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Homework Statement
If X is represented by the Gaussian distribution, that is,
f_{X}(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp{(-\frac{x^2}{2\sigma^2})}
find an expression for the pdf fZ(z) of Z = arctan(x).
The Attempt at a Solution
If Z =g(X), then g(X) is multivalued unless the range of the function is restricted to Z \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
Under this condition, the function is one to one, and so the probability that z will be in some interval (z, z + dz) is equal to the probability that x will be in the corresponding interval (x, x + dx). In other words,
|f_Z(z) dz| = |f_X(x) dx|
f_Z(z) = \left| \frac{dx}{dz} \right| f_X(x)
= \frac{d}{dz} (\tan z) f_X(x)
= (\sec^2(z)) f_X(x)
= (1+\tan^2(z)) \frac{1}{\sigma\sqrt{2\pi}} \exp{(-\frac{\tan^2(z)}{2\sigma^2})}
f_Z(z) = \left| \frac{dx}{dz} \right| f_X(x)
= \frac{d}{dz} (\tan z) f_X(x)
= (\sec^2(z)) f_X(x)
= (1+\tan^2(z)) \frac{1}{\sigma\sqrt{2\pi}} \exp{(-\frac{\tan^2(z)}{2\sigma^2})}
Am I doing this right?
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