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Let $X \\sim N(0, 1)$ (standard Normal distribution)
\nAnd $Y = X^2$
\nThen $Y \\sim \\chi^2(1)$ (Chi-squared distribution with 1 degree of freedom)
\nIf $Z = \\sqrt{Y}$ why is $Z$ not the standard normal distribution?
", "advice": "Taking the square of a standard normal produces a chi-square random variable.
\nBut taking the square root of the chi-square won't produce a normal since the range of the resulting distribution will be $(0, \\infty)$ not the entire set of real numbers and the PDF of $Z$ is $2$ times the PDF of $X$ so it still integrates to $1$.
\nPerforming both of these transformations is a good exercise.
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