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December 1st, 2014, 08:38 AM   #1
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Proof beta distribution = exponential when B =0

Proof beta with b = 1 is exponential distribution

$\displaystyle \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}$

suppose $\displaystyle \beta=1$

then we get

$\displaystyle \alpha x^{\alpha-1}$

If we take $\displaystyle -ln(x)$

we get

$\displaystyle \alpha(-ln(x))^{\alpha-1}$

and the last therm should be equal to

$\displaystyle \alpha e ^{-\alpha x}$

But how do i get there ?
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December 1st, 2014, 08:36 PM   #2
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Quote:
Originally Posted by MathHatesMe View Post
Proof beta with b = 1 is exponential distribution
It can't be since the support of density of the beta-distribution is $(0,1)$, and that of the exponential is $(0,\infty)$

Post the actual question please.

CB
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December 2nd, 2014, 04:35 PM   #3
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December 4th, 2014, 06:54 AM   #4
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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:
$$
P(W<w)=P(-\ln(X)<w)=P(\ln(X)>-w)=1-P(\ln(X)<-w) \\ \phantom{P(W<w)}
=1-P(X<e^{-w}) \\ \phantom{P(W<w)}
=1-F(e^{-w}) \\ \phantom{P(W<w)}
=1-(e^{-w})^{\alpha}=1-e^{-\alpha w}
$$

But $1-e^{-\alpha w}$ is the cdf of the exponential distribution with parameter $\alpha$.

CB

Last edited by CaptainBlack; December 4th, 2014 at 07:05 AM.
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December 6th, 2014, 06:08 AM   #5
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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 $\displaystyle x^{\alpha}$

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December 6th, 2014, 06:25 AM   #6
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But how do we get it in te method i proposed ?
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December 6th, 2014, 07:26 AM   #7
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Quote:
Originally Posted by MathHatesMe View Post
But how do we get it in te method i proposed ?
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
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December 6th, 2014, 07:53 AM   #8
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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 ?
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December 6th, 2014, 07:58 AM   #9
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Quote:
Originally Posted by CaptainBlack View Post
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
Well it took the pdf of a Beta distribution and then i took $\displaystyle \beta = 1$

Which gives me the following pdf

$\displaystyle \alpha x^{\alpha-1}$

and then instead of x i use $\displaystyle -ln(x)$

which gives $\displaystyle \alpha(-ln(x))^{\alpha-1}$

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)
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December 6th, 2014, 08:05 AM   #10
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Why do you do this next part and what exactly do you think you are doing?
Quote:
and then instead of x i use $\displaystyle -ln(x)$

which gives $\displaystyle \alpha(-ln(x))^{\alpha-1}$
CB
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