Question

A surveyor wishes to lay out a square region with each side having length L. However, because of measurement error, he instead lays out a rectangle in which the north-south sides both have length X and the east-west sides both have length Y. Suppose that X and Y are independent and that each is uniformly distributed on the interval [L − A, L + A] (where 0 < A < L). What is the expected area of the resulting rectangle?

Answer 1

Step-by-step explanation:

Area, A = Length,x × Breadth,y

A=xy

When x and y are independent, E(xy) = E(x)E(y)

As x and y have the same distribution, U[L-A,L+A], they have the same mean.

We could argue by symmetry that E(x) = L and E(y) + L, also.

We can also reason this from the fact that, if X ~ U[L-A, L+A], f(x) = 1/(2A) from L-A to L+A

Therefore

$E(x)= \int\limits {f(x)} \, dx \\\\=\int\limits^{(L+A)}_{(L-A)} {\frac{1}{2A}x } \, dx$

$=\int\limits^{(L+A)}_{(L-A)} {\frac{1}{2A}x } \, dx \\\\=[\frac{1}{4A}x^2 ]\limits^{(L+A)}_{(L-A)}$

=1/(4A)(L+A)2 - 1/(4A)(L-A)2

= 1/(4A)(L2 + 2AL + A2) - 1/(4A)(L2 - 2AL + A2)

=1/(4A)(2AL+2AL)

= 1/(4A)(4AL)

= L

Thus, E(xy) = E(x)E(y) = L×L = L²