Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose I am given the joint pdf of $X$, $Y$, and I am asked to find the $\operatorname{cov}(X,Y)$.

I know that $\operatorname{cov}(X,Y)=E(XY)-E(X)E(Y)$ and I know how to find $E(X)$ and $E(Y)$.

My questions are:

  1. What is the definition of $E(XY)$? Is it always equal to $$\int_{R\times R} xyf_X(x)f_Y(y)dxdy\,?$$ Or only if $X$, $Y$ are independent?(from the answer I have, the solution I have did not check the independence of $X$ and $Y$, and the answer $\operatorname{cov}(X,Y)$ is not zero, which proves $X$, $Y$ are not independent.)

  2. I remember, but not very clearly, that if the joint pdf of $X$, $Y$ $f_{X,Y}(x,y)$ can be written as $$f_{X,Y}(x,y)=g(x)h(y),$$ then $X$ and $Y$ are independent. Is it always true or need some conditions? I mean, suppose the region is not, say, $[0,1]\times[0,1]$, but, say, $0<x<1,x<y<2x$, is that saying still true?

Thank you so much!

share|improve this question
    
1. We have $E[XY]=\int_{\mathbb R\times\mathbb R}xyF(x,y)dx dy$ in general, where $F(\cdot,\cdot)$ is the cdf of $(X,Y)$. Your formula is true when $X$ and $Y$ are independent (and of course $X$ and $Y$ have a cdf). 2. You can check that $P(X\leq t_1,Y\leq t_2)=P(X\leq t_1)\cdot P(Y\leq t_2)$ thanks to the hypothesis. –  Davide Giraudo Jan 4 '12 at 20:02
2  
$$E[g(X,Y)]=\int_{-\infty}^\infty\int_{-\infty}^\infty g(x,y)f_{X,Y}(x,y)dx dy$$ holds in general where $f_{X,Y}(x,y)$ is the joint pdf of $X$ and $Y$. Your integral in 1. is inncorrect. The equality $f_{X,Y}(x,y)=g(x)h(y)$ needs to hold at all points $(x,y)$ in the plane, not just at some points, in order for $X$ and $Y$ to be independent random variables. If the joint pdf is nonzero only for $0<x<1,x<y<2x$, then $X$ and $Y$ are dependent random variables; no need to try and see if you can express $f(x,y)$ as $g(x)h(y)$ –  Dilip Sarwate Jan 4 '12 at 20:04
    
@DavideGiraudo You probably meant to write pdf instead of cdf? and also $P(X\leq t_1,Y\leq t_2)=P(X\leq t_1)P(Y\leq t_2)$ –  Dilip Sarwate Jan 4 '12 at 20:06
    
OK, got it ,thank you so much! –  breezeintopl Jan 4 '12 at 20:07
    
@Dilip: You should post your comment as an answer, since that covers the OP's questions. –  Mike Spivey Jan 4 '12 at 20:08
show 2 more comments

1 Answer

up vote 4 down vote accepted

In general, for jointly continuous random variables $X$ and $Y$ with joint pdf $f_{X,Y}(x,y)$, $$E[g(X,Y)]=\int_{-\infty}^\infty\int_{-\infty}^\infty g(x,y)f_{X,Y}(x,y)dx dy.$$ In the special case you are considering, this becomes $$E[XY]=\int_{-\infty}^\infty\int_{-\infty}^\infty xyf_{X,Y}(x,y)dx dy.$$

If $X$ and $Y$ are jointly continuous random variables with joint pdf $f_{X,Y}(x,y)$, and $f_{X,Y}(x,y)$ factors into the product of the marginal pdfs $f_X(x)$ and $f_Y(y)$, then $X$ and $Y$ are said to be independent random variables. More useful is the reverse implication: if we assume that $X$ and $Y$ are independent continuous random variables with known pdfs (e.g. standard normal), then they are jointly continuous with joint pdf $f_{X,Y}(x,y)$ equal to the product $f_X(x)f_Y(y)$ of their individual pdfs.

Your expression $\displaystyle E[XY] = \int_{R\times R} xyf_X(x)f_Y(y)dxdy$ is incorrect in the general case, but is correct when $X$ and $Y$ are independent continuous random variables since $f_{X,Y}=f_X(x)f_Y(y)$ in this case. Indeed, if your expression were correct in general, then we would have $$E[XY] = \int_{R\times R} xyf_X(x)f_Y(y)dxdy = \int_{R} xf_X(x)dx \int_{R} yf_Y(y)dy = E[X]E[Y]$$ so that $\text{cov}(X,Y)=E[XY]-E[X]E[Y] = 0$ for all random variables, which is clearly not true. So we have the following.

If $X$ and $Y$ are independent random variables, then $E[XY]=E[X]E[Y]$.

Note that this holds for all random variables, not just continuous random variables. Also, as you probably know, the converse is not true: uncorrelated random variables need not be independent.

With regard to your second question, $X$ and $Y$ are independent if you can find $g(x)$ and $h(y)$ such that the equality $f_{X,Y}(x,y)=g(x)h(y)$ holds at all points $(x,y)$ in the plane, not just at some points. If the joint pdf is nonzero only for $0<x<1,x<y<2x$, then $X$ and $Y$ are dependent random variables; no need to try and see if you can express $f(x,y)$ as $g(x)h(y)$.

Finally, note that all of the above applies provided the various integrals and expectations are defined or exist. $E[XY]=E[X]E[Y]$ does not apply to independent Cauchy random variables, for example, because $E[X]$ and $E[Y]$ are undefined for Cauchy random variables $X$ and $Y$.

share|improve this answer
    
+1: Nicely said. –  Mike Spivey Jan 4 '12 at 20:56
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.