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Topic Title: Beta distribution
Created On Thu Feb 03, 05 10:39 AM

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FNG
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Thu Feb 03, 05 10:39 AM

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Please could someone explain to me what it means to say that F(v)^n, where n=0.25, is a simple special case of the beta distribution?

If we have the beta function b(A,B,p,q) and n (which presumably is given by one of p or q) must lie between 0 and 1, then am I correct in thinking that the upper and lower bounds A and B, must be set to 0 and 1 respectively? What values must p and q then take, to arrive at the beta distribution above?

I've had a long look at the Beta density function, but my intuition with stats is non-existent, so if anyone could help on this, I'd really appreciate it!

Tom

Aaron
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Fri Feb 04, 05 03:23 AM

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You'll have to give us some more context.

1.25*x^0.25, from x = 0 to 1, is a degenerate Beta distribution with A = 1.25 and B = 0.

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Aaron Brown

FNG
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Fri Feb 04, 05 10:37 AM

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Thanks Aaron

I see now that i made a couple of mistakes in my original post: the beta distibution is actually B(x)=x^n. x is a continuous random variable with support [0,1], whilst n is between 0 and 1, but a specific value of 0.25 is given.

The specific context is a model in which x is a liquidity shock (in the form of withdrawals) to a unit deposit base, and a concave function has been chosen so that smaller shocks are more likely than large ones. I want to understand better exactly what n is doing and where it comes from and what it actually means in this context to vary n.

So, essentially I would like to know what shoud I be plugging into the beta function (or gamma functions) that would imply the beta distribution above? From this i will hopefully get a better idea of what I am actually doing if i change n.

does that make more sense?!

Tom

Aaron
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Fri Feb 04, 05 02:18 PM

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Don't call this a Beta distribution. It is a degenerate one, with B = 0 and A = n, but none of the Beta distribution results will hold because you will find yourself dividing by zero.

(n+1)*x^n, n>0, is a geometric probability density function on [0,1], with cumulative density function x^(n+1). The mean is (n+1)/(n+2). This is always greater than 0.5, so large values are more likely than small ones. But picking a small n makes this effect smaller. At n = 0.25 you have a mean of 1.25/2.25 = 5/9 = 0.5556. The median is 0.5^(1/(n+1)). This is also always larger than 0.5, but close for small n. At n = 0.25 this is 0.5^(1/1.25) = 0.25^0.8 = 0.5743.

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Aaron Brown

FNG
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Joined: Dec 2004

Fri Feb 04, 05 03:38 PM

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Thanks

More light reading to add to the pile!

Tom

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