Question 12.4 (ii)

How E(Xt) = E(Sinwt cosYt + coswt sin Yt) = 0 ?,,given that Yt is a std normal RV.

Thanks & Regards,
Nagesh.

 

Please help me..

 

can any one help on this?

can any one help on this?

 

I think I can

Think my reasoning is correct here , doesnt have much to do with statistics tho

Have you heard of odd and even functions?
http://en.wikipedia.org/wiki/Even_and_odd_functions

normal dist is an even function (symmetric about y axis)
cos is an even function (also symmetric about y axis)
sin is an odd function (symmetry in a point at the origin)

what does this imply about expectations

first part E( sinwtcosYt ) = 0
since cos is even evaluated on an even function i.e. Yt
Cos Yt is even , now this even function is getting multiplied by an odd one sinwt , so the result sinwtcosYt is odd and therefore its expectation is 0

Second E( coswtsinYt) = 0
since sin is odd evaluate on an even function i.e. Yt
sin Yt is therefore odd and this is multiplied by an even function cos
so the result coswtsinYt is odd and therefore its expectation is 0

I think this is a correct way of interpreting whats going on , but I dont have the question to hand so there might be something Im missing and/or this could be completely wrong.

Edit : I now think the normal distributions "even ness" actually has nothing to do with it . so it just rest on the fact sin.cos is odd.
getting increasing worried this is wrong I have to say

 

What is wt?

Sin(wt)Cos(Yt) + Sin(Yt)Cos(wt) = Sin(wt+Yt)
Let Xt = wt+Yt;

E[Sin(Xt)] = integral {sin(Xt) pdf of Xt dXt}
If wt and Yt are normal then Xt is also normal.

and you can split the integral from -infiity to 0 and 0 to +infinity.
use the odd propery of sin, Sin(-Xt) = - Sin (Xt)
and the fact that the pdf is symmetrical
the two halves of the integral cancel each other out.