Using law of total expectation on a [0,x] uniform distribution

Question

I'm currently studying for an upcoming probability exam and I got my hands on one of the previous exams for the course from a friend. I was working my way through it but the following question stumped me :

Suppose that X has a continuous distribution with probability density function given by :$$ f(x) = \begin{cases} \ 2x & 0<x<1,\\ 0 & \text{otherwise}. \end{cases} $$ Suppose that Y is a continuous random variable such that the conditional distribution of Y given X = x is uniform(0, x). Find E(Y).

I've tried using the law of total expectation and finding E(E(Y|X)) to find E(Y) but from what I understand the integral would be $$\int_0^x y \times\frac1x dy = 1$$ However, the possible answers I was given are $\frac12, \frac13, \frac14, \frac23$. I'm not sure where I went wrong since it seemed straightforward enough for me. Could someone explain what I'm missing?

Answer 1

We have \begin{align} E[Y|X] &= \frac{X}2 \end{align}

Hence $E[Y] = \int_0^1 f(x)\cdot \frac{x}{2}\,dx$. Try to evaluate this quantity.

Answer 2

Purely informally we have that $Y\mid X=x$ is distributed like $\text{Uniform}([0,x])$ so that $\mathsf E(Y\mid X=x)=\frac x2$. Since $X=x$ has probability $0$ for any $x$, this argument cannot be made precise directly but instead we will use the proper measure theoretic machinery to formalize this argument.

I understand the exercise like this: For every $x\in\mathbb R$, let $Z_x$ be a uniformly distributed random variable on $[0,x]$. Then it is given that for all measurable $A\subset\mathbb R$, $$\mathsf P(Y\in A\mid X) = \mathsf P(Z_X\in A) \quad\text{almost surely}.$$ Here, $\mathsf P(Z_X\in A)$ is the random variable defined by $\mathsf P(Z_X\in A)(\omega) = \mathsf P(Z_{X(\omega)}\in A)$.

Let $\mathsf E(Z_X)$ be the random variable given by $\mathsf E(Z_X)(\omega)=\mathsf E(Z_{X(\omega)})$ for every $\omega\in\Omega$.

With this, we have $$\mathsf E(Y\mid X) = \mathsf E(Z_X).$$

But since $Z_{x}\sim\text{Uniform}([0,x])$, we have $\mathsf E(Z_X) = \frac{X}2$.

Therefore,

$$\mathsf E(Y\mid X) = \frac X2.$$ Hence, $$\mathsf E(Y)=\mathsf E(\mathsf E(Y\mid X)) = \mathsf E\left(\frac X2\right) = \frac 12\int_{0}^1 2x^2\,\mathrm dx = \frac 12 \frac 23=\frac 13.$$