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December 1st, 2014, 07:38 AM  #1 
Member Joined: Nov 2013 Posts: 47 Thanks: 0  Proof beta distribution = exponential when B =0
Proof beta with b = 1 is exponential distribution $\displaystyle \frac{x^{\alpha1}(1x)^{\beta1}}{B(\alpha,\beta)}$ suppose $\displaystyle \beta=1$ then we get $\displaystyle \alpha x^{\alpha1}$ If we take $\displaystyle ln(x)$ we get $\displaystyle \alpha(ln(x))^{\alpha1}$ and the last therm should be equal to $\displaystyle \alpha e ^{\alpha x}$ But how do i get there ? 
December 1st, 2014, 07:36 PM  #2 
Senior Member Joined: Jan 2012 From: Erewhon Posts: 245 Thanks: 112  
December 2nd, 2014, 03:35 PM  #3 
Member Joined: Nov 2013 Posts: 47 Thanks: 0  
December 4th, 2014, 05:54 AM  #4 
Senior Member Joined: Jan 2012 From: Erewhon Posts: 245 Thanks: 112 
So we need to show that if $X \sim \mathrm{Beta}(\alpha,1)$ then $W=\ln(X)$ is exponentially distributed with parameter $\alpha$. The easiest way of doing this, at least for me, is to work with the cumulative distribution functions. For $\mathrm{Beta}(\alpha,1)$ the cdf is $F(x)=x^{\alpha}$ which you can easilly derive if you do not know it from the definition of the density of the beta distribution. Then: $$ P(W<w)=P(\ln(X)<w)=P(\ln(X)>w)=1P(\ln(X)<w) \\ \phantom{P(W<w)} =1P(X<e^{w}) \\ \phantom{P(W<w)} =1F(e^{w}) \\ \phantom{P(W<w)} =1(e^{w})^{\alpha}=1e^{\alpha w} $$ But $1e^{\alpha w}$ is the cdf of the exponential distribution with parameter $\alpha$. CB Last edited by CaptainBlack; December 4th, 2014 at 06:05 AM. 
December 6th, 2014, 05:08 AM  #5 
Member Joined: Nov 2013 Posts: 47 Thanks: 0 
So we get F(x) = x^{a} $\displaystyle beta distribution (\alpha, 1) = Kumaraswamy distribution with \beta = 1$ and we set beta to 1 we get $\displaystyle x^{\alpha}$ 
December 6th, 2014, 05:25 AM  #6 
Member Joined: Nov 2013 Posts: 47 Thanks: 0 
But how do we get it in te method i proposed ?

December 6th, 2014, 06:26 AM  #7 
Senior Member Joined: Jan 2012 From: Erewhon Posts: 245 Thanks: 112  I can't make much sense of what you proposed, try rewriting it saying what you are doing at each step and why. You might also look at this part of the Wikipedia page on the density of a function of a RV. CB 
December 6th, 2014, 06:53 AM  #8 
Member Joined: Nov 2013 Posts: 47 Thanks: 0 
Question has also a second part If $\displaystyle W =\sum wi $ what is the distribution of $\displaystyle 2\alpha W$. So we know the sum of exponentials = $\displaystyle Gamma (k;1/\alpha)$ And we also know So we get Gamma (k; 2) ? Which is equal to a chi square with paramter = degrees of freedom = 2k correct ? 
December 6th, 2014, 06:58 AM  #9  
Member Joined: Nov 2013 Posts: 47 Thanks: 0  Quote:
Which gives me the following pdf $\displaystyle \alpha x^{\alpha1}$ and then instead of x i use $\displaystyle ln(x)$ which gives $\displaystyle \alpha(ln(x))^{\alpha1}$ and that last therm i should be able to rewrite so i get a pdf from an exponential (Kinda wanna know if this method is useable or a wrong track)  
December 6th, 2014, 07:05 AM  #10  
Senior Member Joined: Jan 2012 From: Erewhon Posts: 245 Thanks: 112 
Why do you do this next part and what exactly do you think you are doing? Quote:
 

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beta, distribution, exponential, proof 
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