Question

Please help at problem 4.2_7

Answer 1

@kirbykirby
I need steps, please.

Answer 2

If \(X\) and \(Y\) are independent, then:
\(f_{X,Y}(x,y)=f_X(x)f_Y(y)\) for all pairs \((x,y) \in S\)
If you find just one pair that doesn't satisfy this, then X and Y are NOT independent.

In your case, you want to check if the following are equal:
\[ f_{X,Y}(0,0)=f_X(0)f_Y(0)\\
f_{X,Y}(1,1)=f_X(1)f_Y(1)\\
f_{X,Y}(1,-1)=f_X(1)f_Y(-1)\\
f_{X,Y}(2,0)=f_X(2)f_Y(0)\]

Now they give you \(f_{X,Y}(x,y)=\dfrac{1}{4}\). What you need to find now are the marginals, \(f_X(x)\) and \(f_Y(y)\).

First: when you find out the marginal of \(X\), you essentially sum over the y-variable:
\[ f_X(x)=\sum_{y}f(x,y)\]
So since \(X\) takes on values 0, 1, 2, we'll look at the marginals for those values.
\[ f_X(0)=\sum_{y}f(0,y)=f(0,0)=1/4\\
f_X(1)=\sum_{y}f(1,y)=f(1,1)+f(1,-1)=1/4+1/4=1/2\\
f_X(2)=\sum_{y}f(2,y)=f(2,0)=1/4\]

For Y now: you sum over the x-variable:
\[ f_Y(y)=\sum_{x}f(x,y)\]
So since \(Y\) takes on values -1, 0, 1 we'll look at the marginals for those values.
\[ f_Y(-1)=\sum_{x}f(x,-1)=f(1,-1)=1/4\\
f_Y(0)=\sum_{x}f(x,0)=f(0,0)+f(2,0)=1/4+1/4=1/2\\
f_Y(1)=\sum_{x}f(x,1)=f(1,1)=1/4\]

SO... back to independence.. check if the following are true using the marginals we just obtained: (notice that \(f(x,y)=1/4\) in all cases)... for example the first case:
\[ f_{X,Y}(0,0)=f_X(0)f_Y(0) ->1/4 =(1/4)(1/2)=1/8... \text{not true!}\\
f_{X,Y}(1,1)=f_X(1)f_Y(1)\\
f_{X,Y}(1,-1)=f_X(1)f_Y(-1)\\
f_{X,Y}(2,0)=f_X(2)f_Y(0)\]
--------------------------
Now for the covariance:
\[Cov(X,Y)=E(XY)-E(X)E(Y) \]
Here, \[E(XY)=\sum_{x}\sum_{y}xy\cdot f(x,y)\]
So basically replace all the x's and y's for all the 4 possible points.
\(E(X)\) and \(E(Y)\) are the same definitions as in the single variable case. Just find the expectation using the marginals used above. \[ E(X)=\sum_{x}x\cdot f_X(x)\] for x=0,1,2
\[ E(Y)=\sum_{y}y\cdot f_Y(y)\] for y=-1,0,1

Finally, the correlation coefficient \(\rho\) :
\[ \rho=\frac{Cov(X,Y)}{\sqrt{Var(X)} \sqrt{Var(Y)}}\]
And for \(Var(X)=E(X^2)-[E(X)]^2\) and recalling that
\[ E(X^2)=\sum_{x}x^2\cdot f_X(x)\]

Answer 3

WOW !! Thanks a tooooon