# How could I find the covariance for $X$ and $Y$ in this case?

If $X \sim U(-1, 1)$ (so $X$ is uniformly distributed between $-1$ and $1$) and $Y = X^2$, what is the covariance between $X$ and $Y$? Are they independent?

So the formula for covariance is: $\operatorname{Cov}(X,Y) = E(XY) - E(X)E(Y)$

Since $Y = X^2$

$E(XY) = E(X^3)$ and $E(Y) = E(X^2)$

and

$\operatorname{Cov}(X, Y) = E(X^3) - E(X)E(X^2)$

is this the correct way to go about solving the problem?

• Yes. That is what you need to do. Do you know what to do next? – Graham Kemp Jul 9 '16 at 0:37
• We find $E(X^3)$. One could say it is obviously $0$ by symmetry. Or else, if you want to compute, it is $\int_{-1}^1 (x^3)(1/2)\,dx$, which turns out to be $0$. Similarly and more simply, $E(X)=0$, so $E(X^3)-E(X)E(X^2)=0$. – André Nicolas Jul 9 '16 at 0:45
• Thanks @GrahamKemp I was thinking to integrate in a similar way to that above done but wasn't too sure, however now seeing integration working out above, I'm getting it as well – silverjoe Jul 9 '16 at 16:55
• Also thanks @AndréNicolas that makes sense – silverjoe Jul 9 '16 at 16:57
• @SamuelOpiyo: You are welcome. Essentially the same argument shows that if $X$ is standard normal and $Y=X^2$ then $X$ and $Y$ are uncorrelated. – André Nicolas Jul 9 '16 at 17:02

Yes. That is what you need to do. Do you know what to do next?

Hint:

Obviously $\mathsf E(X)=0$, so $\mathsf E(X)\mathsf E(X^2)=0$. So can you find $\mathsf E(X^3)$ ? (note: you can do it without calculation).

That is indeed correct. It might also help to draw a simple picture of the curve on which the point $(X,Y)$ is constrained to lie, and see if that tells you something about what to expect the covariance to be. The easier question is whether their independent: Notice that if you know what $X$ is, that tells you what $Y=X^2$ is.

You're correct. $$E[X^3] = E[X] = 0$$ and hence $$\operatorname{Cov}(X,X^2) = 0$$

They are uncorrelated but not independent:

Observe that $$P(X \in (\frac{-1}{2},\frac{1}{2}), X^2 \in (\frac{-1}{2},\frac{1}{2})) \ne P(X \in (\frac{-1}{2},\frac{1}{2})) P(X^2 \in (\frac{-1}{2},\frac{1}{2}))$$

This is a classic example of a counterexample to the converse of independent implies uncorrelated. Uncorrelated means linearly independent: $$E[XY] = E[X] E[Y]$$

Independent means:

$$E[f(X)g(Y)] = E[f(X)] E[g(Y)]$$

where $f$ and $g$ are bounded and Borel-measurable. So we see that uncorrelated is what we get if we assume independence and use identity functions for $f$ and $g$. Independence is linear independence, quadratic independence, cubic independence, ..., trigonometric independence, exponential independence and so on.

• "Observe that $P(X \in (0,1), X^2 \in (0,1)) \ne P(X \in (0,1)) P(X^2 \in (0,1))$" This might be difficult to "observe" since actually, $P(X \in (0,1), X^2 \in (0,1)) =P(X \in (0,1)) P(X^2 \in (0,1))$. – Did Jul 9 '16 at 17:39
• @Did WEll that's what I get for typing that when my meds have worn off. Thanks ^-^ – BCLC Jul 10 '16 at 5:28