Probability Density of Mapped $y=ax^2$ to Normal Distribution Let $x$ be a scalar random variable and let $y$ be a scalar quantity such that $y = ax^2$.
If
$$p(x) = \frac 1 {\sigma\sqrt{2\pi}}\, e^{-x^2/(2\sigma^2)}$$
a) Find the probability density of $y$
b) Find expected value of vector $y$;
c) Find covariance of $y$.
Perhaps I am missing something here, but for part a would we simply $x^2 = y/a$  and put that into the probability density of $p(x)$ to get get $p(y)$.
For part b, would we just take the integral of $\int_{-\infty}^\infty p(x)\,dx$.
For the rest, I am not exactly sure how to tackle the problem.
Thank you very much in advance.
 A: One should not use the same character to refer to both the random variable and the argument to the density or to the c.d.f.  Thus in the expression $F_X(x)=\Pr(X\le x)$, the capital $X$ and the lower-case $x$ have different roles.
\begin{align}
f_Y(y) & =\frac{d}{dy} \Pr(Y\le y) = \frac{d}{dy} \Pr\left(-\sqrt{\frac y a}\le X\le\sqrt{\frac y a}\,\right) \\[10pt]
& = \frac d {dy} 2\Pr\left(0\le X\le\sqrt{\frac y a}\right) = 2\frac d {dy} \int_0^\sqrt{y/a} \frac 1 {\sigma\sqrt{2\pi}} e^{-x^2/(2\sigma^2)} \, dx \\[10pt]
& = 2 \frac d {du} \int_0^u \frac 1 {\sigma\sqrt{2\pi}} e^{-x^2/(2\sigma^2)} \, dx \cdot \frac{du}{dy} = 2 \frac 1 {\sigma\sqrt{2\pi}} e^{-u^2/(2\sigma^2)} \cdot\frac d {dy} \sqrt{\frac y a} \\[10pt]
& = 2 \frac 1 {\sigma\sqrt{2\pi}} e^{-u^2/(2\sigma^2)} \cdot \frac 1 2 \sqrt{\frac a y} \cdot\frac 1 a = 2 \frac 1 {\sigma\sqrt{2\pi}} e^{-y/(2a\sigma^2)} \cdot \frac 1 2 \sqrt{\frac a y} \cdot\frac 1 a
\end{align}
and do routine simplifications from there.
For part $(b)$, you can find
$$
\int_{-\infty}^\infty ax^2 f_X(x)\,dx = 2 \int_0^\infty ax^2 f_X(x)\,dx.
$$
