$X \sim U(0,1)$ and $Y \mid X \sim U(0,X)$ . Find $f_{X, Y}(x,y)$, $\mathbb{E}(Y)$ and $\Pr(X+Y < 1)$

Question

Suppose $X \sim U(0,1)$ and $Y \mid X \sim U(0,X)$.

Find:

1. $f_{X, Y}(x,y)$ (The joint probability density function);
2. $\mathbb{E}(Y)$;
3. $\Pr(X+Y < 1)$.

The only one of these I think I've solved is 2):

$$E(Y)=\int_0^1E(Y \mid x)\,\mathrm{d}x = \int_0^1\frac{0+x}{2}\,\mathrm{d}x = \frac 14.$$

Answer 1

(2) is fine. Maybe a tidier argument would use the property $E(E(Y|X))=E(Y)$. But it's ok (and I suggest that you put a $1$ multiplying $E(Y|x)$ to make clear that you used there the fact that $X\sim U(0,1)$.

For (1), remember that $$f_{XY}(x,y)=f_{Y|X=x}(y)\cdot f_X(x).$$

Once you have the joint density you can work on (3), since $$P(X+Y<1)=\iint_A f_{XY}(x,y) dA,$$ Where $A=\{(x,y)\in\mathbb R^2\colon x+y<1\}$. But have in mind the fact that the joint density is $0$ except for a bounded set.

Answer 2

1)

$f(x,y)$ must be constant on sets like $\{\langle x,y\rangle\mid y\in(0,x)\}$ and this observation leads to:

$f(x,y)=g(x)$ for $0<y<x<1$ and $f(x,y)=0$ otherwise.

Then: $$x=P\left(X\leq x\right)=\int_{0}^{x}\int_{0}^{u}g\left(u\right)dvdu=\int_{0}^{x}g\left(u\right)udu$$ for $x\in(0,1)$ so we conclude that $g(u)=\frac1u$

So:

$f(x,y)=\frac1x$ for $0<y<x<1$ and $f(x,y)=0$ otherwise.

2)

$\mathbb EY=\mathbb E[\mathbb E[Y\mid X]]=\mathbb E\frac12X=\frac12\mathbb EX=\frac12\frac12=\frac14$.

3)

Let $[u+v<1]$ denote the function that takes value $1$ if $u+v=1$ and takes value $0$ otherwise.

$\begin{aligned}P\left(X+Y<1\right) & =\int_{0}^{1}\int_{0}^{u}\left[u+v<1\right]\frac{1}{u}dvdu\\ & =\int_{0}^{1}\int_{0}^{\min(u,1-u)}\frac{1}{u}dvdu\\ & =\int_{0}^{1}\frac{\min\left(u,1-u\right)}{u}du\\ & =\int_{0}^{0.5}du+\int_{0.5}^{1}\left(\frac{1}{u}-1\right)du\\ & =\int_{0.5}^{1}\frac{1}{u}du\\ & =\left[\ln u\right]_{0.5}^{1}\\ & =-\ln0.5\\ & =\ln2 \end{aligned} $