1. Joint Probability Function

Let X and Y have joint probability density function

fX,Y(x,y) = x+y, for 0 <=x<= 1, o<=y<=1
0, otherwise

How do i go about determining the corr(X,Y)?

Could someone outline the steps please.. Any help would be appreciated..

2. One has ${\rm Corr}(X,Y)=\frac{{\rm Cov}(X,Y)}{\sqrt{{\rm Var}(X){\rm Var}(Y)}}=\frac{E[XY]-E[X]E[Y]}{\sqrt{(E[X^2]-E[X]^2)(E[Y^2]-E[Y]^2)}}$, right? So what you need to compute is $E[XY]$, $E[X]$, $E[Y]$, $E[X^2]$ and $E[Y^2]$. In fact, the pdf of $(X,Y)$ is symmetric in $x,y$, so that $X$ and $Y$ have same distribution and the formula reduces to: ${\rm Corr}(X,Y)=\frac{E[XY]-E[X]^2}{E[X^2]-E[X]^2}$.
To compute $E[XY]$, just integrate $xy$ times the pdf of $(X,Y)$ (over the square $0\leq x,y\leq 1$).
You can do the same with the other ones (integrating $x$ and $x^2$), but you may find it quicker to first determine the pdf of $X$. For that, you just have to integrate the pdf of $(X,Y)$ with respect to the variable $y$, keeping $x$ fixed.

3. Originally Posted by Laurent
One has ${\rm Corr}(X,Y)=\frac{{\rm Cov}(X,Y)}{\sqrt{{\rm Var}(X){\rm Var}(Y)}}=\frac{E[XY]-E[X]E[Y]}{\sqrt{(E[X^2]-E[X]^2)(E[Y^2]-E[Y]^2)}}$, right? So what you need to compute is $E[XY]$, $E[X]$, $E[Y]$, $E[X^2]$ and $E[Y^2]$. In fact, the pdf of $(X,Y)$ is symmetric in $x,y$, so that $X$ and $Y$ have same distribution and the formula reduces to: ${\rm Corr}(X,Y)=\frac{E[XY]-E[X]^2}{E[X^2]-E[X]^2}$.
To compute $E[XY]$, just integrate $xy$ times the pdf of $(X,Y)$ (over the square $0\leq x,y\leq 1$).
You can do the same with the other ones (integrating $x$ and $x^2$), but you may find it quicker to first determine the pdf of $X$. For that, you just have to integrate the pdf of $(X,Y)$ with respect to the variable $y$, keeping $x$ fixed.
I was just about to reply when I saw yours. I misread the question as find Cov(X, Y) ..... So it's lucky you replied first (although my reply would have got the OP half way ...... )

4. Thanks so much, Ok this is what i got...

E[XY] = 1/3

E[X]^2 = (1/3 + y/2)^2 = y^2/4 + 2y/6 + 1/9

E[X^2] = 1/4 + y/3

E[Y] = 1/3 + x/2

And as a final answer.. i gained..

y^4/16 - 2y^3/24 - 13y^2/144 + 10y/216 + 10/324

5. Originally Posted by brd_7
Thanks so much, Ok this is what i got...

E[X]^2 = (1/3 + y/2)^2 = y^2/4 + 2y/6 + 1/9

There's no $y$ in the answer (what is $y$?). In order to find the average, you integrate $y$ as well:
$E[X]=\int_0^1\int_0^1 x(x+y)dx\,dy$. The same for $X^2$.

6. Oh right, that makes sense.. Im assuming you would do the same if i needed E(Y)?

Well as a final answer i got -1/11..

Thanks for all the help again.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 47, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9848949909210205, "perplexity": 350.49684914733325}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824733.32/warc/CC-MAIN-20171021095939-20171021115939-00610.warc.gz"}