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This question already has an answer here:

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.

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marked as duplicate by callculus, Cesareo, clathratus, Lee David Chung Lin, Leucippus Mar 27 at 4:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Have you already evaluated the value of $C$? Please provide as much information as possible. $\endgroup$ – callculus Mar 26 at 17:39
  • $\begingroup$ Yes, C = 1/3. P( X >= 0) = .8888 $\endgroup$ – Ryan Mar 26 at 17:42
  • $\begingroup$ You might want to read this section: en.wikipedia.org/wiki/… $\endgroup$ – Benedict W. J. Irwin Mar 26 at 17:46
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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$$f_Y(y){={d\over dy}\Pr\{Y<y\}\\=\begin{cases}0&,\quad y<0\\{\sqrt y\over 3}&,\quad 0<y<1\\{\sqrt y\over 6}&,\quad 1<y<4\\0&,\quad y>4\end{cases}}$$

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

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

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