A random variables $X \in N(0,1)$, i.e. $f_X(x)=\frac{1}{\sqrt{(2\pi)}}e^{-\frac{x^2}{2}}$. If I define a new random variable $Y=-X$. What is the density function of the $Y$? If I then want to find cumulative distribution function of $Y$ how would I go about?
Is this a correct way of going about it: $f_Y(y)=f_{-X}(-x)=\frac{1}{\sqrt{(2\pi)}}e^{-\frac{(-x)^2}{2}}=\frac{1}{\sqrt{(2\pi)}}e^{-\frac{x^2}{2}}=f_X(x)$
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cjkilimanjaro
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I think you would just say $f_Y(y) = f_X(-x)$ instead of $f_{-X}(x)$, but this is otherwise correct. – user6247850 Dec 14 '19 at 23:29
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If your $y$ is $-x$ you should end the proof by $f_Y(y)=f_X(x)=f_X(-y)=f_X(y)$ since $e^{-{(-y)^{2}/2}}$ is same as $e^{-y^{2}/2}$. The final answer should be in terms of one variable. [The functions $f_X$ and $f_Y$ are one and the same].
Kavi Rama Murthy
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