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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?

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The variance of X does not determine the variance of Y. –  Did Mar 26 '13 at 17:01
    
"My idea was that both variances are equal" I wonder how you got that idea. –  leonbloy Mar 26 '13 at 17:22
    
@leonbloy It is an explanation given at the presentation of a paper . I was shocked because nobody revolted. –  Stephen Mar 26 '13 at 18:07

3 Answers 3

up vote 1 down vote accepted

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.

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Variance of $Y$ isn't variance of $X$. Suppose $Y$ were a normal distribution with mean 0 and variance 100. Then $Y$ can take values between $\pm \infty$ and $\sigma_Y^2 = 100$. But $X$ can only take values between 0 and 1, so $\sigma_X^2 < 100$.

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Variances are in general not equal (consider the example when $Y$ is normal $N(0,1)$ then $X$ is uniform on $(0,1)$)

There is no miraculous formula. To get the variance of $Y$, you would need to compute $$ \int_B (y-E[Y])^2 f_X \left( \Phi \left( y \right) \right) \phi \left( y \right) \mathrm{d} y $$ where $f_X$ is the density of $X$, $\phi$ is the standard normal pdf, $B$ is the set obtained by transforming the support of $X$ through the function $\Phi^{- 1}$ (for instance, if the support of $X$ is $\left( 0, 1 \right)$ then $B$ is $\mathbb{R}$) and $E[Y]$ is the expectation of $Y$ $$ \int_B y f_X \left( \Phi \left( y \right) \right) \phi \left( y \right) \mathrm{d} y $$ Of course, in the above, I assumed $X$ is an absolutely continuous random variable in order to be able to talk of $f_X$.

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