91Ó°ÊÓ

Prior distribution for some parameter \(\theta \) is a beta distribution with mean \(\frac{1}{3}\) and variance is \(\frac{1}{{45}}\)

Let\(\alpha \)and\(\beta \)be the parameters of the beta distribution, then

Mean =\(\frac{\alpha }{{\alpha + \beta }}\)

That is\(\frac{\alpha }{{\alpha + \beta }} = \frac{1}{3}\)

Variance =\(\frac{{\alpha \beta }}{{{{\left( {\alpha + \beta } \right)}^2}\left( {\alpha + \beta + 1} \right)}}\)

That is\(\frac{{\alpha \beta }}{{{{\left( {\alpha + \beta } \right)}^2}\left( {\alpha + \beta + 1} \right)}} = \frac{2}{{90}}\)

Since\(\frac{\alpha }{{\alpha + \beta }} = \frac{1}{3}\),it follows that

\(\frac{\beta }{{\alpha + \beta }} = \frac{2}{3}\). Therefore,

\(\begin{aligned}\frac{{\alpha \beta }}{{{{\left( {\alpha + \beta } \right)}^2}}} = \frac{\alpha }{{{{\left( {\alpha + \beta } \right)}^2}}} \times \frac{\beta }{{{{\left( {\alpha + \beta } \right)}^2}}}\\ = \frac{1}{3} \times \frac{2}{3}\\ = \frac{2}{9}\end{aligned}\)

From the equation of variance\(\frac{2}{{9\left( {\alpha + \beta + 1} \right)}} = \frac{2}{{90}}\)

Hence\(\left( {\alpha + \beta + 1} \right) = 10\)

Therefore,\(\alpha + \beta = 9\)

And from equation of mean\(\alpha = 3\)and\(\beta = 6\)

Hence the prior pdf of \(\theta \) is

For \(0 < \theta < 1\) \(\xi \left( \theta \right) = \frac{{\left| \!{\overline {\, 9 \,}} \right. }}{{\left| \!{\overline {\, {\left( 3 \right)} \,}} \right. \left| \!{\overline {\, {\left( 6 \right)} \,}} \right. }}{\theta ^2}{\left( {1 - \theta } \right)^5}\)

Hence Prior pdf of \(\theta \) is \(\xi \left( \theta \right) = \frac{{\left| \!{\overline {\, 9 \,}} \right. }}{{\left| \!{\overline {\, {\left( 3 \right)} \,}} \right. \left| \!{\overline {\, {\left( 6 \right)} \,}} \right. }}{\theta ^2}{\left( {1 - \theta } \right)^5}\)