correct it so the question reads as (pdf) fX,Y(x,y) as 8xy

Thank you for the clarification. Let's revise the solution with the correct joint probability density function (pdf):

### Given:
- Joint pdf:  
  \[
  f_{X,Y}(x,y) = 
  \begin{cases} 
  8xy, & 0 < x < 1, \, 0 < y < 1 \\
  0, & \text{otherwise}
  \end{cases}
  \]

### (a) Find the marginal distributions of \(X\) and \(Y\).

**Marginal pdf of \(X\):**  
The marginal pdf of \(X\) is found by integrating the joint pdf over \(y\):
\[
f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dy = \int_{0}^{1} 8xy \, dy
\]
Evaluating the integral:
\[
f_X(x) = 8x \int_{0}^{1} y \, dy = 8x \cdot \left[ \frac{y^2}{2} \right]_0^1 = 8x \cdot \frac{1}{2} = 4x, \quad 0 < x < 1
\]
Thus, 
\[
f_X(x) = 
\begin{cases} 
4x, & 0 < x < 1 \\
0, & \text{otherwise}
\end{cases}
\]

**Marginal pdf of \(Y\):**  
Similarly, the marginal pdf of \(Y\) is found by integrating the joint pdf over \(x\):
\[
f_Y(y) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dx = \int_{0}^{1} 8xy \, dx
\]
Evaluating the integral:
\[
f_Y(y) = 8y \int_{0}^{1} x \, dx = 8y \cdot \left[ \frac{x^2}{2} \right]_0^1 = 8y \cdot \frac{1}{2} = 4y, \quad 0 < y < 1
\]
Thus, 
\[
f_Y(y) = 
\begin{cases} 
4y, & 0 < y < 1 \\
0, & \text{otherwise}
\end{cases}
\]

### (b) Determine whether \(X\) and \(Y\) are independent.

Two random variables \(X\) and \(Y\) are independent if and only if \(f_{X,Y}(x,y) = f_X(x)f_Y(y)\) for all \(x, y\). 

Let's check:
\[
f_X(x)f_Y(y) = (4x)(4y) = 16xy
\]
But the joint pdf \(f_{X,Y}(x,y)\) is \(8xy\), not \(16xy\). Therefore, \(X\) and \(Y\) are **not independent**.

### (c) Conditional expectation \(E[X | Y = y]\).

The conditional pdf \(f_{X|Y}(x|y)\) is given by:
\[
f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} = \frac{8xy}{4y} = 2x
\]
Now, the conditional expectation is:
\[
E[X | Y = y] = \int_{0}^{1} x \cdot 2x \, dx = 2 \int_{0}^{1} x^2 \, dx
\]
The integral is:
\[
2 \cdot \left[ \frac{x^3}{3} \right]_0^1 = 2 \cdot \frac{1}{3} = \frac{2}{3}
\]
So, \(E[X | Y = y] = \frac{2}{3}\).

### (d) Joint pdf of \(U = X/Y\) and \(V = Y\), and marginal pdf of \(U\).

Let \(U = \frac{X}{Y}\) and \(V = Y\). The transformation yields:
\[
f_{U,V}(u,v) = f_{X,Y}(uv, v) \cdot \left|\text{Jacobian}\right|
\]
The Jacobian of the transformation \((x,y) \rightarrow (u,v)\) is:
\[
J = \begin{vmatrix} 
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} 
\end{vmatrix} 
= \begin{vmatrix} 
v & u \\
0 & 1 
\end{vmatrix} 
= v
\]
So,
\[
f_{U,V}(u,v) = 8(uv)v \cdot v = 8u v^3, \quad 0 < u < 1, \, 0 < v < 1
\]
The marginal pdf of \(U\) can be found by integrating over \(v\):
\[
f_U(u) = \int_{0}^{1} 8u v^3 \, dv = 8u \cdot \int_{0}^{1} v^3 \, dv
\]
The integral is:
\[
8u \cdot \left[ \frac{v^4}{4} \right]_0^1 = 8u \cdot \frac{1}{4} = 2u, \quad 0 < u < 1
\]

---

Do you want further details or have any questions?

**Relative Questions:**
1. How does the form of the joint pdf influence the marginal distributions?
2. What conditions must hold for two variables to be considered independent?
3. Why is it important to understand the relationship between conditional expectations and joint distributions?
4. How do transformations like \(U = X/Y\) and \(V = Y\) help in simplifying complex random variables

Explore the corrected joint probability density function (pdf) fX,Y(x,y) = 8xy and its solutions. Understand marginal distributions, conditional expectations, and transformations of random variables. Ideal for advanced undergraduate studies in probability theory.

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