Conditional density function of $Z_1\mid Z_1+Z_2$

Question

Say that we have two random variables $Z_1,Z_2$ such that $Z_1$ has ${\mathcal{B}}_{1/2,1}$ and $Z_1+Z_2$ follows ${\mathcal{B}}_{1,1/2}$ where ${\mathcal{B}}_{\alpha,\beta}$ is the beta density function. Is it possible to find the conditional density function of $Z_1\mid Z_1+Z_2$?

Answer 1

Once you assume independence between $Z_1,Z_2$ the problem is not difficult. Otherwise you cannot solve it unless you have further information to find the joint density $f_{Z_1,Z_2}(x,y)$

So let's assume independence and, just to simplify the notation, let's indicate $X$ and $Y$ instead of $Z_1$, $Z_2$

Setting

$$\begin{cases} u=x+y \\ v=x, \end{cases}$$

That is

$$\begin{cases} x=v \\ y=u-v, \end{cases}$$

it is self evident that the Jacobian $|J|=1$ thus

$$f_{UV}(u,v)=\frac{1}{4\sqrt{v(1+v-u)}}\mathbb{1}_{(0;1)}(v)\mathbb{1}_{(v;v+1)}(u)$$

As you can see, the joint density of $(U,V)$ is defined on the following parallelogram

and thus it is easy to integrate it in $V$ to derive $f_U(u)$

The conditional density you are looking for is obviously

$$f_{X|X+Y}(u,v)=\frac{f_{UV}(u,v)}{f_U(u)}$$

---

I am trying to compute $Pr(v<1/3,u<2/3)$ with the joint distribution function

Answering to your question, take a look at the integration area you are interested in (yellow area)

$$\mathbb{P}\Bigg[V<\frac{1}{3};U<\frac{2}{3}\Bigg]=\int_0^{\frac{1}{3}}\frac{1}{2\sqrt{v}}\Bigg[\int_{v}^{\frac{2}{3}}\frac{1}{2\sqrt{1+v-u}} du \Bigg]dv\approx0.19475$$