for the variance formula $\text{var}(X) = E[X^2]-(E[X])^2$
how are you suppose to work out $E[X^2]$ given the interval $[0,1]$ and $E[x]= 1/2$
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You cannot determine $E[X^{2}]$ without more information. For $f_{1}(x) = 1$, we have that $$ E[X] = \int_{0}^{1} x f_{1}(x) \, dx = \int_{0}^{1} x \, dx = \left. \frac{1}{2} x^{2} \right|_{x = 0}^{x = 1} = \frac{1}{2}, $$ and $$ E[X^{2}] = \int_{0}^{1} x^{2} f_{1}(x) \, dx = \int_{0}^{1} x^{2} \, dx = \left. \frac{1}{3} x^{3} \right|_{x = 0}^{x = 1} = \frac{1}{3}. $$ For $f_{2}(x) = 12 (x - \frac{1}{2})^{2}$, we have that $$ E[X] = \int_{0}^{1} x f_{2}(x) \, dx = \int_{0}^{1} 12 x (x - \tfrac{1}{2})^{2} \, dx = \left. \frac{x^{2} (6 x^{2} - 8 x + 3)}{2} x^{2} \right|_{x = 0}^{x = 1} = \frac{1}{2}, $$ and $$ E[X^{2}] = \int_{0}^{1} x^{2} f_{2}(x) \, dx = \int_{0}^{1} 12 x^{2} (x - \tfrac{1}{2})^{2} \, dx = \left. \frac{x^{3} (12 x^{2} - 15 x + 5)}{5} \right|_{x = 0}^{x = 1} = \frac{2}{5}. $$ |
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