Question

Let (X,Y) be a uniformly random point in the triangle in the plane with vertices (0,0), (0, 1),(1, 0). Find Cov(X, Y ) and Corr(X, Y ).

Answer 1

Denote the triangle by $T$, so that

$T=\{(x,y)\mid 0\le x\le1,0\le y\le1-x\}$

We have

$\displaystyle\iint_T\mathrm dA=\int_0^1\int_0^{1-x}\mathrm dy\,\mathrm dx=\frac12$

(Alternatively, $T$ is a triangle with base and height 1, so its area is 1/2.)

Then the joint PDF of $X,Y$ is

$f_{X,Y}(x,y)=\begin{cases}\frac12&\text{for }(x,y)\in T\\0&\text{otherwise}\end{cases}$

Covariance is defined as

$\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]$

We can get the expectations we need from the joint PDF:

$E[X]=\displaystyle\iint_{\Bbb R^2}xf_{X,Y}(x,y)\,\mathrm dA=\frac12\iint_Tx\,\mathrm dA=\frac12\int_0^1\int_0^{1-x}x\,\mathrm dy\,\mathrm dx=\frac1{12}$

$E[Y]=\displaystyle\iint_{\Bbb R^2}yf_{X,Y}(x,y)\,\mathrm dA=\frac1{12}$

$E[XY]=\displaystyle\iint_{\Bbb R^2}xyf_{X,Y}(x,y)\,\mathrm dA=\frac1{48}$

Then the covariance is

$\mathrm{Cov}[X,Y]=\dfrac1{48}-\dfrac1{12^2}=\boxed{\dfrac1{72}}$

Correlation is defined as

$\mathrm{Corr}[X,Y]=\dfrac{\mathrm{Cov}[X,Y]}{\sqrt{\mathrm{Var}[X]\mathrm{Var}[Y]}}$

We have for any random variable $X$,

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

Now,

$E[X^2]=\displaystyle\iint_{\Bbb R^2}x^2f_{X,Y}(x,y)\,\mathrm dA=\frac1{24}$

$E[Y^2]=\displaystyle\iint_{\Bbb R^2}y^2f_{X,Y}(x,y)\,\mathrm dA=\frac1{24}$

so that the correlation is

$\mathrm{Corr}[X,Y]=\dfrac{\frac1{72}}{\sqrt{\left(\frac1{24}-\frac1{12^2}\right)^2}}=\boxed{\dfrac25}$