I have referred to quite a few questions being asked here but there aren't any solutions that point towards how to solve this (either that or I'm bad at searching).
Suppose we have 3 random variables, $X_1, X_2, X_3$, with joint pdf:
$$\pi(x_1,x_2,x_3) \propto x_1^{\alpha_1 -1}x_2^{\alpha_2 -1}x_3^{\alpha_3 -1}(1-x_1-2x_2-3x_3)^{\alpha_4-1}$$
where $0 < x_1 < 1$, $0 < x_2 < \frac 12$, $0< x_3 < \frac 13$, $0 < x_1+2x_2+3x_3 < 1$, $\alpha_i ,\forall i = 1, 2, 3, 4$ are known positive parameters.
What is the conditional distribution of $\pi(x_1|x_2,x_3)?$
I know that
$\pi(x_1,x_2,x_3) = \pi(x_1)\pi(x_2|x_1)\pi(x_3|x_1,x_2)$
$\pi(x_3|x_1,x_2) = \frac {\pi(x_1,x_2,x_3)} {\pi(x_1)\pi(x_2|x_1)}$ Eq(1)
$\pi(x_1|x_2,x_3) = \pi(x_1)\pi(x_2,x_3|x_1)$
Since $\pi(x_2,x_3|x_1) = \pi(x_3|x_1,x_2)$ (I found this online so I don't really know the validity of this)
$\pi(x_1|x_2,x_3) = \pi(x_1)\pi(x_3|x_1,x_2)$
Substituting Eq(1):
$\pi(x_1|x_2,x_3) = \pi(x_1)\frac {\pi(x_1,x_2,x_3)} {\pi(x_1)\pi(x_2|x_1)} = \frac {\pi(x_1,x_2,x_3)} {\pi(x_2|x_1)}$
Now, it leaves me to find $\pi(x_2|x_1)$ but I am not sure how to do it. I am not even sure if my above steps are correct.
Any help is really, really appreciated!
$\pi(x_1,x_2,x_3) = \pi(x_1|x_2,x_3)\pi(x_2,x_3) \implies \pi(x_1|x_2,x_3) = \frac{\pi(x_1,x_2,x_3)}{\pi(x_2,x_3)}$. To find $\pi(x_2,x_3)$, you need to marginalize the joint distribution over $x_1$ i.e. $\pi(x_2,x_3)=\int\limits_{x_1=0}^{x_1 = 1} \pi(x_1,x_2,x_3) dx_1$. Use integration by parts maybe to get the integral. Just like one done here. May require more simplification but I hope the idea is clear.
$x_2^{\alpha_2}x_3^{\alpha_3-1}\int\limits_{0}^{1}x_1^{\alpha_1}(1-x_1-2x_2-3x_3)^{\alpha_4 -1} = x_2^{\alpha_2-1}x_3^{\alpha_3-1}\Big\{\big[-x_1^{\alpha_1-1}\frac{(1-x_1-2x_2-3x_3)^{\alpha_4}}{\alpha_4}\big]_{0}^{1} +\int\limits_{0}^{1} \frac{(1-x_1-2x_2-3x_3)^{\alpha_4}}{\alpha_4}dx_1\Big\} = x_2^{\alpha_2}x_3^{\alpha_3-1}\Big\{(-1)^{(\alpha_4+1)}\frac{(2x_2+3x_3)^{\alpha_4}}{\alpha_4}+\big[-\frac{(1-x_1-2x_2-3x_3)^{\alpha_4+1}}{\alpha_4(\alpha_4+1)}\big]_0^1\Big\} = x_2^{\alpha_2}x_3^{\alpha_3-1}\Big\{(-1)^{(\alpha_4+1)}\frac{(2x_2+3x_3)^{\alpha_4}}{\alpha_4}+\big[(-1)^{\alpha_4+2}\frac{(2x_2+3x_3)^{\alpha_4+1}}{\alpha_4(\alpha_4+1)}+\frac{(1-2x_2-3x_3)^{\alpha_4+1}}{\alpha_4(\alpha_4+1)}\big]\Big\} $