You-have-a-1×1-cm-square-If-you-pick-two-random-points-inside-the-square-what-is-the-average-distance-between-those-points-

Question Number 6609 by Temp last updated on 05/Jul/16

You have a 1 \times 1 cm square . If you pick

two random points inside the square,

what is the average distance between

those points ?

Commented by prakash jain last updated on 05/Jul/16

Let X and Y be two random variable .

f_{X} (x) and f_{Y} (y) are probability density function

of X and Y .

In the above case we can assume uniform .

We define a new random variable

Z = \sqrt{X^{2} + Y^{2}}

Then mean value of Z will be given by :

E (Z) = \int {Z f}_{Z} (z) d z

We need to find f_{Z} (z) from f_{X} and f_{Y} .

will continue to find f_{Z} (z)

Commented by prakash jain last updated on 05/Jul/16

f_{X} (x) = {\begin{cases} 1 & 0 ⩽ x ⩽ 1 \\ 0 & e l s e w h e r e \end{cases}

f_{Y} (y) = {\begin{cases} 1 & 0 ⩽ y ⩽ 1 \\ 0 & e l s e w h e r e \end{cases}

for uniform density

Commented by prakash jain last updated on 05/Jul/16

Since X and Y are independent variable

we can write joint probablity distribution

as

f_{X Y} (x, y) = f_{X} (x) \cdot f_{Y} (y)

Expected value of distance

z = g (x, y) = \sqrt{x^{2} + y^{2}}

E (g (x, y)) = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \sqrt{x^{2} + y^{2}} f_{X} (x) f_{Y} (y) d x d y

= \int_{0}^{1} \int_{0}^{1} \sqrt{x^{2} + y^{2}} d x d y = \frac{1}{2} (\sqrt{2} + \sinh^{- 1} 1)

Please check . I am not 100 % sure .

Commented by Temp last updated on 05/Jul/16

What if you integrate for polar coordinates

over half of the square, and double the

integral ?

y = r \sin θ

x = r \cos θ

e t c \dots

Commented by prakash jain last updated on 07/Jul/16

If we tranform to polar the limits for r

will dependent on θ .

For half square (triangle)

r = 1 / \cos θ

= \int_{0}^{π / 4} \int_{1}^{1 / \cos θ} r^{2} d r d θ