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Let $X$ be a random variable with a pdf

$$f_X(x) = \begin{cases}1 & \text{if }0\le x\le 1\\0 & \text{otherwise} \end{cases}$$

Let $Y = X^n$.

How do I compute $p_{X,Y}$? What is an intuitive explanation to the sign of the correlation?

How do I find the limit of the correlation, and what is an intuitive explanation for it?

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If by $p_{X,Y}$ you mean the correlation of $X$ and $Y$, then you need to compute $$E[XY]=E[X^{n+1}], E[X], E[Y] = E[X^n], E[X^2], E[Y^2] = E[X^{2n}]$$ all of which are readily computed from the density of $X$. Then you can compute $\text{cov}(X,Y)$, $\text{var}(X)$ and $\text{var}(Y)$ from these numbers and thus get $p_{X,Y}$. The intuitive explanation of positive correlation is that $Y$ increases as $X$ increases. –  Dilip Sarwate Oct 7 '12 at 2:11
    
@DilipSarwate, is my solution correct? Can I assume $n$ is nonnegative? –  idealistikz Oct 10 '12 at 3:33
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2 Answers

The joint distribution function of $(X,Y)$ has no density since $(X,Y)\in D$ almost surely, where $D=\{(x,x^n)\mid x\in[0,1]\}$ has zero Lebesgue measure.

The correlation of $X$ and $Y$ has the sign of $n$ since $Y=u_n(X)$ where the function $u_n:x\mapsto x^n$ is increasing when $n\gt0$ and decreasing when $n\lt0$ (the second case being restricted to $n\gt-1$ since, otherwise $Y$ is not integrable hence the correlation does not exist). The computation of the actual value of the correlation is standard, hence I suggest you signal which specific difficulties, if any, you encounter when performing it.

More generally, one can mention that, for every pair of nondecreasing functions $u$ and $v$ and every random variable $Z$ such that the expectations exist, $\mathbb E(u(Z)v(Z))\geqslant\mathbb E(u(Z))\mathbb E(v(Z))$, hence $\mathrm{Cov}(u(Z),v(Z))\geqslant0$. Obviously, the same conclusion holds if the functions $u$ and $v$ are both nonincreasing, and it is reversed if one function is nondecreasing and the other is nonincreasing.

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The correlation coefficient is the following. $$p_{X,Y} = \frac{\frac{n}{2(n+1)(n+2)}}{\sqrt{\frac{1}{12}\frac{n^2}{(n+1)^2(2n+1)}}}$$

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