Mean & Variance of X: Solving Confusing Problem w/sin(θ)

  • Context: Undergrad 
  • Thread starter Thread starter SomeRandomGuy
  • Start date Start date
  • Tags Tags
    Confusing
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
Sign up Log in
2 replies · 2K views
SomeRandomGuy
Messages
55
Reaction score
0
Let X be the sin of the angle theta in radians chosen uniformly from (-pi/2,pi/2). Find the mean and variance of X. HINT: X = sin(theta). Specify the support of X and check to see if your result describes a p.d.f.

Anyone got any idea's? I managed to solve the majority of other problems and don't even know where to begin this one. Thanks for any help.
 
Physics news on Phys.org
Since the sine of any angle is between -1 and 1 the "support of X" is
[-1,1]. The integral of 1 from [itex]-2\pi[/itex] to [itex]2\pi[/itex] is [itex]4\pi[/itex] so the "uniform distribution" density is [itex]\frac{1}{4\pi}[/itex]. The mean value of X will, of course, be
[tex]\int_{-1}^1X P(X)dX= \frac{1}{4\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}sin(\theta)cos(\theta)d\theta[/tex].
 
HallsofIvy said:
The integral of 1 from [itex]-2\pi[/itex] to [itex]2\pi[/itex] is [itex]4\pi[/itex] so the "uniform distribution" density is [itex]\frac{1}{4\pi}[/itex].

First, i'd like to thank you for your response. I understand everything you said except what I left in quotes. Why did you integrate 1 from -2pi to 2pi? Other than that, I already had the support being [-1,1] and I know how to find mean and variance when given some P(X).