I'm given a random variable X n </msub> that is wrt the following measure:

Michelle Mendoza 2022-07-13 Answered
I'm given a random variable X n that is wrt the following measure:
lim n # { X n [ a , b ] } n = a b sin 2 θ d θ .
with an error term O ( log n / n ).
Now I'm trying to find the distribution of cos 2 X n . The text I'm reading says this is a b cos 2 θ d θ, which I'm already not sure of, and I also want to figure out what the error term of this is.
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Answers (1)

gozaderaradiox5
Answered 2022-07-14 Author has 19 answers
(1) lim n # { X n [ a , b ] } n = k a b sin 2 θ d θ     w h e r e   k = 2 π
Please note that I have added a factor k = 2 π . See later.
First of all, as the values taken by the LHS of (1) are between 0 and 1, the same must be true for the RHS of (1).
This is why I have added factor k in order that
2 π 0 π sin 2 θ d θ = 1.
Let f X and F X be the pdf and cdf resp. of random variable X.
The LHS of (1) is plainly a b f X ( x ) d x under the classical "subjectivist" assumptions of probability (I think that the error term O ( log ( n ) / n ) comes from a theorem of De Finetti, not sure, but we don't need it ; I would even say that I find its presence a little strange).
Therefore, if identity (1) is valid for any a , b, they define a same measure ; therefore:
f X ( x ) = k sin 2 ( x ) = k 1 2 ( 1 cos ( 2 x ) ) = 1 π ( 1 cos ( 2 x ) )
giving, by integration:
F X ( x ) = { 0 for   x < 0 1 π ( 1 2 x 1 4 sin ( 2 x ) ) for   x [ 0 , π ] 1 for   x > π
Let Y := cos 2 X.
Let us compute the cdf of Y which can take values y [ 0 , 1 ].
F Y ( y ) = P ( Y < y ) = P ( y < cos X < y ) = 2 P ( cos X < y )
(last equality by symmetry of the pdf of X wrt π / 2)
F Y ( y ) = 2 P ( X < cos 1 ( y ) c ) = 2 F X ( c )
(2) F Y ( y ) = 2 1 π [ 1 2 c 1 4 sin ( 2 c ) ]
It remains to express (2) in a simpler form, and finally to take the derivative of cdf F Y with respect to y in order to get the pdf f Y .

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