Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

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.

share|cite|improve this answer
up vote 0 down vote accepted

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.