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For a random variable $X$, $(X^3-1)$ is uniformly distributed in the interval $[0,7]$

I need to find the expected value and variance of $X$

and I know that:

cumulative distribution function: $$F_{X}(x)= \begin{cases} 0&;x < 1\\ \frac{x^3-1}{7}&;1\leq x \leq 2\\ 1&;x>2\\ \end{cases}$$

probability density function:

$$f_{X}(t)= \begin{cases} 0&;t \not\in (1,2)\\ \frac{3x^2}{7}&; t \in (1,2)\\ \end{cases}$$

My attempt:

$(1)$

$$E(X)=\int_{-\infty}^{\infty}t\cdot f_{_X}(t)dt$$

$$=\int_{-\infty}^{1}t\cdot 0dt+\int_{1}^{2}\frac{3t^2}{7}\cdot dt+\int_{2}^{\infty}t\cdot 0dt$$

$$=0+\frac{3}{28}x^4\bigg|_{1}^{2}+0=\boxed{\frac{45}{28}}$$

$(2)$

$$\text{Var}(X)=E(X^2)-(E(X))^2$$

$$E(X^2)=$$

$$E(X)^2=\bigg(\frac{2025}{784}\bigg)$$

Is it correct so far ? how to calculate $E(X^2)?$

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    $\begingroup$ Hint: $$E\big[g(X)\big] = \int_{-\infty}^{\infty} g(x)\cdot f(x) \ dx$$ $\endgroup$ Commented Aug 3, 2015 at 18:33
  • $\begingroup$ @jameselmore Your hint is correct but I don't know how to use it $\endgroup$ Commented Aug 3, 2015 at 20:52
  • $\begingroup$ You do the exact same thing as you did for $E[X]$, but the integrand is $t^2\cdot f_X(t)$, instead of $t\cdot f_X(t)$. $\endgroup$ Commented Aug 3, 2015 at 20:54

1 Answer 1

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$$E[g(x)]=\int_{-\infty}^{\infty}g(x)\cdot f(x)dx$$

$$E[X^2]=\int_{-\infty}^{\infty}t^2\cdot f(t)dt$$

$$=\int_{1}^{2}t^2\cdot \frac{3t^2}{7}dt$$ $$=\frac{3}{7}\int_{1}^{2}t^4dt$$ $$=\frac{3}{7}\frac{t^5}{5}\bigg|_{1}^{2}=\frac{31}{5}\frac{3}{7}=\boxed{\frac{93}{35}}$$

$$\text{Var}(X)=\frac{93}{35}-\frac{2025}{784}=\boxed{\frac{291}{3920}}$$

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