Finding covariance between profit and quality

Question

The quality $X$ of an item is uniformly distributed on the interval $[0,1]$ and the profit $Y$ is given by $Y = X^5$. Find the covariance between $X$ and $Y$ .

Can someone interpret this question for me?

Answer 1

The covariance of $X$ and $Y$ is $E(XY)-E(X)E(Y)$.

The expectation of $X$ is, by symmetry, $\frac{1}{2}$.

For the expectation of $Y$, we calculate $\int_0^1 x^5\,dx$. This is $\frac{1}{6}$.

We have $XY=X^6$ and $E(X^6)=\int_0^1 x^6\,dx=\frac{1}{7}$.

Now all we need to do is put the pieces together.

Remark: If the covariance of two random variables $X$ and $Y$ is not far from $1$ or $-1$, then there is a strong linear relationship between $X$ and $Y$. If the covariance is not far from $1$, then $Y$ grows roughly linearly as $X$ grows. If the covariance is not far from $-1$, then $Y$ shrinks roughly linearly as $X$ grows. In this case, there is of course a very strong relationship between $X$ and $Y$: if we know $X$, then we know $Y$. However, the relationship is highly non-linear.

Answer 2

The covariance between any two random variables $X$ and $Y$ measures the linear dependence between them and is defined by $$ Cov(X,Y) = {\bf E}[XY]-{\bf E}[X]{\bf E}[Y]. $$ Thus in your case \begin{align} Cov(X,Y) &= Cov(X,X^5) \\ &= {\bf E}[X^6]-{\bf E}[X]{\bf E}[X^5] \\ &= \int_0^1 x^6 dx-\int_0^1 x dx \int_0^1 x^5 dx \\ &= \frac{1}{7}-\frac{1}{2} \cdot \frac{1}{6} \\ &= \frac{5}{84}. \end{align}

So it appears that there is a linear dependence between the quality of this item and the profit it generates, though (perhaps sadly) not a very strong one. Maybe there are other dependencies ...