Working on part(b) and the correct answer is 1/8. Don't know what is my error on R.H.D..

Question

So, I worked part(a) which I already checked that it is right but stuck on part(b).

But I don't get the correct answer for R.H.D. Can anyone help me out with my error?

Thank you.

Answer 1

It looks like your L.H.S. is correct. For the R.H.S. you should have this:

$$\int_{\frac{1}{2}}^{\frac{3}{4}} \frac{\frac{1}{2}-(1-y)}{1-y} \cdot 2(1-y)dy + \int_{\frac{3}{4}}^{1} 1 \cdot 2(1-y)dy$$

Looks like you had this second integral right, but the first one is slightly different. You found $f_Y(y)$ correctly.

Consider $P(X \leq \frac{1}{2} | Y=y)$ when $\frac{1}{2}<y<\frac{3}{4}$.

$$P \bigg(X \leq \frac{1}{2} \bigg| Y=y \bigg) = \frac{\textrm{width of }x\textrm{'s less than }\frac{1}{2} \textrm{ when }Y=y}{\textrm{width of all possible }x\textrm{'s when }Y=y}$$

The $x$ values less than $\frac{1}{2}$ will always be between the line $x=1-y$ and $x=\frac{1}{2}$.

The entire domain that $x$ could be, given that $\frac{1}{2}<y<\frac{3}{4}$, would be anywhere between the lines $x=1-y$ and $x=2(1-y)$.

Therefore, $$P \bigg(X \leq \frac{1}{2} \bigg| Y=y \bigg) = \frac{\frac{1}{2}-(1-y)}{2(1-y)-(1-y)} = \frac{\frac{1}{2}-(1-y)}{1-y}$$

Going back to the full R.H.S. $$\int_{\frac{1}{2}}^{\frac{3}{4}} \frac{\frac{1}{2}-(1-y)}{1-y} \cdot 2(1-y)dy + \int_{\frac{3}{4}}^{1} 1 \cdot 2(1-y)dy$$

$$=\int_{\frac{1}{2}}^{\frac{3}{4}} \bigg[ \frac{1}{2}-(1-y) \bigg] \cdot 2dy + \int_{\frac{3}{4}}^{1} 1 \cdot 2(1-y)dy$$

$$=\frac{1}{16} + \frac{1}{16} = \frac{1}{8}$$

I know I skipped a few steps at the end, but I think you can evaluate those integrals.