Find Covariance of $X$ and $Y$ and show whether $X$ and $Y$ are independent.

Question

Given the following Joint PDF

\begin{equation*} f(x,y) = \left\{ \begin{array}{ll} c & \quad -1< x \leq 1 ; \lvert x \rvert <y<1 \\ 0 & otherwise \end{array} \right. \end{equation*} $1.$ Find the value of $c$.

$2.$ Find marginal PDF $f_Y(y)$ of $y$

$3.$ Find conditional PDF of $X$ given $Y$

$4.$ Find covriance of $X$ and $Y$

$5.$ Are $X$ and $Y$ independent?

My attemept:

$part1$

I have found value of $c$ to be $c=1$

$Part2$

Marginal PDF of $Y$ is also found to be: $ f_Y(y) = \int_{-1}^1(1)dx= 2$

$Part3$

The conditional PDF is also found to be $f(x|y)= f(x,y)/f_Y(y)=1/2$

$Part4$

The covariance is given by:

$Cov(x,y)= E(XY)-E(X)E(Y)$

I don't know how to proceed further.

$Part5$

$X$ and $Y$ are independent when:

$P(X=x,Y=y)=P(X=x)P(Y=y)$ (Rest I don't know how to proceed further)

Please Help me in parts $4$ and $5$. How do I solve them

Answer 1

Hints:

For part 2: $f(y) =\int\limits_{-\infty}^\infty f(x,y)\, dx$ which is $0$ when $y<0$ or $y\ge 1$ and is $\int\limits_{-y}^y c\, dx$ when $0 < y < 1$. It is not $2$

For part 3: $f(x \mid y)= \frac{f(x,y)}{f(y)}$ as you have said, though not $\frac12$. Note it is $0$ when $|x| \ge y$ and $0<y <1$, and is undefined when $y<0$ or $y\ge 1$

For part 4: $E(XY) = \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty x\,y\,f(x,y) \, dx\, dy = \int\limits_0^1 \int\limits_{-y}^y x\,y\,c \, dx\, dy$ while $E(X) = \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty x\,f(x,y) \, dx\, dy = \int\limits_0^1 \int\limits_{-y}^y x\,c \, dx\, dy$ and $E(Y) = \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty y\,f(x,y) \, dx\, dy = \int\limits_0^1 \int\limits_{-y}^y y\,c \, dx\, dy$

For part 5: Is $f(x \mid y) = f(x)$ for all $x,y$ where they can be compared? Is the conditional support of $X$ independent of the value of $Y$?

Answer 2

Henrys answer has everything but I will include a little bit of intuition behind some these topics incase they are new to you. I would advise a nice little picture of everything, particularly the "triangle of support."

Firstly if you ever have a constant PDF, in any dimension (2 in your case) , the "c" will always be $\frac{1}{\text{Volume}}$, this is because we are uniformly random.

Moving onto Part $2$ your answer is wrong but lets think about why. You have given a constant for the PDF of $Y$ i.e something uniform. Can this be correct? Are " larger" (relatively speaking ) values of $y$ more likely than smaller ones? We know that $X$ is uniform and $Y$ must be greater than $|X|$ so surely $Y$ is more likely to be large than it is small? i.e non uniform.

Moving onto Part $3$ there is not much intuition I could give you except when we learn later that $X$ and $Y$ are not independent.

For part $4$ we can see that everything is symmetrical about the $y$ axis and hence $X$ is centred around $0$ so has zero mean. Similarly the probability $XY = t$ is the same as the probability $XY = -t$ (well strictly speaking they are both zero as these random variables are continuous) however it should be clear that the symmetry of $X$ gives symmetry to $XY$ and so also $\mathbb{E}[XY] = 0 $

For part $5$ you should not think about "probability $X$ equals something as these are continuous random variables. Instead think about "probability $X$ less than / greater than something. Really quickly as the PDF of (X,Y) is only non-zero for $|x| < y $ it should be clear that the value of $X$ influences that of $Y.$ Maybe try to work out $\mathbb{P}[Y> \frac{1}{2}]$ conditioned on the event $\{ \frac{-1}{2} < X < \frac{1}{2} \}$ and then conditioned on $\{ \frac{1}{2} \leq X\}$