Vector problem(probability)

Geometry Level pending

A point is randomly selected with uniform probability in the X-Y plane within the rectangle with corners at (0,0),(1,0),(1,2), and (0,2). If p is the length of the position vector of the point, the expected value of p^2 is

4/3 2/3 1 5/3

Dinesh Kumar by Dinesh Kumar , 25, India

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1 solution

Brian Moehring
Aug 8, 2018

The probability density for the point ( X , Y ) (X,Y) is f X , Y ( x , y ) = { 1 2 0 ≤ x ≤ 1 , 0 ≤ y ≤ 2 0 otherwise f_{X,Y}(x,y) = \begin{cases} \frac{1}{2} & 0 \leq x \leq 1, 0 \leq y \leq 2 \\ 0 & \text{otherwise}\end{cases} and P 2 = X 2 + Y 2 P^2 = X^2 + Y^2 so E [ P 2 ] = ∫ 0 1 ∫ 0 2 1 2 ( x 2 + y 2 ) d y d x = 5 3 \mathbb{E}[P^2] = \int_0^1 \int_0^2 \frac{1}{2}(x^2 + y^2)\,dy\,dx = \boxed{\frac{5}{3}}

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