Exercise 5.5.4

Answers

$\int \int {(x - m_{1})}^{2} p (x, y) dxdy = \int \int (x^{2} - 2 m_{1} x + m_{1}^{2}) p (x) p (y) dxdy = \int_{x} x^{2} p (x) \int_{y} p (y) dydx - 2 m_{1} \int_{x} xp (x) \int_{y} p (y) dydx + m_{1}^{2} \int \int p (x, y) dxdy = \int_{x} x^{2} p (x) dx - 2 m_{1} \int_{x} xp (x) dx + m_{1}^{2} = \int_{x} x^{2} p (x) dx - 2 m_{1} \int_{x} xp (x) dx + \int_{x} m_{1}^{2} p (x) dx = \int_{x} {(x - m_{1})}^{2} p (x) dx = σ_{1}^{2}$
$\int \int (x - m_{1}) (y - m_{2}) p (x, y) dxdy = \int \int xyp (x, y) dxdy - m_{2} \int \int xp (x, y) dxdy - m_{1} \int \int yp (x, y) dxdy + m_{1} m_{2} \int \int p (x, y) dxdy = \int_{x} xp (x) \int_{y} yp (y) dydx - m_{2} m_{1} - m_{1} m_{2} + m_{1} m_{2} = m_{1} m_{2} - m_{1} m_{2} = 0$

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2020-03-20 00:00