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Let the pair (X,Y) be uniformly distributed on the unit disc, so that

$f_{X,Y}(x,y)=\begin{cases}\frac{1}{\pi}&\text{if }x^2+y^2\leq1,\\0&\text{otherwise}.\end{cases}$

Find $\mathbb E\sqrt{X^2+Y^2}$ and $\mathbb E(X^2+Y^2)$.

We are not familiar with coordinate transformations, and my teacher told us to simply look carefully at the volume we're trying to calculate.

We know that

$$\mathbb E(g(X,Y))=\int_{-\infty}^\infty\int_{-\infty}^\infty g(x,y)f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$$.

Applying this to $g(X,Y)=\sqrt{X^2+Y^2}$ we get

$\begin{aligned}\mathbb E(\sqrt{X^2+Y^2})=&\int_{-\infty}^\infty\int_{-\infty}^\infty \sqrt{x^2+y^2}f_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx\\ =&\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{x^2+y^2}\pi^{-1}\,\mathrm dy\,\mathrm dx. \end{aligned}$

Now here I would need to apply some useful transformation. Could someone help me out from here?

Sha Vuklia
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2 Answers2

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This $$\int_{-1}^1\int_{-1}^{\sqrt{1-x^2}}\sqrt{x^2+y^2}\pi^{-1}\,\mathrm dy\,\mathrm dx$$ Should be this $$\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{x^2+y^2}\pi^{-1}\,\mathrm dy\,\mathrm dx$$ Then change to polar coordinates $x=r\cos\theta$ and $y=r\sin \theta$, and remember that the jacobian for this transformation is $r$, the integral becomes $$\int_{0}^{2\pi}\int_{0}^{1}r^2\pi^{-1}\,\mathrm dr\,\mathrm d\theta$$

GuPe
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Another way to do this, which I think is much more elegant, is to do the transformation at the outset. Instead of picking $X$ and $Y$ you pick $r \sim [0, 1]$ and $\theta \sim [0, 2\pi)$ and do the transformation $$ \begin{align} x &= \sqrt r \cos \theta\\ y &= \sqrt r \sin \theta \end{align} $$ Then $\sqrt{x^2+y^2} = \sqrt r$ and $x^2 + y^2 = r$ so you have $$ \mathbb E[r] = \frac{1}{2} $$ and $$ \mathbb E[\sqrt r] = \frac{2}{3} $$ Also check out : http://mathworld.wolfram.com/DiskPointPicking.html