpha + \beta)}{ \Gamma(\alpha + \beta + 1)} \cdot \frac{\Gamma(\alpha + 1)}{\Gamma(\alpha)}


    = \frac{\alpha}{\alpha + \beta} (using the well known property of the Gamma function).


    The calculation of E(X^2) and hence Var(X) = E(X^2) - [E(X)]^2 is left for you.
    --Can you derive this same thing with the generalized beta distribution (that has 4 parameters: alpha, beta; a, b; where a and b are the values specifying the range over which the distribution lies)
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  8. #8
    MHF Contributor matheagle's Avatar
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    Instead try the k^{th} moment....then let k=1,2...

    Quote Originally Posted by mr fantastic View Post
    I will do \mu = E(X):

    E(X^k) = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \int_0^1 x^k \cdot x^{\alpha - 1} (1 - x)^{\beta - 1} \, dx


    = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \int_0^1 x^{\alpha+k-1} (1 - x)^{\beta - 1} \, dx


    = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \cdot \frac{\Gamma(\alpha + k) \, \Gamma(\beta)}{\Gamma(\alpha + \beta + k)}


    = \frac{\Gamma(\alpha + \beta)}{ \Gamma(\alpha + \beta + k)} \cdot \frac{\Gamma(\alpha + k)}{\Gamma(\alpha)}
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    Quote Originally Posted by matheagle View Post
    Instead try the k^{th} moment....then let k=1,2...
    --I don't think that is what I am searching for. Let me try to explain the problem more clearly...

    standard beta distribution is defined from 0 to 1 and has the pdf_1 =

    f(x;a,b) = [1 / B(a,b)] * x^(a - 1) * (1 - x)^(b - 1)

    -where 'a' and 'b' are the alpha and beta parameters respectively

    but the generalized beta function has the pdf_2=

    f(x;a,b,c,d) = [1 / B(a,b)] * [1 / (d - c)^(a + b - 1)] * (x - c)^(a - 1) * (d - x)^(b - 1)

    -where 'c' and 'd' are the min and max scale parameters respectively. I believe they are supposed to scale the original distribution (having 0 to 1 range) to the specified c to d range distribution.

    The question I have is in two parts...

    1) How is the pdf_2 derived?

    2) How to derive the expected value using pdf_2?
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    Quote Originally Posted by mr fantastic View Post
    I will do \mu = E(X):

    E(X) = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \int_0^1 x \cdot x^{\alpha - 1} (1 - x)^{\beta - 1} \, dx


    = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \int_0^1 x^{\alpha} (1 - x)^{\beta - 1} \, dx


    = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \cdot \frac{\Gamma(\alpha + 1) \, \Gamma(\beta)}{\Gamma(\alpha + \beta + 1)} (using the proven result at the link I gave you)


    = \frac{\Gamma(\alpha + \beta)}{ \Gamma(\alpha + \beta + 1)} \cdot \frac{\Gamma(\alpha + 1)}{\Gamma(\alpha)}


    = \frac{\alpha}{\alpha + \beta} (using the well known property of the Gamma function).


    The calculation of E(X^2) and hence Var(X) = E(X^2) - [E(X)]^2 is left for you.

    The link you gave is broken , i cannot open it. so i could not understand how did you derive from this integral of beta function
    = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \int_0^1 x^{\alpha} (1 - x)^{\beta - 1} \, dx

    this gamma function result

    = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \cdot \frac{\Gamma(\alpha + 1) \, \Gamma(\beta)}{\Gamma(\alpha + \beta + 1)} (using the proven result at the link I gave you)
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  11. #11
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    Quote Originally Posted by uniquebatgirl View Post
    The link you gave is broken , i cannot open it. so i could not understand how did you derive from this integral of beta function
    = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \int_0^1 x^{\alpha} (1 - x)^{\beta - 1} \, dx

    this gamma function result

    = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \cdot \frac{\Gamma(\alpha + 1) \, \Gamma(\beta)}{\Gamma(\alpha + \beta + 1)} (using the proven result at the link I gave you)
    Use repeated integration by parts (do it \beta - 1 times):

    \displaystyle u = (1 - x)^{\beta - 1} \Rightarrow du = -(\beta - 1) (1 - x)^{\beta - 2}

    \displaystyle dv = x^{\alpha} \Rightarrow v = \frac{1}{\alpha + 1} x^{\alpha + 1}

    etc.

    until you end up integrating \displaystyle x^{\alpha + \beta - 1} (with appropriate factors out the front). It is a standard technique found in many subject-appropriate textbooks. And I'm sure a Google search will also turn up a proof.
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  12. #12
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    Quote Originally Posted by uniquebatgirl View Post
    The link you gave is broken , i cannot open it.
    The links are have now been fixed

    CB
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