Random point from $[0,1] \times [0,1]$ - probability distributions and independences Let $Q=(X,Y)$ be a random point from square $[0,1] \times[0,1]$ with $\lambda$ (Lebesgue's) measure. Find probability distribution of $X$ and $Y$. Are $X, Y$ independent? Find probability distribution of $X+Y$. Is $X, X+Y$ independent?
So, my answer is as follows. Let $B \in \mathcal{B}(\mathbb{[0,1]})$. The distribution of X is $\mathbb{P}_X(B) = \mathbb{P}(X \in B) = \mathbb{P}(X^{-1}(B)) = \lambda(B)$. The distribution of Y will be the same.
Now, $X$ and $Y$ seem independent to me, because:
$$
\mathbb{P}(X \in [0,1], Y \in [0,1])=\mathbb{P}((X,Y) \in [0,1]\times [0,1]) = \lambda_{[0,1] \times [0,1]}((X,Y))=\lambda(X)\lambda(Y)=\mathbb{P}(X \in [0,1]) \mathbb{P}(Y\in[0,1])
$$
Is that a correct reasoning? What does the $X+Y$ distribution look like? Are $X, X+Y$ independent?
 A: From $P(X^{-1}(B))$ to $\lambda(B)$, there are too many steps hidden.
First of all, $X$ is a random variable from $[0,1]^2 \to [0,1]$. It is given by $X(x,y) = x$ (the first coordinate map)
What should you be doing? We need a description for $X^{-1}(B)$(for $B \subset [0,1]$). But that is obvious : it is clear that $X^{-1}(B) = B \times [0,1]$ (first coordinate in $B$, second anywhere).
(Let $\lambda_i$ denote the Lebesgue measure in dimension $i$)
So it follows that $P(X^{-1}(B)) = P(B \times [0,1]) = \lambda_2(B\times [0,1]) = \lambda_1(B)$, for $B \subset [0,1]$. Consequently, the probability distribution function of $X$ is described by that of the uniform distribution on $[0,1]$.$Y$ has the same distribution.

Now, for independence, your intuition is right, but somewhere along the line you got lost with meaningless expressions like $\lambda(X)$ etc.
What you should be doing is trying to prove $P(X \in A, Y \in B) = P(X \in A) P(Y \in B)$. For this , start with the LHS : note that $P(X \in A, Y \in B)$ is the same as $P((X,Y) \in A \times B)$ which equals $\lambda_2(A \times B)$.
The right hand side equals, by the distributions of $X,Y$ , the quantity $\lambda_1(A)\lambda_1(B)$. 
But these two are equal by the fact that $\lambda_2$ is the product measure $\lambda_1 \times \lambda_1$. Hence, we are done.

For the distribution of $X+Y$, we must find the probability that $X+Y \leq t$ for each $x$.
This is the same as $P((X,Y) \in A)$ ,where $A = \{(a,b) : 0 \leq a+b \leq t\}$.  So we must find the Lebesgue measure of this set.
We break into two cases : that of $t \leq 1$ and $1 < t \leq 2$. Note that in other cases the set $A$ would either have full or null measure.
For $t \in [0,1]$ the measure of $A$ is $\int_0^t \int_{0}^{t-x}1 dydx = \frac{t^2}{2}$. For $t\in [1,2]$ the measure is $\frac 12 + \int_{0}^{t-1} \int_{0}^{t-1-x} 1dydx = \frac 12 + \frac{(t-1)^2}{2}$. (I leave how I derive these iterated integrals as exercises : try to draw thw set $A$ and use standard double integrals to express its measure) .
Thus, we have $P(X+Y \leq t) = \frac{t^2}{2}$ for $t \in [0,1]$ and $\frac{(t-1)^2}2 + \frac 12$ for $t \in [1,2]$. For $t \geq 2$ it is $1$, for $t \leq 0$ it is zero.

For independence, I leave you to see what happens, try to play around.
