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there's a something that I though about that Disturbing me for quite some time now:

Let $X$ , $Y $ be a random variables. I call random variables independent if for every possible value of $x_{i}$ that $X$ might get and $y_{j}$ that $Y$ might get:

$p(X=x_{i} \cap Y=y_{j})=P(X=x_{i})\cdot P(Y=y_{j})$.

In addition two random variables are Non-correlated if $ E[X\cdot Y]=E[X]\cdot E[Y]$,

when $E$ represents the expected value.

I know that if $X$ and $Y$ are independent so they are Non-correlated.

when $X$ and $Y$ are dependent, I can think of only cases which $E[X\cdot Y]=0$ because one of the expected value of $X$ or $Y$ is o.

my QUESTION is: Are they any dependent random varaibles and non-correlated so that $E[X]\neq0$ and $E[Y]\neq0$, and if not, how does one prove that?

Thank you for the help.

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1 Answer 1

up vote 4 down vote accepted

Yes. Just take your favourite example of $X, Y$ dependent and uncorrelated with expectation 0, and consider the new variables $X+1$ and $Y+1$. They're still dependent and uncorrelated, but now they have expectation 1.


Explicitly: a simple example with $E[X]=E[Y]=0$ is for $(X,Y)$ to be uniformly distributed in the set $\{(0,1),(1,0),(0,-1),(-1,0)\}$. Adding $1$ to $X$ and $Y$, we obtain new random variables $U=X+1$ and $V=Y+1$ such that $(U,V)$ is uniformly distributed in $\{(1,2),(2,1),(1,0),(0,1)\}$.

We have $E[U]=\frac{1+2+1+0}{4}=1$, $E[V]=\frac{2+1+0+1}{4}=1$, $E[UV]=\frac{2+2+0+0}{4}=1=E[U]E[V]$, so $U$ and $V$ are uncorrellated. But $P(U=1 \cap V=1)=0 \neq \frac{1}{2} \cdot \frac{1}{2}=P(U=1)P(V=1)$, so $U$ and $V$ are dependent.

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Do you mean by $X+1$ to add to every $x_{i}$ $1$? –  henry11 Apr 6 '11 at 18:02
    
@henry11: That is what he means: you get $Pr(X+1 = x_i +1) = Pr(X = x_i)$. Note that two random variables are uncorrelated if their covariance $E[(X-\mu_X)(Y-\mu_Y)] = 0$, and that adding 1 to $X$ and to $Y$ also adds 1 to their means so leaving the covariance unchanged. –  Henry Apr 6 '11 at 18:41
    
I'll need much more simple explanation, I don't know what is covariance yet, and what the expression you wrote means. My level is just up to what I wrote. –  henry11 Apr 6 '11 at 18:44

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