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Let $X$ have the density $f(x)=2 x$ if $0 \leq x \leq 1$ and $f(x)=0$ otherwise. Show that $X$ has the mean $2 / 3$ and the variance $1 / 18$. Find the mean and the variance of the random variable $Y=-2 X+3$.
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To find expected values for continuous random variables we integrate e.g.
$$E(X)=\int_{-\infty}^{\infty} x f(x) d x=\int_{0}^{1} x \cdot 2 x d x=2 / 3$$
and similarly
$$\operatorname{Var}(X)=\int_{0}^{1}(x-2 / 3)^{2} \cdot 2 x d x=1 / 2-8 / 9+4 / 9=1 / 18$$
You could also use the formula $\operatorname{Var}(X)=E\left(X^{2}\right)-E(X)^{2}$ where
$$E\left(X^{2}\right)=\int_{0}^{1} x^{2} \cdot 2 x d x=1 / 2$$
so that $\operatorname{Var}(X)=1 / 2-(2 / 3)^{2}=1 / 18$. Use the linearity of expectation for the last bits to get
$$E(Y)=-2 \times 2 / 3+3=5 / 3 \quad \text { and } \quad \operatorname{Var}(Y)=(-2)^{2} \times 1 / 18=2 / 9$$
by Diamond (74,866 points)

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