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I was looking at the Bernoulli Distribution and its relation to the Prior Belief Distribution. The equation is written as $$ \frac{x^{\alpha - 1}(1-x)^{\beta -1}}{B(\alpha, \beta)}. $$

I've also studied the properties of the of the Beta function which is define as $$ B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} \, \mathrm dt. $$

The Beta function is basically the integral of the Bernoulli Distribution from what I can see.

I graphed the Prior Belief Distribution for a couple of variations for $\alpha$ and $\beta$ and noticed that dividing by the Beta function squashes the height of the distribution by some amount.

My question is what is the Beta function doing exactly. Is it squashing the distribution to within some range? If so, can someone show me a proof as to why this works so that I can internalize it better?

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  • $\begingroup$ Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. I have edited your question to reflect this principle by replacing them with inline LaTeX. (This is particular silly since the images you used were from an online LaTeX render). $\endgroup$ – Brian Apr 3 at 0:13
  • $\begingroup$ @Brian thanks, sorry about that. I am used to using stackoverflow which does not allow using latex. old habit $\endgroup$ – Bolboa Apr 3 at 0:19
  • $\begingroup$ In general, $\frac{f(x)}{\int_{t=-\infty}^\infty f(t)\,dt}$ normalizes $f(x)$ so that the area under the curve is $1$, which goes toward making it a probability distribution. (And similarly for two, three, etc. variables, adjusting for the support of $f(x)$.) $\endgroup$ – Brian Tung Apr 3 at 0:31
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So the main job of the $B(\alpha, \beta)$ function here is to normalise the distribution, i.e. ensure that it has integral $1$ when integrated over the whole range. Can you see why that is?

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  • $\begingroup$ So in other words, if I were to graph $\frac{x^{\alpha - 1}(1-x)^{\beta -1}}{B(\alpha, \beta)}$, we always want to ensure that the area under the curve is equal to 1 between interval 1 and 0? $\endgroup$ – Bolboa Apr 3 at 0:29
  • $\begingroup$ Yep that's exactly it $\endgroup$ – George Dewhirst Apr 3 at 0:30

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