# 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!

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seems cool!!! ))) – Seyhmus Güngören Aug 14 '12 at 20:32
a) The sketch should be something like a square with a shaded region which is a band of width $1/3$ (in $1$-norm) from the diagonal. (I think the answer is $11/36$.) b) This has nothing to do with meeting, right? You can either derive $\text{Cov}(X, Y)$ from the definition, or use the property that independence implies no correlation. c) Uh, this answers (b) too. – Tunococ Aug 14 '12 at 20:35
Seyhmus essentially repeated what I did except that he explicitly evaluated E(XY). – Michael Chernick Aug 14 '12 at 21:00
@Panda: Note that for (a) you can find answer without explicit integration. Just use geometry to find area of the region, divide by $4$. – André Nicolas Aug 14 '12 at 21:04
I still stand for $11/36$. Please check your integrals. (I did not integrate. I just used geometry. I think you need to split into 3 integrals if you don't do any transformation.) – Tunococ Aug 14 '12 at 21:53

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

$$E[X]=E[Y]=\int_0^2f_X(x)xdx=\int_0^2\frac{1}{2}x \, dx=1$$

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I don't argree with part a. X and Y must be within 20 minutes (1/3 of a hour) of each other. That means -1/3<=X-Y<=1/3. So x must be integrate from max(0,y-1/3) to min (2,y+1/3). – Michael Chernick Aug 14 '12 at 22:03
No it is not correct. your agument that $-1/3<=X-Y<=1/3$ is correct but you have to take this integral over a triangle which starts from $-2$ and ends at $+2$ and has maximum of $0.5$ at $x=0$. This is the distribution of $Z=X-Y$. You can get it via convolution of two pdfs one is $1_{\{0,2\}}$ and other is $1_{\{-2, 0\}}$. Then you can find the area between $-1/3$ and $1/3$. This will give you exacty $11/36$. Your way is correct but the integration should be done with respect to $Z=X-Y$. – Seyhmus Güngören Aug 14 '12 at 22:10
My limits of integration are for the direct calculation of the area when they meet rather than computing the probability that they don't meet and subtravting it from 1. – Michael Chernick Aug 14 '12 at 22:14
I see but $P(|X-Y|<1/3)$ can be simply evaluated as I said. As you agree that $P(|X-Y|<1/3)$ will give the correct answer and my solution to $P(|X-Y|<1/3)$ is definitely correct then there should be some problem in your solution.. – Seyhmus Güngören Aug 14 '12 at 22:17
ok then complete your solution please. – Seyhmus Güngören Aug 14 '12 at 22:37

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).

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