Finding E[Y] and V(Y) for Continuous Random Variable

Question

I've been at this one for a while:

A random variable $X$ follows the distribution $$f_X(x) = \begin{cases} Cx^2&,\quad -2\leq x \leq 1\\0&,\quad \text{Otherwise}\end{cases}$$and $Y\triangleq X^2$.

Find E[Y] and Var[Y].

I calculated C = 1/3.

I've tried $\int_b^a yf(y) dy$ for E[Y] and a similar formula using $y^2$ instead of $y$. The integrals were executed piecewise.

I could use some help, as all of my attempts (variations on the above two strategies) have been dead wrong. The similar question on the board does not appear to address the calculation of these variables.

Have you already evaluated the value of $C$? Please provide as much information as possible. — callculus42, Mar 26 '19 at 17:39
You might want to read this section: https://en.wikipedia.org/wiki/Random_variable#Functions_of_random_variables — Benedict W. J. Irwin, Mar 26 '19 at 17:46

score 0 · Answer 1 · answered Mar 26 '19 at 17:50

Hint

Since $\int_{-2}^1 f_X(x)dx=1$ we have$$C={1\over 3}$$also $$\Pr\{Y<y\}{=\Pr\{X^2<y\}\\=\Pr\{-\sqrt y<X<\sqrt y\}\\=\begin{cases}0&,\quad y<0\\\int_{-\sqrt y}^{\sqrt y}{x^2\over 3}dx&,\quad 0<y<1\\\int_{-\sqrt y}^{1}{x^2\over 3}dx&,\quad 1<y<4\\1&,\quad y>4\end{cases}\\=\begin{cases}0&,\quad y<0\\{2\over 9}y^{3\over 2}&,\quad 0<y<1\\{1+y^{3\over 2}\over 9}&,\quad 1<y<4\\1&,\quad y>4\end{cases}}$$therefore$$f_Y(y){={d\over dy}\Pr\{Y<y\}\\=\begin{cases}0&,\quad y<0\\{\sqrt y\over 3}&,\quad 0<y<1\\{\sqrt y\over 6}&,\quad 1<y<4\\0&,\quad y>4\end{cases}}$$

score 0 · Answer 2 · answered Mar 27 '19 at 01:54

Use: $\mathsf E(Y)=\mathsf E(X^2)$, the "Law of the Unconscious Statistician" and $\mathsf {Var}(Y)=\mathsf E(X^4)-\mathsf E^2(X^2)$

$$\begin{align}\mathsf E(Y)&=\int_{-2}^1 x^2\cdot\frac {x^2}3~\mathrm d x\\[3ex]\mathsf{Var}(Y) &= \int_{-2}^1 x^4\cdot\frac {x^2}3~\mathrm d x-\mathsf E^2(Y)\end{align}$$

Finding E[Y] and V(Y) for Continuous Random Variable

2 Answers2