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For the bivariate beta RV $(X,Y)$ with PDF $f (x,y) = { \frac{\Gamma(p_1+p_2+p_3) }{\Gamma(p_1) \Gamma(p_2) \Gamma (p_3) }} x^{p_1-1} y^{p_2-1 }(1-x-y)^{p_3-1} , x \ge 0 , y \ge 0$ and $x + y \le 1$ ,

where $p_1, p_2$, and $p_3$ are positive real numbers. find the marginal PDFs of $X$ and $Y $and the conditional PDFs. Find also the conditional PDF of $\frac{Y}{(1-X)}$, given $X = x$.

My claim is that $X$ follows $ \text{Beta}(p_1,p_2+p_3)$, But I can't prove this.

1 Answers1

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Well, for $0\leq z\leq 1$ and $b>0, c>0$ we have: $$\int_0^{z} y^{b-1} (z-y)^{c-1}\mathrm d y = \dfrac{\Gamma(b)\Gamma(c) z^{b+c-1}}{\Gamma(b+c)}$$

So for $0\leq x\leq 1$ and $p_1,p_2,p_3\in\Bbb R^+$

$$\int_0^{1-x} \dfrac{\Gamma(p_1+p_2+p_3)x^{p_1-1}y^{p_2-1} (1-x-y)^{p_3-1}}{\Gamma(p_1)\Gamma(p_2)\Gamma(p_3)}\mathrm d y = \dfrac{\Gamma(p_1+p_2+p_3) x^{p_1-1}(1-x)^{p_2+p_3-1}}{\Gamma(p_1)\Gamma(p_2+p_3)}$$

Which is the probability density function for $\mathcal{Beta}(p_1,p_2+p_3)$

Note: that requires the denominator to be a product of gamma functions rather than a sum, as I suspect you should have.

Graham Kemp
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  • What about the conditional pdf of $X$, and that of $\frac{Y}{1-X}$ –  Aug 04 '17 at 04:59
  • $f_{Y\mid X}(y\mid x) =\dfrac{ f_{X,Y}(x,y)}{f_X(x)}$, and so $f_{Y/(1-X)\mid X}(z\mid x) = \dfrac{(1-x),f_{X,Y}(x,z(1-x))}{f_X(x)}$ – Graham Kemp Aug 04 '17 at 05:23
  • Can you elaborate on the second one that is can you show that $\frac{Y}{1-X} $ follows $\text{Beta}(p_{2},p_{3})$? –  Aug 04 '17 at 05:34