Commit c53d49d9 authored by Konrad Völkel's avatar Konrad Völkel
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fixed variance of continuous uniform distribution

parent 8ba0a2b5
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......@@ -64,13 +64,13 @@ $$
Das zweite Moment ist
$$
\mathbb{E}(X^2) = \int_{-\infty}^\infty x^2 f(x)dx = \left[ \frac{x^3}{3(b-a)} \right]_a^b = \frac{b^3-a^3}{3(b-a)} = \frac{(b-a)(a+b)^2}{3(b-a)} = \frac{(a+b)^2}{3}
\mathbb{E}(X^2) = \int_{-\infty}^\infty x^2 f(x)dx = \left[ \frac{x^3}{3(b-a)} \right]_a^b = \frac{b^3-a^3}{3(b-a)} = \frac{a^2+ab+b^2}{3}
$$
Die Varianz ist
$$
\mathbb{V}(X) = \mathbb{E}(X^2) - {\left(\mathbb{E}(X)\right)}^2 = \frac{(a+b)^2}{3} - \frac{(a+b)^2}{4} = \frac{4-3}{12}(a+b)^2 = \frac{(a+b)^2}{12}
\mathbb{V}(X) = \mathbb{E}(X^2) - {\left(\mathbb{E}(X)\right)}^2 = \frac{a^2+ab+b^2}{3} - \frac{(a+b)^2}{4} = \frac{(a-b)^2}{12}
$$
:::
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