In Exercise 5.13, the joint density function of Y1 and Y2

Step 1 of 4

(a)

Given the Joint probability density function of \({Y_1}\;{\rm{and}}\;{Y_2}\) is,

\(f\left( {{y_1},{y_2}} \right) = \left\{ \begin{array}{l}30{y_1}y_2^2;{y_1} - 1 \le {y_2} \le 1 - {y_1},0 \le {y_1} \le 1\\0;{\rm{otherwise}}\end{array} \right)\).

Let \({Y_1}\;{\rm{and}}\;{Y_2}\) be jointly continuous random variables with joint density function \(f\left( {{y_1},{y_2}} \right)\). Then, the marginal density of \({Y_1}\;{\rm{and}}\;{Y_2}\), respectively, are given by,

\({f_{{Y_1}}}\left( {{y_1}} \right) = \int\limits_{{y_2}} {f\left( {{y_1},{y_2}} \right)d{y_2}} \)

\({f_{{Y_2}}}\left( {{y_2}} \right) = \int\limits_{{y_1}} {f\left( {{y_1},{y_2}} \right)d{y_1}} \)

Hence,

\({f_{{Y_1}}}\left( {{y_1}} \right) = \int_{{y_1} - 1}^{1 - {y_1}} {30{y_1}y_2^2d{y_2}} \)

\({f_{{Y_1}}}\left( {{y_1}} \right) = 30{y_1}\int_{{y_1} - 1}^{1 - {y_1}} {y_2^2d{y_2}} \)

\({f_{{Y_1}}}\left( {{y_1}} \right) = 20{y_1}{\left( {1 - {y_1}} \right)^3},0 \le {y_1} \le 1\)

By the definition,

\(D\left( {\alpha ,\beta } \right) = \frac{{\left| \!{\overline {\,  \alpha  \,}} \right. \left| \!{\overline {\,  \beta  \,}} \right. }}{{\left| \!{\overline {\,  {\alpha  + \beta } \,}} \right. }}\)

Now,

\(D\left( {2,4} \right) = \frac{{\left| \!{\overline {\,  2 \,}} \right. \left| \!{\overline {\,  4 \,}} \right. }}{{\left| \!{\overline {\,  {2 + 4} \,}} \right. }}\)

\(D\left( {2,4} \right) = \frac{{\left| \!{\overline {\,  2 \,}} \right. \left| \!{\overline {\,  4 \,}} \right. }}{{\left| \!{\overline {\,  6 \,}} \right. }}\)

\(D\left( {2,4} \right) = \frac{1}{{20}}\)

So,\(\frac{1}{{\beta \left( {2,4} \right)}}y_1^{2 - 1}{\left( {1 - {y_1}} \right)^{4 - 1}} = 20{y_1}{\left( {1 - {y_1}} \right)^3}\).  

Hence, the marginal density of \({Y_1}\) is a \(\beta \)-density with \(\alpha  = 2\), and \(\beta = 4\).