The relationship of two random variables is given by
$$ X = \Phi(Y) \Leftrightarrow Y = \Phi^{-1}(X),$$
where $\Phi(\bullet)$ is the standard normal cdf and $\Phi^{-1}(\bullet)$ the inverse of the standard normal cdf.
The variance of X is $\mathbf{V}(X)=\sigma_x^2.$
What is the variance of Y?
My idea was that both variances are equal, so that $\sigma^2_y = \sigma^2_x$.
Is this right or is it necessary to transform the variance of y?
Best Answer
May be I'm wrong, but I think not necessary that they are equal. You said nothing about the distribution of Y. $\Phi$ and $\Phi^{-1}$ are just operators of functional transform. I'll split it into two parts and call the second result not $Y$ but $Z$ (in order not to get lost). So you have: 1. $X = \Phi(Y) $ or $Y = \Phi^{-1}(X)$ 2. $Z = \Phi^{-1}(X)$ or $X = \Phi(Z) $
I'll assume that the pdf of $Y$ is $w_y(\xi)$.
1). Then the pdf of $X$ will be: $$w_{x}(\xi)=w_{y}(\Phi^{-1}(\xi))\left| {\frac{d \Phi^{-1}(\xi) }{d\xi }} \right|$$
2). The pdf of $Z$ will be: $$w_{z}(\xi)=w_{x}(\Phi(\xi))\left| {\frac{d \Phi(\xi) }{d\xi }} \right|$$ I guess that after using sequential transformation of pdf in general you will not get the same and so the variances will not be equal. But you can find cases when they will be.