for the first part, could u tell me why the lower bound is 0 rather than x? We can see from the question that 0

1. Expressing $E(Y|X)$ in terms of $X$:

The conditional expectation of $Y$ given $X$ is defined as:

E(Y∣X)= y⋅f(y∣X)dy

Where $f(y|X)$ is the conditional probability density function (PDF) of $Y$ given $X$.

In this case, the joint PDF is given by:

To find the conditional PDF $f(y|X)$, we need to find the marginal PDF of $X$. To do this, integrate the joint PDF over the possible values of $Y$:

= 15 x²  = 15 x²(1-x²) / 2

Now, the conditional PDF $f(y|X)$ is given by:

f(y∣X)=fX,Y​(x,y) / f(x)​

= 2y / (1 - x²) ; x < y < 1

= 0 ; otherwise​

Now, the conditional expectation $E(Y|X)$ is:

E(Y∣X)= 

=  

= 2(1 - x³) / 3(1 - x²)