Variance of Conditional Expected value

Question

I don't understand the highlighted line. How was 1/4 calculated? E(Z|Y=y) is in part (a)

Help would be appreciated, thanks

Answer 1

So $\mathbb E[Z|Y]$ has expectation equals to $$\frac{1}{2} \times 1 + \frac{1}{2} \times \frac{1}{2}=\frac{3}{4}$$

The above is because we know from (c) that it has value of $1$ with probability of $0.5$, and value $0.5$ with probability of $0.5$.

For the variance calculation, the following shows where $\frac{1}{4}$ comes from

Case when we take value of $1$ $$(1-\frac{3}{4})^2=(\frac{1}{4})^2$$

Other case when we take value of $\frac{1}{2}$ $$(\frac{1}{2}-\frac{3}{4})^2=(\frac{1}{4})^2$$

When we apply probability to the above two cases (which are both $\frac{1}{2}$), we get the result.