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I know that β=α is what will give a symmetric standard beta pdf, but why is this so?

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I guess by symmetric, you mean symmetric about $x=\frac{1}{2}$, where $$f(x;\alpha,\beta) = \frac{x^{\alpha-1} (1-x)^{\beta-1}}{B(\alpha,\beta)}$$ is the beta pdf.

If $\alpha=\beta$ the numerator becomes $$x^{\alpha-1} (1-x)^{\alpha-1}=\Big(x(1-x)\Big)^{\alpha-1}.$$ Now $x(1-x)=x-x^2 = \frac{1}{4} - (x- \frac{1}{2})^2$ is obviously symmetric about $x=\frac{1}{2}$ and so is $$f(x;\alpha,\alpha) =\frac{\Big(\frac{1}{4} - (x- \frac{1}{2})^2\Big)^{\alpha-1}}{B(\alpha,\alpha)}$$

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