Expected value of an expected value of a joint density function

Question

I had a question I was hoping for some help on:

Let $Y_1$ and $Y_2$ be continuous random variables with joint density function:

$$f(x,y) = \begin{cases} 6(1-y_2) & \text{if $0 <= y_1 <= y_2 <= 1$ } \\ 0 & \text{elsewhere} \end{cases}$$

a) Find E[$Y_1|Y_2 = y_2$]

b) Use the answer you found in part a) to find E[E[$Y_1|Y_2 = y_2$]]

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I did the work to find that a) is $$\int_{0}^{1}\frac{6(1-y_2)}{6y_2 - 6y_2^2}\ dy_1 = \dfrac{1}{y_2}$$ (If you wouldn't mind checking that answer as seeing my question relies on part b), which in turn relies on a) ). However, I'm not sure at all how to approach b). Would someone assist me? Thank you so much in advance, I really appreciate it!

Answer 1

(a) The marginal density function for $Y_2$:

\begin{eqnarray*} f_{Y_2}(y_2) &=& \int_0^{y_2} 6(1-y_2)\;dy_1 \\ &=& \bigg[6(1-y_2)y_1 \bigg]_0^{y_2} \\ &=& 6y_2(1-y_2). \end{eqnarray*}

So now,

\begin{eqnarray*} f_{Y_1|Y_2}(y_1|y_2) &=& \dfrac{f_{Y_1,Y_2}(y_1,y_2)}{f_{Y_2}(y_2)} \\ &=& \dfrac{6(1-y_2)}{6y_2(1-y_2)} \\ &=& \dfrac{1}{y_2}. && \\ && \\ \therefore\quad E(Y_1\mid Y_2=y_2) &=& \int_0^{y_2} y_1\dfrac{1}{y_2}\;dy_1 \\ &=& \left[\dfrac{y_1^2}{2y_2} \right]_0^{y_2} \\ &=& \dfrac{y_2}{2}. \end{eqnarray*}

(b)

\begin{eqnarray*} E(E(Y_1\mid Y_2=y_2)) &=& \int_0^{1} E(Y_1\mid Y_2=y_2)\;f_{Y_2}(y_2)\;dy_2 \\ &=& \int_0^{1} \dfrac{y_2}{2}\;6y_2(1-y_2)\;dy_2 \\ &=& \left[y_2^3 - \frac{3}{4}y_2^4 \right]_0^1 \\ &=& 1-\dfrac{3}{4} = \dfrac{1}{4}. \end{eqnarray*}