Problem 1 - Probability

1.

What is the expectation of the square of Euclidean distance between two points that are distributed independently and uniformly within the unit square $[0,1]^2$?

2.

What is the expectation of Euclidean distance between the origin and a point that is distributed uniformly within the unit square $[0,1]^2$

1.

$\int_{x_1=0}^1 \int_{x_2=0}^1 \int_{y_1=0}^1 \int_{y_2=0}^1 (x_1-x_2)^2 + (y_1-y_2)^2 dx_1dx_2dy_1dy_2$

$=\int_{x_1=0}^1 \int_{x_2=0}^1 \int_{y_1=0}^1 \int_{y_2=0}^1 (x_1^2+x_2^2+y_1^2+y_2^2-2x_1x_2-2y_1y_2)dx_1dx_2dy_1dy_2$

$=\sum_{i=1}^2\int_{x_i=0}^1 x_i^2 dx_i+\sum_{j=1}^2 \int_{y_j=0}^1 y_j^2dy_1 -... ...1 \int_{x_2=0}^1 x_1x_2dx_1dx_2 -2 \int_{y_1=0}^1 \int_{y_2=0}^1 y_1y_2dy_1dy_2$

$=4 \times \frac{1}{3} -2\Pi_{k=1}^2 \int_{x_k=0}^1 x_kdx_k -2 \Pi_{l=1}^2 \int_{y_l=0}^1 y_ldy_l$

$=\frac{4}{3}-2\times\frac{1}{2}\times\frac{1}{2}-2\times\frac{1}{2}\times\frac{1}{2}$

$=\frac{4}{3}-1=\frac{1}{3}$

2.

$\int_{x=0}^{1} \int_{y=0}^{1}\sqrt{x^2+y^2}dydx$

When $\theta \in [0,\frac{\pi}{4}]$, the radius $r$ is given by $r = \frac{1}{\cos \theta}$. We thus choose the following change of variables:

$$\Phi:\mathbb{R}^2\rightarrow\mathbb{R}^2, (r,\theta)\mapsto(r\cos\theta, r\sin\theta)$$

$$\hbox{Jac}(\Phi)(r,\theta) = \left\vert \begin{tabular}{ll} $\cos... ...& $-r\sin\theta$ \\ $\sin\theta$ & $r\cos\theta$ \end{tabular} \right\vert = r$$

We know that, if $g:[0,1]\times[0,1]\rightarrow \mathbb{R}$ and $f:[0,\sqrt{2}]\times[0,\frac{\pi}{2}]\rightarrow \mathbb{R},t\mapsto [g\;\hbox{o}\;\Phi(t)]\times\vert\hbox{Jac}(\Phi)(t)\vert$, we have:

$$\int_{[0,1]^2} g(x,y)dxdy = \int_{[0,\sqrt{2}]\times[0,\frac{\pi}{2}]} f(r,\theta)drd\theta$$

Because of the symmetry in $x$ and $y$, we just have to integrate on half of the unit square.

$$\int_{x=0}^{1} \int_{y=0}^{1} \sqrt{x^2+y^2}dxdy = 2 \int_{\theta=0}^{\frac{\pi}{4}} \int_{r=0}^{\frac{1}{\cos\theta}} r^2drd\theta$$

$$= 2 \int_{\theta=0}^{\frac{\pi}{4}} \left[ \frac{r^3}{3}\right]_{r=0}^{\frac{1}{\cos\theta}} d\theta$$

$$= \frac{2}{3} \int_{\theta=0}^{\frac{\pi}{4}} \frac{1}{\cos^3 \theta} d\theta$$

We concentrate on $\int \frac{1}{\cos^3\theta}d\theta$.

$$\int \frac{1}{\cos^3\theta} d\theta = \int \frac{\cos \theta}{\cos^4 \theta} d\theta = \int \frac{\cos\theta}{(1-\sin^2\theta)^2}$$

We use the following change of variables: $\phi:u\mapsto\sin\theta$.

$$\int \frac{1}{\cos^3\theta} d\theta = \int \frac{1}{(1-u^2)^2}du = \int \frac{1}{(1+u)^2}\frac{1}{(1-u)^2}du$$

We decompose $\frac{1}{(1+u)^2}\frac{1}{(1-u)^2}$ into simpler elements:

$$\frac{1}{(1+u)^2}\frac{1}{(1-u)^2} = \frac{\alpha}{(1-u)^2}+\frac{\beta}{1-u}+\frac{\gamma}{1+u}+\frac{\delta}{(1+u)^2}\quad (1)$$

If we form $(1)\times(1-u)^2$ and then take $u=1$, then $\frac{1}{4}=\alpha$.

If we form $(1)\times(1+u)^2$ and then take $u=-1$, then $\frac{1}{4}=\gamma$.

If we take $u=0$ in $(1)$, then $1 = \alpha + \beta + \gamma + \delta = \frac{1}{2} + \beta + \delta$.

If we take $u=2$ in $(1)$, then $\frac{1}{9} = \alpha - \beta + \frac{\gamma}{9} + \frac{\delta}{3}$.

That is:

$$\begin{tabular}{l} $\frac{1}{2} = \beta + \delta$ \\ $-\frac{1}{6} = -\beta + \frac{\delta}{3}$ \\ \end{tabular}$$

Thus, $\alpha=\beta=\gamma=\delta=\frac{1}{4}$.

$$\int \frac{1}{\cos^3\theta} d\theta = \frac{1}{4} \int \frac{1}{(1-u)^2}+\frac{1}{1-u}+\frac{1}{1+u}+\frac{1}{(1+u)^2}du$$

$$= \frac{1}{4} \left( -\frac{1}{1+u} + \frac{1}{1-u} + \ln(1+u) - \ln(1-u) \right)$$

$$= \frac{1}{4} \left( \frac{1}{1-\sin\theta} - \frac{1}{1+\sin\theta} + \ln(1+\sin\theta) - \ln(1-\sin\theta) \right)$$

Therefore:

$$\int_{x=0}^{1} \int_{y=0}^{1}\sqrt{x^2+y^2}dydx = \frac{1}{6} \le... ...c{1}{\sqrt{2}}} + \ln(1+\frac{1}{\sqrt{2}}) - \ln(1-\frac{1}{\sqrt{2}}) \right)$$

$$=\frac{\sqrt{2}+\hbox{argth}(\sqrt{2})}{3}$$