find the conditional probability,but i stuck somewhere,and dont know how to solve it

Question

The conditional probability density of $X$ provided that the continuous event V has values between $ y $ and $y+dy$ is given by

$$f_{X|Y}(x|y)=\frac{2xy+a}{a^2(y+1)}\ \text{for} \ 0\le x \le a,$$

where $a$ is a constant and the probability density of $Y$ is given by

$$f_{Y}(y)= \frac{2(y+1)}{b^2+2b}\ \text{for} \ 0\le y \le b.$$

Find the conditional probability density $f_{Y|X}(y|x)$ ,i.e., the conditional probability density function of $Y$ provided that the continuous event U has values between $x $ and $ x+dx$

My idea is: Use the definition

$$f_{X|Y}(x|y)=\frac{f_{X Y}(x,y)}{f_{Y}(y)}.$$

So $${f_{X Y}(x,y)}=\frac{2xy+a}{a^2(y+1)}\frac{2(y+1)}{b^2+2b}=\frac{2(2xy+a)}{a^2(b^2+2b)}.$$

And now find the $$f_{X}(x)=2\int^{b}_{0}\frac{2xy+a}{a^2(y+1)}.dy$$

Now, I have three questions

1. Are my ideas right? I wouldn't use "continuous event V has values between $ y $ and $y+dy$" and "continuous event U has values between $x $ and $ x+dx$" ,in fact, I don't know what they are for this question.

2. Assuming that I am right, I still have a problem. I don't know how to evaluate $$\int^{b}_{0}\frac{2xy+a}{a^2(y+1)}dy.$$

Can anyone help me?

Answer 1

Your ideas are OK.

The correct wording for "continuous event" would be continuous random variable.

As far as the integral:

$$\int^{b}_{0}\frac{2xy+a}{a^2(y+1)}dy,$$

write it in the form

$$\int^{b}_{0}(2xy+a)\frac1{a^2(y+1)}dy$$

and integrate by part with with $v'=\frac1{a^2(y+1)}$ and $u=(2xy+a)$.

For $\int \frac1{y+1}\ dy$ look here.