# Math Help - Independence

1. ## Independence

First I would like to clear up some misunderstanding I have about independence. In my text I have:
If $F(x,y)=F_X(x)F_Y(y)$, then X and Y are indep.
However, in my notes from class I have:
If $f(x,y)=f_X(x)f_Y(y)$, then X and Y are indep.
Which is correct?

Is this the only way to show independence? Does it have anything to do with expectation values? I had a question in which I had:
$X=Z, Y=Z^2; Z \sim N(\mu,\sigma^2)$
I was first asked to find the followings: $,,$ and here is what I did:
$==\mu$
$==Var(Z)+^2=\sigma^2+\mu^2$
$=$, for this I took the 3rd derivative of the moment generating function and got this expression: $2\sigma^2\mu+\mu^3$

I was then asked about whether X and Y are independent. Naturally I attempted to use what I just did: $\neq$ therefore dependent. Is this correct?

2. $Y=X^2$ so they are clearly dependent.
It certainly doesn't matter if you have normality or not.
If you differentiate $F_X(x)F_Y(y)=F(x,y)$ wrt x and y you get $f_X(x)f_Y(y)=f(x,y)$
Likewise if $f_X(x)f_Y(y)=f(x,y)$, then
$F(x,y)=\int_{-\infty}^x\int_{-\infty}^yf(s,t)dtds =\int_{-\infty}^xf_X(s)ds\int_{-\infty}^yf_Y(t)dt=F_X(x)F_Y(y)$

3. Okay, I can see why they are dependent, but how can I show it? I can write:
$f_X(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$

But what is $f_Y(y)$?
Is it just $[f_X(x)]^2$, so $f_Y(y)=\frac{1}{\sigma^22\pi}e^{-2\frac{(y-\mu)^2}{2\sigma^2}}$ ?
Or since $Y=X^2$, is it just $f_Y(y)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(\sqrt{y}-\mu)^2}{2\sigma^2}}$ ?

And worse, how can I combine them and get $f(x,y)$?
I'm quite lost.

4. You can't square the density of X to get the density of $X^2$
Try showing that the covariance is not zero.
Show that $E(XY)-E(X)E(Y)=E(X^3)-E(X)E(X^2)$ is not zero.

5. I know that is $X$ and $Y$ are independent, then $Cov(X,Y)=0$. But I thought it doesn't work the other way around, namely, even if $Cov(X,Y)=0$, it doesn't necessary mean that they are independent. Then even if I show that $Cov(X,Y)\neq0$, how can I say that they are indeed dependent?

6. Originally Posted by synclastica_86
I know that is $X$ and $Y$ are independent, then $Cov(X,Y)=0$. But I thought it doesn't work the other way around, namely, even if $Cov(X,Y)=0$, it doesn't necessary mean that they are independent. Then even if I show that $Cov(X,Y)\neq0$, how can I say that they are indeed dependent?

Yes, that's called the contrapositive.
A implies B, and not B implies not A.

7. Originally Posted by matheagle
You can't square the density of X to get the density of $X^2$
But in principle, with the information given, can I find $f_Y(y)$ and $f(x,y)$ , and therefore, their respective distribution functions?

8. Originally Posted by synclastica_86
But in principle, with the information given, can I find $f_Y(y)$ and $f(x,y)$ , and therefore, their respective distribution functions?

The formula for obtaining the density of $Y=X^2$ is right under

Derivation of the pdf for one degree of freedom
on Chi-square distribution - Wikipedia, the free encyclopedia
But X and Y do not have a joint distribution in $R^2$.