Q1: Let $X : \Omega \rightarrow \mathbb{R}^2 $ be a random vector which is uniformly distributed on the unit square $[0,1] \times [0,1]$. What is the correct marginal probability density function?

A1: $$ f_{X_1}(t) = \begin{cases} 1 &\text{ for } t \in [0,1] \ 0 &\text{ else } \end{cases} $$

A2: $$ f_{X_1}(t) = \begin{cases} 1 - t &\text{ for } t \in [0,1] \ 0 &\text{ else } \end{cases} $$

A3: $$ f_{X_1}(t) = \begin{cases} t^2 &\text{ for } t \in [0,1] \ 0 &\text{ else } \end{cases} $$

A4: $$ f_{X_1}(t) = \begin{cases} 1+t &\text{ for } t \in [0,1] \ 0 &\text{ else } \end{cases} $$

Q2: Let $X : \Omega \rightarrow \mathbb{R}^2 $ be a random vector with components $X_1: \Omega \rightarrow \mathbb{R}$ and $X_2: \Omega \rightarrow \mathbb{R}$. If $X_1$ and $X_2$ are (absolutely) continuous random variables, what is a correct implication for the marginal density functions?

A1: $$ f_{X}(s,t) = f_{X_1}(s) f_{X_2}(t) $$ for all $s,t$ implies that $X_1$ and $X_2$ are independent.

A2: $$ f_{X}(s,t) = f_{X_1}(s) + f_{X_2}(t) $$ for all $s,t$ implies that $X_1$ and $X_2$ are independent.

A3: $$ f_{X}(s,t) = f_{X_1}(s) f_{X_2}(t) $$ for all $s,t$ implies that $X_1$ and $X_2$ are not independent.

A4: $$ f_{X}(s,t) = f_{X_1}(s) + f_{X_2}(t) $$ for all $s,t$ implies that $X_1$ and $X_2$ are not independent.