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  #1  
Old 09-20-2016, 10:10 AM
jchaney jchaney is offline
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Default Finan 38.7

$ f(x) = \frac{1}{2\pi} $
$y = cos x $

find pdf of y.

1. arccos y = x
2. d/dx arcos y = - 1/((1-x^2)^.5)
3. (1/2pi)*| - 1/((1-x^2)^.5)| = 1/(2pi*(1-x^2)^.5)
But, solution is 1/(pi*(1-x^2)^.5)
Where did I go wrong?
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Old 09-20-2016, 11:16 AM
Katuarial Katuarial is offline
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Over the interval cos(x) is 2 to 1, so you have to multiply the pdf by 2 which will then give you the correct answer.
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Last edited by Katuarial; 09-20-2016 at 11:26 AM..
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Old 09-20-2016, 12:41 PM
toesockshoe toesockshoe is offline
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Quote:
Originally Posted by jchaney View Post
$ f(x) = \frac{1}{2\pi} $
$y = cos x $

find pdf of y.

1. arccos y = x
2. d/dx arcos y = - 1/((1-x^2)^.5)
3. (1/2pi)*| - 1/((1-x^2)^.5)| = 1/(2pi*(1-x^2)^.5)
But, solution is 1/(pi*(1-x^2)^.5)
Where did I go wrong?
Note that your method only works perfectly if the function is one to one.... Note cos(x) is not one to one over the interval [0,2pi] so you need to multiple by a constant.....To figure out the constant, it might be more intuitive to solve pdf transformations of non one-to-one functions using the CDF method.... drawing a picture might help.... this problem however is a common transformation and you can memorize to multiple it by 2 over the interval.
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Old 09-21-2016, 02:46 PM
jchaney jchaney is offline
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Thank you for the response. I think I understand what you are explaining. Because the inverse of f(x) range and domain do not match with a 1 to 1, but rather 1 to 2, then the solution is doubled. Aside from trigonometric identities and square transformation, ( Y=X^2), what other problems will have non 1 to 1 relations?

Now I understand why I was missing so many of these problems. Looks like the range of the original function will also determine where to consider the negative value of the square in the inverse.

I spend days hitting my head against a wall trying to figure out this stuff. -- There is new crack in the wall today.

Last edited by jchaney; 09-21-2016 at 03:09 PM..
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  #5  
Old 09-21-2016, 09:42 PM
toesockshoe1 toesockshoe1 is offline
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Quote:
Originally Posted by jchaney View Post
Thank you for the response. I think I understand what you are explaining. Because the inverse of f(x) range and domain do not match with a 1 to 1, but rather 1 to 2, then the solution is doubled. Aside from trigonometric identities and square transformation, ( Y=X^2), what other problems will have non 1 to 1 relations?

Now I understand why I was missing so many of these problems. Looks like the range of the original function will also determine where to consider the negative value of the square in the inverse.

I spend days hitting my head against a wall trying to figure out this stuff. -- There is new crack in the wall today.
trig functions and y=x^2 were the only ones ive seen in my practice..
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