91Ó°ÊÓ

The probability density function (pdf) of a Beta distribution is given by \[ f(x;\alpha,\beta) = \frac{x^{\alpha - 1}(1 - x)^{\beta - 1}}{B(\alpha, \beta)} \] where \( B(\alpha, \beta) \) is the Beta function, defined as \( B(\alpha, \beta) = \int_0^1 t^{\alpha - 1}(1 - t)^{\beta - 1} dt \). In this exercise, we are interested in the case where \( \alpha = \beta \).

For a function to be symmetric about a vertical line, the function's values should be the same at equidistant points from the line of symmetry. If we have a vertical line \( x = \frac{1}{2} \), then the function will be symmetric if \( f(\frac{1}{2} - h) = f(\frac{1}{2} + h) \) for all \( h > 0 \).

By substituting \( \alpha = \beta \) into the pdf, we get \[ f(x;\alpha,\beta) = \frac{x^{\alpha - 1}(1 - x)^{\alpha - 1}}{B(\alpha, \alpha)} \].Given that \( f(\frac{1}{2} - h) = f(\frac{1}{2} + h) \), let's substitute \( x = \frac{1}{2} - h \) and \( x = \frac{1}{2} + h \) into the equation and we also need to check that these two are equal to each other. \[ f(\frac{1}{2} - h;\alpha,\beta) = \frac{(\frac{1}{2} - h)^{\alpha - 1}(1 - \frac{1}{2} + h)^{\alpha - 1}}{B(\alpha, \alpha)} = \frac{h^{\alpha - 1}(1 - h)^{\alpha - 1}}{B(\alpha, \alpha)} \], \[ f(\frac{1}{2} + h;\alpha,\beta) = \frac{(\frac{1}{2} + h)^{\alpha - 1}(1 - \frac{1}{2} - h)^{\alpha - 1}}{B(\alpha, \alpha)} = \frac{h^{\alpha - 1}(1 - h)^{\alpha - 1}}{B(\alpha, \alpha)} \]. As we can see the these two values are equal when \( \alpha = \beta \), therefore the graph of the Beta pdf is symmetric about the vertical line \( x = \frac{1}{2} \) when \( \alpha = \beta \).