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.
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 $$ \int_{-\infty}^\infty ax^2 f_X(x)\,dx = 2 \int_0^\infty ax^2 f_X(x)\,dx. $$