Dear students,

Here is the solution of problem 2 (sheet2)

f(x') = Int g(x)h(x'-x)dx

E(x') = Int x' (Int  g(x) h(x'-x) dx) dx' =

      = Int g(x) (Int x' h(x'-x) dx') dx  = [the order of integration
                                             can be changed, since there
                                             is no preference for x or x']

      =  Int g(x) (Int (u +x) h(u) du) dx = [now, du = dx', since x is
                                             is a constant (parameter)
                                             with respect to the inner 
					     integral over x', and the 
					     integration ranges remain at 
					     infinity]
      
      =  Int g(x) (E_h(u) + x*1) dx =       [h(u) is normalized, and x
                                             can be put in front of the 
					     inner integral]

      = E_h(u) + E_g(x)                     [g(x) is normalized]
      
q.e.d. 

The expectation value E(x'^2) can be calculated in analogy!

The only 'trick' is to replace (inside the double integral)
x'2 by (x+u)^2 = x^2 + 2 x u + u^2

and then to perform the integrations as for problem a. In result, you
will find

E(x'^2) = E(x^2) + E(u^2) + 2 E(x) E(u)

which in combination with (E(x'))^2 = (E(x))^2 + (E(u))^2 + 2 E(x) E(u) 

results in

Var(x') = Var(x) + Var(u)