Let $X$ and $Y$ be independent random variables each having the uniform distribution on $[0,1]$. Let $U = \min\{X,Y\}$ and $V = \max\{X,Y\}$. Find $\textbf{E}(U)$, and hence calculate $\textbf{Cov}(U,V)$.


In the first place, let us determine the distribution of $U$ and $V$:

\begin{align*} \textbf{P}(U\leq u) & = \textbf{P}(\min\{X,Y\}\leq u) = 1 - \textbf{P}(\min\{X,Y\} > u)\\\\ & = 1 - \textbf{P}(X > u,Y > u) = 1 - \textbf{P}(X > u)\textbf{P}(Y > u)\\\\ & = 1 - (1 - F(u))(1 - F(u)) = F(u)(2 - F(u)) \Rightarrow\\\\ f_{U}(u) & = 2f(u) - 2f(u)F(u) = 2f(u)(1-F(u)) \end{align*}

Analogously, we have \begin{align*} \textbf{P}(V\leq v) & = \textbf{P}(\max\{X,Y\}\leq v) = \textbf{P}(X\leq v, Y\leq v)\\\\ & = \textbf{P}(X\leq v)\textbf{P}(Y\leq v) = F(v)^{2}\Rightarrow\\\\ f_{V}(v) & = 2F(v)f(v) \end{align*}

where \begin{cases} f(x) = 1\quad\text{for}\quad 0 \leq x \leq 1\\\\ F(x) = x\quad\text{for}\quad 0 \leq x \leq 1 \end{cases}

Which means that $f_{U}(u) = 2(1-u)$ and $f_{V}(v) = 2v$. Based on this, it is routine to calculate $\textbf{E}(U)$ and $\textbf{Cov}(U,V)$ according to \begin{cases} \textbf{E}(U) = \displaystyle\int_{0}^{1}uf_{U}(u)\mathrm{d}u\\\\ \textbf{Cov}(U,V) = \textbf{E}(UV) - \textbf{E}(U)\textbf{E}(V) \end{cases}

In the first place, I'd be grateful if someone could double-check my results. Thenceforth, I'd like to know if there is another approach to the same problem. Thanks in advance!

  • 2
    $\begingroup$ "Based on this, it is routine to calculate ... Cov(U,V)=E(UV)−E(U)E(V)" Hmmm... How do you deduce E(UV) from the PDFs of U and V? "I'd be grateful if someone could double-check my results" Where are "your results" for E(U) and for Cov(U,V)? $\endgroup$ – Did Jan 30 at 17:42

Here is a direct way:

\begin{align} \mathsf{E}U &=\mathsf{E}U1\{X\le Y\}+\mathsf{E}U1\{X> Y\} \\ &=\int_0^1\int_x^1 x\, dydx+\int_0^1\int_0^x y\, dydx=\frac{1}{3}. \end{align}

To calculate the expectation of $UV$, notice that $UV=XY$. Therefore, $\mathsf{E}UV=\mathsf{E}X\mathsf{E}Y=1/4.$


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