Probability distributions - Exam paper question - $\mathrm{Cov}(X,Y)$, PDF Two people have decided to meet at a certain point in a forest sometime between noon and 2pm.
Their respective independent arrival times are $X$ and $Y$ such that $X \sim \mathrm{Unif}(0,2)$ and $Y \sim \mathrm{Unif}(0,2)$. 
Hence the joint density of $X$ and $Y$ is
$$f_{X,Y} {(x,y)} = \begin{cases} 1/4, &  0< x <2 , 0< y <2 \\ 0, & \text{otherwise.} \end{cases} $$
They have agreed that whoever arrives first will wait for at most $20$ minutes for the arrival of the other.
a) Sketch the region in the $xy$ plane of times values for which they will meet and specify precisely the appropriate bounds (in terms of $x$ and $y$) for this region; then find the probabilty that they will meet by integrating the joint PDF $f_{X,Y} {(x,y)}$ over this region.
b) Since $X$ and $Y$ are independent, what value must $\mathrm{Cov}(X,Y)$ have?
c) Calculate explicitly $\mathrm{Cov}(X,Y)$ starting from its definition. Recall that
$$\mathrm{Cov}(X,Y) = E[(X - E(X))(Y - E(Y))].$$
I know this is quite a long question but I didn't know how to break it down into smaller parts without just having to type it into three different questions to ask.
If you could give me an idea about the sketch great! I'm not sure how to integrate the PDF as it's only $1/4$? Would it not just be $x/4 + C$?
Also for $\mathrm{Cov}(X,Y)$, I don't seem to have any notes on this, so detail would be good too.
Test is in the morning and you guys have been a big help so far!
 A: I forgot part a)
a- When you shade the square in $\{[0,2],[0,2]\}$ then $20 min=1/3$ and they can not meet only if you start waiting from $2-1/3=5/3$ and it follows that you have the probability that thay cannot meet
$$\int_0^{5/3}\int_0^{5/3}\frac{1}{4} \, dx \, dy=\frac{25}{36}$$ 
and they meet with probability
$$1-\frac{25}{36}=\frac{11}{36}$$
b- If $X$ and $Y$ are independent then their covariance should be zero as follows
$$E[XY]=E[X]E[Y]$$ 
$$Cov(X,Y)=E[(X-E(X))(Y-E(Y))]=E[XY]-E[X]E[Y]=0$$
c- $$E[XY]=\int_0^2\int_0^2 \frac{1}{4}xy \, dx \, dy=1$$
and you have
$$f_X(x)=\int_{-\infty}^\infty\frac{1}{4}f_{XY}(x,y)\,dy=\int_0^2 \frac{1}{4}1 \, dy=\frac{1}{2} \in [{0,2}]$$
you get
$$E[X]=E[Y]=\int_0^2f_X(x)xdx=\int_0^2\frac{1}{2}x \, dx=1$$
A: If X and Y are independent then they must be uncorrelated which means Cov(X,Y)=0.  Another way to show that is by using the fact that if X and Y are independent then E(XY) =E(X) E(Y).
Now Cov(X,Y) = E[(X-EX) (Y-EY)]= E(X-EX) E(Y-EY) by independence
and then=0 since both E(X-EX) and E(Y-EY)=0.
The integral will depend on the region of integration and you would be integrating out both x and y. Also keep in mind to determine the region of integration that X and Y must be within 20 minutes of each other.  So -20/60<=X-Y<=20/60.
The answer is obtained by taking ∫$_0$$^2$∫$_a$ $^b$  1/4 dx dy where a=max(0, y-1/3) and b=min(2, y+1/3).
